Chapter 14 Statistics And Probability

Given Data:

The marks obtained by 17 students in a mathematics test are:

91, 82, 100, 100, 96, 65, 82, 76, 79, 90, 46, 64, 72, 68, 66, 48, 49

To Find:

The range of the given data.

Solution:

The range of a data set is defined as the difference between the maximum value and the minimum value present in the data set.

Range = Maximum Value - Minimum Value

We need to find the smallest (minimum) and largest (maximum) values from the given set of marks.

The given data points are: 91, 82, 100, 100, 96, 65, 82, 76, 79, 90, 46, 64, 72, 68, 66, 48, 49

Let's examine the data to find the minimum and maximum values.

The minimum value in the data set is 46.

The maximum value in the data set is 100.

Now, we can calculate the range using the formula:

$Range = \text{Maximum Value} - \text{Minimum Value}$

$Range = 100 - 46$

$Range = 54$

Final Answer:

The range of the data is 54.

This corresponds to option (B).

Given:

The class interval is 130-150.

Lower limit of the class = 130

Upper limit of the class = 150

To Find:

The class-mark of the class 130-150.

Solution:

The class-mark of a class interval is the average of its lower and upper limits.

The formula for class-mark is:

$Class\text{-}mark = \frac{\text{Lower Limit} + \text{Upper Limit}}{2}$

Substituting the given values into the formula:

$Class\text{-}mark = \frac{130 + 150}{2}$

$Class\text{-}mark = \frac{280}{2}$

$Class\text{-}mark = 140$

Final Answer:

The class-mark of the class 130-150 is 140.

This corresponds to option (C).

Given:

Total number of times the die is thrown = 1000.

The frequency of each outcome is given in the table:

Frequency of outcome 1 = 180

Frequency of outcome 2 = 150

Frequency of outcome 3 = 160

Frequency of outcome 4 = 170

Frequency of outcome 5 = 150

Frequency of outcome 6 = 190

To Find:

The probability that the die shows 5 if thrown once more.

Solution:

The empirical probability of an event is given by the formula:

$P(\text{Event}) = \frac{\text{Number of times the event occurred}}{\text{Total number of trials}}$

In this case, the event is getting a 5 when the die is thrown.

From the given data:

Number of times the die shows 5 = Frequency of outcome 5 = 150

Total number of trials = Total number of times the die was thrown = 1000

So, the probability of getting a 5 is:

$P(5) = \frac{\text{Number of times 5 occurred}}{\text{Total number of trials}}$

$P(5) = \frac{150}{1000}$

Now, we simplify the fraction:

$P(5) = \frac{\cancel{150}^{3}}{\cancel{1000}_{20}}$

$P(5) = \frac{3}{20}$

Final Answer:

The probability that the die shows 5 is $\frac{3}{20}$.

This corresponds to option (B).

Given:

The class interval is 90-120.

Lower limit of the class = 90

Upper limit of the class = 120

To Find:

The class-mark of the class 90-120.

Solution:

The class-mark of a class interval is the average of its lower and upper limits.

The formula for class-mark is:

$Class\text{-}mark = \frac{\text{Lower Limit} + \text{Upper Limit}}{2}$

Substituting the given values into the formula:

$Class\text{-}mark = \frac{90 + 120}{2}$

$Class\text{-}mark = \frac{210}{2}$

$Class\text{-}mark = 105$

Final Answer:

The class-mark of the class 90-120 is 105.

This corresponds to option (B).

Given Data:

The given data set is:

25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, 20

To Find:

The range of the given data.

Solution:

The range of a data set is the difference between the maximum value and the minimum value in the data set.

$Range = \text{Maximum Value} - \text{Minimum Value}$

Let's examine the given data set to find the minimum and maximum values.

The data points are: 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, 20.

By inspecting the data, the minimum value is 6.

By inspecting the data, the maximum value is 32.

Now, we calculate the range:

$Range = 32 - 6$

$Range = 26$

Final Answer:

The range of the data is 26.

This corresponds to option (D).

Given:

Mid value (Class-mark) of the class = 10

Width of the class = 6

To Find:

The lower limit of the class.

Solution:

Let the lower limit of the class be $L$ and the upper limit be $U$.

The class-mark is the average of the lower and upper limits:

$Class\text{-}mark = \frac{L + U}{2}$

We are given that the class-mark is 10, so:

$\frac{L + U}{2} = 10$

$L + U = 2 \times 10$

$L + U = 20$

The width of the class is the difference between the upper and lower limits:

$Width = U - L$

We are given that the width is 6, so:

$U - L = 6$

We now have a system of two linear equations with two variables $L$ and $U$:

(1) $L + U = 20$

(2) $U - L = 6$

We can solve this system. From equation (2), we can express $U$ in terms of $L$:

$U = L + 6$

Substitute this expression for $U$ into equation (1):

$L + (L + 6) = 20$

$2L + 6 = 20$

Subtract 6 from both sides:

$2L = 20 - 6$

$2L = 14$

Divide both sides by 2:

$L = \frac{14}{2}$

$L = 7$

Thus, the lower limit of the class is 7.

Final Answer:

The lower limit of the class is 7.

This corresponds to option (B).

Given:

Number of continuous classes = 5

Width of each class = 5

Lower class-limit of the lowest class = 10

To Find:

The upper class-limit of the highest class.

Solution:

Let the lower class-limit of the first class (lowest class) be $L_1$. We are given $L_1 = 10$.

Let the width of each class be $w$. We are given $w = 5$.

In a continuous frequency distribution, the upper limit of a class is the lower limit of the next class.

Let the classes be $[L_1, U_1), [L_2, U_2), [L_3, U_3), [L_4, U_4), [L_5, U_5)$.

The width of a class $[L_i, U_i)$ is $U_i - L_i$. So, $U_i - L_i = 5$ for all $i = 1, 2, 3, 4, 5$.

Also, since the classes are continuous, $U_i = L_{i+1}$ for $i = 1, 2, 3, 4$.

The lower limit of the first class is $L_1 = 10$.

The upper limit of the first class is $U_1 = L_1 + w = 10 + 5 = 15$.

The lower limit of the second class is $L_2 = U_1 = 15$.

The upper limit of the second class is $U_2 = L_2 + w = 15 + 5 = 20$.

The lower limit of the third class is $L_3 = U_2 = 20$.

The upper limit of the third class is $U_3 = L_3 + w = 20 + 5 = 25$.

The lower limit of the fourth class is $L_4 = U_3 = 25$.

The upper limit of the fourth class is $U_4 = L_4 + w = 25 + 5 = 30$.

The lower limit of the fifth class is $L_5 = U_4 = 30$.

The upper limit of the fifth class is $U_5 = L_5 + w = 30 + 5 = 35$.

The five continuous classes are [10, 15), [15, 20), [20, 25), [25, 30), and [30, 35).

The lowest class is [10, 15), and its lower limit is 10.

The highest class is [30, 35), and its upper limit is 35.

Alternatively:

Let the lower limit of the lowest class be $L_{min} = 10$.

Let the width of each class be $w = 5$.

There are 5 classes.

In a continuous distribution, the lower limits of consecutive classes are separated by the width.

Lower limit of 1st class = 10

Lower limit of 2nd class = $10 + 5 = 15$

Lower limit of 3rd class = $15 + 5 = 20$

Lower limit of 4th class = $20 + 5 = 25$

Lower limit of 5th class = $25 + 5 = 30$

The highest class starts at 30. Since the width is 5, the upper limit of this class is $30 + 5 = 35$.

Final Answer:

The upper class-limit of the highest class is 35.

This corresponds to option (C).

Given:

Mid-point (Class-mark) of the class = $m$

Upper class limit of the class = $l$

To Find:

The lower class limit of the class.

Solution:

Let the lower class limit be denoted by $x$.

In a continuous frequency distribution, the mid-point (class-mark) of a class is the average of its lower and upper limits.

The formula for the mid-point is:

$Mid\text{-}point = \frac{\text{Lower Limit} + \text{Upper Limit}}{2}$

Substituting the given values and our variable $x$ into the formula:

$m = \frac{x + l}{2}$

We need to solve this equation for $x$.

Multiply both sides of the equation by 2:

$2 \times m = 2 \times \frac{x + l}{2}$

$2m = x + l$

Subtract $l$ from both sides of the equation to isolate $x$:

$2m - l = x + l - l$

$x = 2m - l$

So, the lower class limit of the class is $2m - l$.

Final Answer:

The lower class limit of the class is $2m - l$.

This corresponds to option (B).

Given:

The class marks of a frequency distribution are 15, 20, 25, ...

The class mark we are interested in is 20.

To Find:

The class interval corresponding to the class mark 20.

Solution:

The difference between consecutive class marks gives the class width.

Class width ($w$) = $20 - 15 = 5$

Class width ($w$) = $25 - 20 = 5$

So, the class width is $w = 5$.

Let $m$ be the class mark, $L$ be the lower limit, and $U$ be the upper limit of a class.

The class mark is defined as the average of the lower and upper limits:

$m = \frac{L + U}{2}$

The width of the class is the difference between the upper and lower limits:

$w = U - L$

From the second equation, $U = L + w$. Substitute this into the first equation:

$m = \frac{L + (L + w)}{2}$

$m = \frac{2L + w}{2}$

$m = L + \frac{w}{2}$

Rearranging to find the lower limit $L$:

$L = m - \frac{w}{2}$

Similarly, from $w = U - L$, we have $L = U - w$. Substitute this into the first equation:

$m = \frac{(U - w) + U}{2}$

$m = \frac{2U - w}{2}$

$m = U - \frac{w}{2}$

Rearranging to find the upper limit $U$:

$U = m + \frac{w}{2}$

For the class mark $m = 20$ and class width $w = 5$, we can find the lower and upper limits.

Lower Limit ($L$) = $m - \frac{w}{2} = 20 - \frac{5}{2} = 20 - 2.5 = 17.5$

Upper Limit ($U$) = $m + \frac{w}{2} = 20 + \frac{5}{2} = 20 + 2.5 = 22.5$

The class interval corresponding to the class mark 20 is [17.5, 22.5), which is written as 17.5 – 22.5 in the options.

Final Answer:

The class corresponding to the class mark 20 is 17.5 – 22.5.

This corresponds to option (B).

Given:

The class intervals are 10-20 and 20-30.

To Find:

Which class interval includes the number 20.

Solution:

In a continuous frequency distribution, the convention is that the upper limit of a class is excluded from that class, while the lower limit of the next class is included in that class.

This means the interval 10-20 typically represents the range $[10, 20)$, which includes values from 10 up to, but not including, 20.

The interval 20-30 typically represents the range $[20, 30)$, which includes values from 20 up to, but not including, 30.

Following this convention, the number 20 is the upper limit of the class 10-20 and is not included in this class.

The number 20 is the lower limit of the class 20-30 and is included in this class.

Final Answer:

The number 20 is included in the class interval 20-30.

This corresponds to option (B).

Given Data:

The data points are:

268, 220, 368, 258, 242, 310, 272, 342, 310, 290, 300, 320, 319, 304, 402, 318, 406, 292, 354, 278, 210, 240, 330, 316, 406, 215, 258, 236.

The class intervals are of equal size, and one interval is 250-270 (270 not included).

This implies a continuous frequency distribution where the class intervals are of the form [lower limit, upper limit).

To Find:

The frequency of the class 310-330.

Solution:

The class interval 250-270 (270 not included) is represented as $[250, 270)$. The width of this class is $270 - 250 = 20$.

Since the class intervals have equal sizes, the width of every class interval is 20.

The class 310-330 represents the interval $[310, 330)$. This means we need to count the number of data points that are greater than or equal to 310 and strictly less than 330.

Let's go through the data points and count those that fall within the interval $[310, 330)$: ($310 \leq x < 330$)

• 268 (No)
• 220 (No)
• 368 (No)
• 258 (No)
• 242 (No)
• 310 (Yes, $310 \geq 310$)
• 272 (No)
• 342 (No)
• 310 (Yes, $310 \geq 310$)
• 290 (No)
• 300 (No)
• 320 (Yes, $320 < 330$)
• 319 (Yes, $319 < 330$)
• 304 (No)
• 402 (No)
• 318 (Yes, $318 < 330$)
• 406 (No)
• 292 (No)
• 354 (No)
• 278 (No)
• 210 (No)
• 240 (No)
• 330 (No, $330$ is not included)
• 316 (Yes, $316 < 330$)
• 406 (No)
• 215 (No)
• 258 (No)
• 236 (No)

The data points within the interval $[310, 330)$ are 310, 310, 320, 319, 318, and 316.

There are 6 such data points.

Therefore, the frequency of the class 310-330 is 6.

Final Answer:

The frequency of the class 310-330 is 6.

This corresponds to option (C).

Given Data:

The data points are:

30, 32, 45, 54, 74, 78, 108, 112, 66, 76, 88, 40, 14, 20, 15, 35, 44, 66, 75, 84, 95, 96, 102, 110, 88, 74, 112, 14, 34, 44.

A grouped frequency distribution is to be constructed with classes of equal sizes.

One of the class intervals is 63-72, with 72 included in this interval.

To Find:

The number of classes in the distribution.

Solution:

The class interval 63-72 includes 63 and 72. This is an inclusive class interval, represented as $[63, 72]$.

The class width ($w$) for an inclusive interval $[L, U]$ is given by the formula $w = U - L + 1$.

Using the given class interval [63, 72]:

$w = 72 - 63 + 1$

$w = 9 + 1$

$w = 10$

So, the width of each class in the distribution is 10.

Next, we need to find the minimum and maximum values in the given data set.

The data points are:

30, 32, 45, 54, 74, 78, 108, 112, 66, 76, 88, 40, 14, 20, 15, 35, 44, 66, 75, 84, 95, 96, 102, 110, 88, 74, 112, 14, 34, 44.

By examining the data, the minimum value is 14.

By examining the data, the maximum value is 112.

The class intervals must cover the entire range of the data, from the minimum value to the maximum value.

Since one class is [63, 72] and the class width is 10, the lower limits of the classes will follow a pattern with a difference of 10. The lower limits would be ..., $63-10$, 63, $63+10$, ... i.e., ..., 53, 63, 73, ...

The classes are inclusive, so they are of the form $[L, L+w-1]$, which is $[L, L+9]$ since $w=10$.

Starting from the given class [63, 72], we can determine the sequence of class intervals:

• Working downwards from [63, 72]: The lower limit before 63 would be $63 - 10 = 53$. The class is [53, $53+9$] = [53, 62].
• Before [53, 62]: The lower limit before 53 would be $53 - 10 = 43$. The class is [43, 52].
• Before [43, 52]: The lower limit before 43 would be $43 - 10 = 33$. The class is [33, 42].
• Before [33, 42]: The lower limit before 33 would be $33 - 10 = 23$. The class is [23, 32].
• Before [23, 32]: The lower limit before 23 would be $23 - 10 = 13$. The class is [13, 22].

The minimum data value is 14, which falls within the class [13, 22]. So, this must be the first class.

Now, working upwards from [63, 72]:

• After [63, 72]: The lower limit after 63 would be $63 + 10 = 73$. The class is [73, $73+9$] = [73, 82].
• After [73, 82]: The lower limit after 73 would be $73 + 10 = 83$. The class is [83, 92].
• After [83, 92]: The lower limit after 83 would be $83 + 10 = 93$. The class is [93, 102].
• After [93, 102]: The lower limit after 93 would be $93 + 10 = 103$. The class is [103, 112].

The maximum data value is 112, which falls within the class [103, 112]. The next class would start at $103 + 10 = 113$, which is beyond the maximum value, so [103, 112] is the last class.

The complete list of classes is:

• [13, 22]
• [23, 32]
• [33, 42]
• [43, 52]
• [53, 62]
• [63, 72]
• [73, 82]
• [83, 92]
• [93, 102]
• [103, 112]

Counting these intervals, we find there are 10 classes.

Final Answer:

The number of classes in the distribution will be 10.

This corresponds to option (B).

Given:

The frequency distribution table is:

The class intervals are of unequal widths.

To Find:

The adjusted frequency for the class 25-45 for drawing a histogram.

Solution:

When the class intervals in a frequency distribution are unequal, the frequencies need to be adjusted before drawing a histogram. The height of the rectangle in the histogram is proportional to the frequency density, not just the frequency.

The formula for adjusted frequency (which is proportional to frequency density) is:

$Adjusted\ Frequency = \frac{\text{Frequency of the Class}}{\text{Width of the Class}} \times \text{Minimum Class Width}$

First, let's find the width of each class interval:

• Class 5-10: Width = $10 - 5 = 5$
• Class 10-15: Width = $15 - 10 = 5$
• Class 15-25: Width = $25 - 15 = 10$
• Class 25-45: Width = $45 - 25 = 20$
• Class 45-75: Width = $75 - 45 = 30$

The class widths are 5, 5, 10, 20, and 30.

The minimum class width among these is 5.

We need to find the adjusted frequency for the class 25-45.

For the class 25-45:

• Frequency = 8
• Width of the class = 20
• Minimum class width = 5

Using the formula for adjusted frequency:

$Adjusted\ Frequency\ (25\text{-}45) = \frac{\text{Frequency of 25-45}}{\text{Width of 25-45}} \times \text{Minimum Class Width}$

$Adjusted\ Frequency\ (25\text{-}45) = \frac{8}{20} \times 5$

$Adjusted\ Frequency\ (25\text{-}45) = \frac{\cancel{8}^{2}}{\cancel{20}_{5}} \times 5$

$Adjusted\ Frequency\ (25\text{-}45) = \frac{2}{5} \times 5$

$Adjusted\ Frequency\ (25\text{-}45) = 2$

Final Answer:

The adjusted frequency for the class 25-45 is 2.

This corresponds to option (D).

Given:

Number of initial observations ($n_1$) = 5

Mean of initial observations ($\bar{x}_1$) = 30

Number of observations after excluding one = 4

Mean of remaining observations ($\bar{x}_2$) = 28

To Find:

The value of the excluded number.

Solution:

The mean of a set of numbers is calculated by dividing the sum of the numbers by the count of the numbers.

Mean = $\frac{\text{Sum of observations}}{\text{Number of observations}}$

Let $S_1$ be the sum of the initial five numbers.

The mean of the initial five numbers is given by:

$\bar{x}_1 = \frac{S_1}{n_1}$

Substitute the given values:

$30 = \frac{S_1}{5}$

To find the sum $S_1$, multiply both sides by 5:

$S_1 = 30 \times 5$

$S_1 = 150$

Let the excluded number be $x$.

After excluding one number, the number of observations is $n_2 = 5 - 1 = 4$.

The sum of the remaining four numbers is $S_2 = S_1 - x$.

The mean of the remaining four numbers is given by:

$\bar{x}_2 = \frac{S_2}{n_2}$

Substitute the given mean and the sum of the remaining numbers:

$28 = \frac{150 - x}{4}$

To solve for $x$, multiply both sides by 4:

$28 \times 4 = 150 - x$

$112 = 150 - x$

Rearrange the equation to find $x$:

$x = 150 - 112$

$x = 38$

The excluded number is 38.

Final Answer:

The excluded number is 38.

This corresponds to option (D).

Given:

The five observations are: $x, x+3, x+5, x+7, x+10$.

The mean of these five observations is 9.

To Find:

The mean of the last three observations: $x+5, x+7, x+10$.

Solution:

The mean of a set of observations is calculated as the sum of the observations divided by the number of observations.

Let the sum of the five observations be $S_5$.

$S_5 = x + (x+3) + (x+5) + (x+7) + (x+10)$

$S_5 = x+x+3+x+5+x+7+x+10$

$S_5 = (x+x+x+x+x) + (3+5+7+10)$

$S_5 = 5x + 25$

The number of observations is 5.

The mean of these five observations is given as 9.

Mean = $\frac{\text{Sum of observations}}{\text{Number of observations}}$

$9 = \frac{5x + 25}{5}$

Multiply both sides by 5:

$9 \times 5 = 5x + 25$

$45 = 5x + 25$

Subtract 25 from both sides:

$45 - 25 = 5x$

$20 = 5x$

Divide both sides by 5:

$x = \frac{20}{5}$

$x = 4$

Now that we have the value of $x$, we can find the values of the last three observations.

The last three observations are: $x+5, x+7, x+10$.

Substitute $x=4$ into these expressions:

First observation (of the last three) = $x+5 = 4+5 = 9$

Second observation (of the last three) = $x+7 = 4+7 = 11$

Third observation (of the last three) = $x+10 = 4+10 = 14$

The last three observations are 9, 11, and 14.

Now, we calculate the mean of these three observations.

Sum of the last three observations = $9 + 11 + 14 = 34$

Number of these observations = 3

Mean of the last three observations = $\frac{\text{Sum of last three observations}}{\text{Number of last three observations}}$

Mean = $\frac{34}{3}$

To express this as a mixed fraction, we divide 34 by 3.

$34 \div 3 = 11$ with a remainder of 1.

So, $\frac{34}{3} = 11 \frac{1}{3}$.

Final Answer:

The mean of the last three observations is $11 \frac{1}{3}$.

This corresponds to option (C).

Given:

The observations are $x_1, x_2, ..., x_n$.

The number of observations is $n$.

The mean of these observations is $\overline{x}$.

To Find:

The value of $\sum\limits_{i=1}^{n} (x_i - \overline{x})$.

Solution:

The mean $\overline{x}$ of $n$ observations $x_1, x_2, ..., x_n$ is defined as:

$\overline{x} = \frac{\sum\limits_{i=1}^{n} x_i}{n}$

From this definition, the sum of the observations is:

$\sum\limits_{i=1}^{n} x_i = n \overline{x}$

We need to evaluate the expression $\sum\limits_{i=1}^{n} (x_i - \overline{x})$.

Using the property of summation, $\sum (a_i - b_i) = \sum a_i - \sum b_i$, we can write:

$\sum\limits_{i=1}^{n} (x_i - \overline{x}) = \sum\limits_{i=1}^{n} x_i - \sum\limits_{i=1}^{n} \overline{x}$

The first term on the right side is the sum of the observations, which is $\sum\limits_{i=1}^{n} x_i = n \overline{x}$.

The second term is the sum of the constant value $\overline{x}$ repeated $n$ times. The sum of a constant $c$ repeated $n$ times is $n \times c$. In this case, the constant is $\overline{x}$.

So, $\sum\limits_{i=1}^{n} \overline{x} = n \times \overline{x} = n \overline{x}$.

Substitute these values back into the expression:

$\sum\limits_{i=1}^{n} (x_i - \overline{x}) = (n \overline{x}) - (n \overline{x})$

$\sum\limits_{i=1}^{n} (x_i - \overline{x}) = 0$

The sum of the deviations of a set of observations from their mean is always zero.

Final Answer:

The value of $\sum\limits_{i=1}^{n} (x_i - \overline{x})$ is 0.

This corresponds to option (B).

Given:

A set of $n$ observations: $x_1, x_2, ..., x_n$.

Each observation is increased by 5.

To Find:

How the mean of the data changes when each observation is increased by 5.

Solution:

Let the original observations be $x_1, x_2, ..., x_n$.

The number of observations is $n$.

The original mean, denoted by $\overline{x}_{old}$, is given by:

$\overline{x}_{old} = \frac{x_1 + x_2 + ... + x_n}{n} = \frac{\sum\limits_{i=1}^{n} x_i}{n}$

Now, each observation is increased by 5. The new observations are:

$y_1 = x_1 + 5$

$y_2 = x_2 + 5$

...

$y_n = x_n + 5$

The new mean, denoted by $\overline{x}_{new}$, is the sum of the new observations divided by the number of observations ($n$).

$\overline{x}_{new} = \frac{y_1 + y_2 + ... + y_n}{n} = \frac{\sum\limits_{i=1}^{n} y_i}{n}$

Substitute $y_i = x_i + 5$ into the formula for the new mean:

$\overline{x}_{new} = \frac{\sum\limits_{i=1}^{n} (x_i + 5)}{n}$

Using the property of summation $\sum (a_i + b_i) = \sum a_i + \sum b_i$, we can split the sum in the numerator:

$\overline{x}_{new} = \frac{\sum\limits_{i=1}^{n} x_i + \sum\limits_{i=1}^{n} 5}{n}$

The term $\sum\limits_{i=1}^{n} x_i$ is the sum of the original observations.

The term $\sum\limits_{i=1}^{n} 5$ is the sum of the constant 5 repeated $n$ times, which is $5 \times n = 5n$.

So, the equation becomes:

$\overline{x}_{new} = \frac{(\sum\limits_{i=1}^{n} x_i) + 5n}{n}$

Now, we can separate the terms in the numerator:

$\overline{x}_{new} = \frac{\sum\limits_{i=1}^{n} x_i}{n} + \frac{5n}{n}$

We know that $\frac{\sum\limits_{i=1}^{n} x_i}{n}$ is the original mean $\overline{x}_{old}$.

$\frac{5n}{n} = 5$ (assuming $n \neq 0$, which is true for calculating a mean).

Therefore, the new mean is:

$\overline{x}_{new} = \overline{x}_{old} + 5$

This shows that when each observation of the data is increased by a constant value (in this case, 5), the mean of the data is also increased by the same constant value.

Final Answer:

If each observation of the data is increased by 5, then their mean is increased by 5.

This corresponds to option (D).

Given:

The mean of $n$ observations $x_1, x_2, ..., x_n$ is $\overline{x}$.

The mean of $n$ observations $y_1, y_2, ..., y_n$ is $\overline{y}$.

The mean of the combined set of $2n$ observations $x_1, x_2, ..., x_n, y_1, y_2, ..., y_n$ is $\overline{z}$.

To Find:

The value of $\overline{z}$ in terms of $\overline{x}$ and $\overline{y}$.

Solution:

The mean of a set of observations is the sum of the observations divided by the number of observations.

For the first set of observations $x_1, x_2, ..., x_n$, the mean is $\overline{x}$.

$\overline{x} = \frac{\sum\limits_{i=1}^{n} x_i}{n}$

The sum of the first set of observations is:

$\sum\limits_{i=1}^{n} x_i = n \overline{x}$

For the second set of observations $y_1, y_2, ..., y_n$, the mean is $\overline{y}$.

$\overline{y} = \frac{\sum\limits_{i=1}^{n} y_i}{n}$

The sum of the second set of observations is:

$\sum\limits_{i=1}^{n} y_i = n \overline{y}$

The combined set of observations consists of $x_1, x_2, ..., x_n$ and $y_1, y_2, ..., y_n$.

The total number of observations in the combined set is $n + n = 2n$.

The sum of the observations in the combined set is the sum of the observations in the first set plus the sum of the observations in the second set.

Sum of combined set = $\sum\limits_{i=1}^{n} x_i + \sum\limits_{i=1}^{n} y_i$

Substitute the expressions for the sums in terms of the means:

Sum of combined set = $n \overline{x} + n \overline{y}$

The mean of the combined set, $\overline{z}$, is the sum of the combined set divided by the total number of observations (2n).

$\overline{z} = \frac{\text{Sum of combined set}}{\text{Total number of observations}}$

$\overline{z} = \frac{n \overline{x} + n \overline{y}}{2n}$

We can factor out $n$ from the numerator:

$\overline{z} = \frac{n (\overline{x} + \overline{y})}{2n}$

Assuming $n \neq 0$ (which must be true for the mean to be defined), we can cancel out the factor of $n$ from the numerator and the denominator.

$\overline{z} = \frac{\cancel{n} (\overline{x} + \overline{y})}{2\cancel{n}}$

$\overline{z} = \frac{\overline{x} + \overline{y}}{2}$

Thus, the mean of the combined set is the average of the two individual means, but only because the number of observations in each set is the same ($n$). If the number of observations were different, say $n_1$ and $n_2$, the combined mean would be $\frac{n_1 \overline{x} + n_2 \overline{y}}{n_1 + n_2}$.

Final Answer:

The mean of the combined set is $\frac{\overline{x} + \overline{y}}{2}$.

This corresponds to option (B).

Given:

The original observations are $x_1, x_2, ..., x_n$.

The mean of these $n$ observations is $\overline{x}$.

A new set of observations is formed by combining $ax_1, ax_2, ..., ax_n$ and $\frac{x_1}{a}, \frac{x_2}{a}, ..., \frac{x_n}{a}$, where $a \neq 0$.

To Find:

The mean of the new combined set of observations.

Solution:

The mean of the original $n$ observations is given by:

$\overline{x} = \frac{\sum\limits_{i=1}^{n} x_i}{n}$

From this, the sum of the original observations is:

$\sum\limits_{i=1}^{n} x_i = n\overline{x}$

The new set of observations consists of two groups:

Group 1: $ax_1, ax_2, ..., ax_n$ (n observations)

Group 2: $\frac{x_1}{a}, \frac{x_2}{a}, ..., \frac{x_n}{a}$ (n observations)

The total number of observations in the new combined set is $n + n = 2n$.

The sum of the observations in the new combined set is the sum of Group 1 plus the sum of Group 2.

Sum of Group 1 = $\sum\limits_{i=1}^{n} (ax_i) = a \sum\limits_{i=1}^{n} x_i$

Sum of Group 2 = $\sum\limits_{i=1}^{n} (\frac{x_i}{a}) = \frac{1}{a} \sum\limits_{i=1}^{n} x_i$

Sum of the new combined set = (Sum of Group 1) + (Sum of Group 2)

Sum of the new combined set = $a \sum\limits_{i=1}^{n} x_i + \frac{1}{a} \sum\limits_{i=1}^{n} x_i$

Factor out $\sum\limits_{i=1}^{n} x_i$:

Sum of the new combined set = $\left(a + \frac{1}{a}\right) \sum\limits_{i=1}^{n} x_i$

Substitute the value of $\sum\limits_{i=1}^{n} x_i$ from the original mean formula:

Sum of the new combined set = $\left(a + \frac{1}{a}\right) (n\overline{x})$

The mean of the new combined set, $\overline{z}$, is the sum of the new combined set divided by the total number of observations ($2n$).

$\overline{z} = \frac{\text{Sum of the new combined set}}{\text{Total number of new observations}}$

$\overline{z} = \frac{\left(a + \frac{1}{a}\right) (n\overline{x})}{2n}$

Cancel out the common factor $n$ from the numerator and the denominator (since $n$ must be greater than 0 for the concept of mean to be meaningful).

$\overline{z} = \frac{\left(a + \frac{1}{a}\right) \cancel{n}\overline{x}}{2\cancel{n}}$

$\overline{z} = \frac{\left(a + \frac{1}{a}\right) \overline{x}}{2}$

This can also be written as:

$\overline{z} = \left( a + \frac{1}{a} \right)\frac{\overline{x}}{2}$

Final Answer:

The mean of the new combined set of observations is $\left( a + \frac{1}{a} \right)\frac{\overline{x}}{2}$.

This corresponds to option (B).

Given:

There are $n$ groups of observations.

For the $i$-th group ($i = 1, 2, ..., n$):

• Number of observations = $n_i$
• Mean of observations = $\overline{x_i}$

The mean of all the groups taken together is $\overline{x}$.

To Find:

The formula for the combined mean $\overline{x}$.

Solution:

The mean of a group is defined as the sum of observations divided by the number of observations in that group.

For the $i$-th group, the mean $\overline{x_i}$ is given by:

$\overline{x_i} = \frac{\text{Sum of observations in group } i}{\text{Number of observations in group } i}$

Let $S_i$ be the sum of observations in the $i$-th group.

$\overline{x_i} = \frac{S_i}{n_i}$

From this, the sum of observations in group $i$ is:

$S_i = n_i \overline{x_i}$

The mean of all the groups taken together ($\overline{x}$) is the total sum of observations from all groups divided by the total number of observations from all groups.

Total sum of observations = Sum of sums of all groups

Total sum $= S_1 + S_2 + ... + S_n = \sum\limits_{i=1}^{n} S_i$

Substitute the expression for $S_i$:

Total sum $= \sum\limits_{i=1}^{n} (n_i \overline{x_i})$

Total number of observations = Sum of the number of observations in all groups

Total number $= n_1 + n_2 + ... + n_n = \sum\limits_{i=1}^{n} n_i$

The combined mean $\overline{x}$ is:

$\overline{x} = \frac{\text{Total sum of observations}}{\text{Total number of observations}}$

$\overline{x} = \frac{\sum\limits_{i=1}^{n} n_i\overline{x_i}}{\sum\limits_{i=1}^{n} n_i}$

This formula is used to calculate the combined mean (or weighted mean) of several groups.

Final Answer:

The mean $\overline{x}$ of all the groups taken together is given by $\frac{\sum\limits_{i=1}^{n} n_i\overline{x_i}}{\sum\limits_{i=1}^{n} n_i}$.

This corresponds to option (C).

Given:

Number of original observations ($n$) = 100

Original mean ($\overline{x}_{old}$) = 50

Value of the observation replaced = 50

Value of the new observation = 150

To Find:

The resulting mean ($\overline{x}_{new}$) after replacement.

Solution:

The mean of a set of observations is the sum of the observations divided by the number of observations.

Original Mean = $\frac{\text{Sum of original observations}}{\text{Number of observations}}$

Let $S_{old}$ be the sum of the original 100 observations.

$\overline{x}_{old} = \frac{S_{old}}{n}$

Substitute the given values:

$50 = \frac{S_{old}}{100}$

To find the sum $S_{old}$, multiply both sides by 100:

$S_{old} = 50 \times 100$

$S_{old} = 5000$

When one observation (value 50) is removed and a new observation (value 150) is added, the number of observations remains the same (100). The sum of the observations changes.

The new sum, $S_{new}$, is calculated as the original sum minus the value of the excluded observation plus the value of the included observation.

$S_{new} = S_{old} - (\text{Value of replaced observation}) + (\text{Value of new observation})$

$S_{new} = 5000 - 50 + 150$

$S_{new} = 5000 + (150 - 50)$

$S_{new} = 5000 + 100$

$S_{new} = 5100$

The number of observations for the new set is still $n = 100$.

The resulting mean, $\overline{x}_{new}$, is the new sum divided by the number of observations:

$\overline{x}_{new} = \frac{S_{new}}{n}$

$\overline{x}_{new} = \frac{5100}{100}$

$\overline{x}_{new} = 51$

Final Answer:

The resulting mean will be 51.

This corresponds to option (B).

Given:

Number of original observations ($n$) = 50

Each observation is subtracted from a constant value = 53

Mean of the new numbers obtained = –3.5

To Find:

The mean of the original numbers.

Solution:

Let the original observations be $x_1, x_2, ..., x_{50}$.

The number of observations is $n = 50$.

Let the mean of the original numbers be $\overline{x}_{old}$.

By definition, the original mean is:

$\overline{x}_{old} = \frac{x_1 + x_2 + ... + x_{50}}{50} = \frac{\sum\limits_{i=1}^{50} x_i}{50}$

This implies that the sum of the original observations is:

$\sum\limits_{i=1}^{50} x_i = 50 \times \overline{x}_{old}$

According to the problem, each original number $x_i$ is subtracted from 53 to obtain a new number $y_i$.

The new observations are: $y_1 = 53 - x_1$, $y_2 = 53 - x_2$, ..., $y_{50} = 53 - x_{50}$.

The number of new observations is still 50.

The mean of these new observations, denoted by $\overline{y}$, is given as –3.5.

The mean of the new numbers is:

$\overline{y} = \frac{y_1 + y_2 + ... + y_{50}}{50} = \frac{\sum\limits_{i=1}^{50} y_i}{50}$

Substitute the expression for $y_i$ into the formula for $\overline{y}$:

$\overline{y} = \frac{\sum\limits_{i=1}^{50} (53 - x_i)}{50}$

We are given $\overline{y} = -3.5$. So,

$-3.5 = \frac{\sum\limits_{i=1}^{50} (53 - x_i)}{50}$

Using the properties of summation ($\sum (a - b_i) = \sum a - \sum b_i$), we can split the sum in the numerator:

$-3.5 = \frac{\sum\limits_{i=1}^{50} 53 - \sum\limits_{i=1}^{50} x_i}{50}$

The sum of a constant (53) repeated 50 times is $50 \times 53$.

$\sum\limits_{i=1}^{50} 53 = 50 \times 53 = 2650$

The sum $\sum\limits_{i=1}^{50} x_i$ is the sum of the original observations, which is equal to $50 \times \overline{x}_{old}$.

Substitute these values back into the equation for the new mean:

$-3.5 = \frac{2650 - (50 \times \overline{x}_{old})}{50}$

We can divide each term in the numerator by 50:

$-3.5 = \frac{2650}{50} - \frac{50 \times \overline{x}_{old}}{50}$

$-3.5 = 53 - \overline{x}_{old}$

Now, we solve this equation for $\overline{x}_{old}$. Add $\overline{x}_{old}$ to both sides and add 3.5 to both sides:

$\overline{x}_{old} = 53 + 3.5$

$\overline{x}_{old} = 56.5$

Thus, the mean of the original numbers is 56.5.

Alternatively (Using property of mean):

If each observation $x_i$ is transformed linearly to $y_i = a x_i + b$, then the new mean $\overline{y}$ is related to the original mean $\overline{x}_{old}$ by the formula $\overline{y} = a \overline{x}_{old} + b$.

In this problem, the new observation $y_i$ is obtained by subtracting the original observation $x_i$ from 53: $y_i = 53 - x_i$.

This can be written in the form $y_i = a x_i + b$ as $y_i = (-1)x_i + 53$.

Here, $a = -1$ and $b = 53$.

The new mean $\overline{y} = -3.5$.

Using the property:

$\overline{y} = a \overline{x}_{old} + b$

$-3.5 = (-1) \overline{x}_{old} + 53$

$-3.5 = -\overline{x}_{old} + 53$

Rearranging the equation to solve for $\overline{x}_{old}$:

$\overline{x}_{old} = 53 + 3.5$

$\overline{x}_{old} = 56.5$

Both methods yield the same result.

Final Answer:

The mean of the given numbers is 56.5.

This corresponds to option (D).

Given:

Total number of observations ($N$) = 25

Mean of 25 observations ($\overline{x}_{total}$) = 36

Number of first observations ($n_1$) = 13

Mean of first 13 observations ($\overline{x}_1$) = 32

Number of last observations ($n_2$) = 13

Mean of last 13 observations ($\overline{x}_2$) = 40

To Find:

The value of the 13th observation.

Solution:

The mean of a set of observations is given by the formula:

$Mean = \frac{\text{Sum of observations}}{\text{Number of observations}}$

This can be rearranged to find the sum of observations:

Sum of observations = Mean $\times$ Number of observations

Step 1: Calculate the total sum of 25 observations.

The mean of 25 observations is 36.

Total sum of 25 observations ($S_{total}$) = $\overline{x}_{total} \times N$

$S_{total} = 36 \times 25$

$S_{total} = 900$

Step 2: Calculate the sum of the first 13 observations.

The mean of the first 13 observations is 32.

Sum of first 13 observations ($S_1$) = $\overline{x}_1 \times n_1$

$S_1 = 32 \times 13$

$S_1 = 416$

Step 3: Calculate the sum of the last 13 observations.

The mean of the last 13 observations is 40.

Sum of last 13 observations ($S_2$) = $\overline{x}_2 \times n_2$

$S_2 = 40 \times 13$

$S_2 = 520$

Step 4: Determine the value of the 13th observation.

The total set of 25 observations can be considered as the first 12 observations, followed by the 13th observation, followed by the remaining 12 observations (from 14th to 25th).

The sum of the first 13 observations ($S_1$) includes the sum of the first 12 observations and the 13th observation.

The sum of the last 13 observations ($S_2$) includes the sum of the 13th observation and the sum of the observations from the 14th to the 25th.

The sum of the first 13 observations ($S_1$) and the sum of the last 13 observations ($S_2$) together sum up the observations from 1 to 12, the observations from 14 to 25, and the 13th observation *twice* (since it is in both sets).

So, the sum of $S_1$ and $S_2$ is equal to the total sum of 25 observations ($S_{total}$) plus the value of the 13th observation (which was counted twice).

Let the 13th observation be $x_{13}$.

$S_1 + S_2 = S_{total} + x_{13}$

Substitute the calculated sums:

$416 + 520 = 900 + x_{13}$

$936 = 900 + x_{13}$

To find $x_{13}$, subtract 900 from both sides:

$x_{13} = 936 - 900$

$x_{13} = 36$

Final Answer:

The 13th observation is 36.

This corresponds to option (B).

Given Data:

The data set is: 78, 56, 22, 34, 45, 54, 39, 68, 54, 84

To Find:

The median of the given data.

Solution:

To find the median of a data set, we first need to arrange the data in ascending order (from least to greatest).

The given data points are: 78, 56, 22, 34, 45, 54, 39, 68, 54, 84.

Arranging the data in ascending order:

22, 34, 39, 45, 54, 54, 56, 68, 78, 84

Next, we need to determine the number of observations in the data set.

The number of observations is $n = 10$.

Since the number of observations ($n=10$) is an even number, the median is the average of the two middle values.

The positions of the two middle values are $\frac{n}{2}$ and $\frac{n}{2} + 1$.

$\frac{n}{2} = \frac{10}{2} = 5$

$\frac{n}{2} + 1 = \frac{10}{2} + 1 = 5 + 1 = 6$

So, the median is the average of the value at the 5th position and the value at the 6th position in the ordered data.

Ordered data: 22, 34, 39, 45, 54 (5th), 54 (6th), 56, 68, 78, 84

The value at the 5th position is 54.

The value at the 6th position is 54.

Median = $\frac{\text{Value at 5th position} + \text{Value at 6th position}}{2}$

Median = $\frac{54 + 54}{2}$

Median = $\frac{108}{2}$

Median = 54

Final Answer:

The median of the data is 54.

This corresponds to option (C).

Given:

We are drawing a frequency polygon for a continuous frequency distribution.

The ordinates (y-values) are the frequencies of the respective classes.

To Find:

What the abscissae (x-values) represent when plotting points for a frequency polygon.

Solution:

A frequency polygon is a graphical representation of a frequency distribution. It can be drawn in two ways:

1. By joining the mid-points of the tops of the rectangles in a histogram.
2. By plotting points with coordinates (class mark, frequency) and joining these points by line segments.

In both methods, the points that are joined to form the polygon are located at the class marks of the intervals along the horizontal axis (abscissa) and at the corresponding frequencies along the vertical axis (ordinate).

The class mark of a class interval is the mid-point of that interval. It represents the center of the class.

Therefore, when plotting the points for a frequency polygon with frequencies as ordinates, the abscissae must be the class marks of the corresponding classes.

Final Answer:

For drawing a frequency polygon, the points are plotted with frequencies as ordinates and class marks of the classes as abscissae.

This corresponds to option (C).

Given Data:

The data set is: 4, 4, 5, 7, 6, 7, 7, 12, 3

To Find:

The median of the given data.

Solution:

To find the median of a data set, we first need to arrange the data in ascending order (from least to greatest).

The given data points are: 4, 4, 5, 7, 6, 7, 7, 12, 3.

Arranging the data in ascending order:

3, 4, 4, 5, 6, 7, 7, 7, 12

Next, we need to determine the number of observations in the data set.

The number of observations is $n = 9$.

Since the number of observations ($n=9$) is an odd number, the median is the value at the middle position.

The position of the median is given by the formula $\frac{n+1}{2}$.

Median position = $\frac{9+1}{2} = \frac{10}{2} = 5$-th position.

So, the median is the value at the 5th position in the ordered data.

Ordered data: 3, 4, 4, 5, 6 (5th), 7, 7, 7, 12

The value at the 5th position is 6.

Median = 6

Final Answer:

The median of the data is 6.

This corresponds to option (C).

Given Data:

The data set is: 15, 14, 19, 20, 14, 15, 16, 14, 15, 18, 14, 19, 15, 17, 15

To Find:

The mode of the given data.

Solution:

The mode of a data set is the observation that occurs most frequently.

To find the mode, we can count the frequency of each distinct value in the data set.

Let's list the distinct values and their frequencies:

• Value 14: Occurs 4 times (14, 14, 14, 14)
• Value 15: Occurs 5 times (15, 15, 15, 15, 15)
• Value 16: Occurs 1 time (16)
• Value 17: Occurs 1 time (17)
• Value 18: Occurs 1 time (18)
• Value 19: Occurs 2 times (19, 19)
• Value 20: Occurs 1 time (20)

Let's summarize the frequencies:

Frequency of 14 = 4

Frequency of 15 = 5

Frequency of 16 = 1

Frequency of 17 = 1

Frequency of 18 = 1

Frequency of 19 = 2

Frequency of 20 = 1

The value with the highest frequency is 15, which occurs 5 times.

Therefore, the mode of the data is 15.

Final Answer:

The mode of the data is 15.

This corresponds to option (B).

Given:

Total number of people in the sample study = 642.

Number of people who have a high school certificate = 514.

To Find:

The probability that a randomly selected person has a high school certificate.

Solution:

The empirical probability of an event is calculated as the ratio of the number of favorable outcomes to the total number of possible outcomes.

$P(\text{Event}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

In this case:

The event is selecting a person who has a high school certificate.

The number of favorable outcomes is the number of people who have a high school certificate, which is 514.

The total number of possible outcomes is the total number of people in the sample study, which is 642.

So, the probability is:

$P(\text{High school certificate}) = \frac{514}{642}$

Now, we calculate the decimal value of this fraction:

$P(\text{High school certificate}) = 0.800623...$

Rounding this value to one decimal place, we get 0.8.

Final Answer:

The probability that the person has a high school certificate is approximately 0.8.

This corresponds to option (D).

Given:

Total number of children surveyed = 364.

Number of children who liked to eat potato chips = 91.

To Find:

The probability that a randomly selected child does not like to eat potato chips.

Solution:

First, we need to find the number of children who do not like to eat potato chips.

Number of children who do not like chips = Total number of children - Number of children who like chips

Number of children who do not like chips = $364 - 91 = 273$

The empirical probability of an event is given by the formula:

$P(\text{Event}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

In this case, the event is selecting a child who does not like to eat potato chips.

The number of favorable outcomes is the number of children who do not like chips, which is 273.

The total number of possible outcomes is the total number of children surveyed, which is 364.

So, the probability is:

$P(\text{Does not like chips}) = \frac{273}{364}$

We can simplify this fraction. Both 273 and 364 are divisible by several numbers. We can divide both by their greatest common divisor or simplify step-by-step.

Divide by 7: $\frac{273 \div 7}{364 \div 7} = \frac{39}{52}$

Divide by 13: $\frac{39 \div 13}{52 \div 13} = \frac{3}{4}$

As a decimal, $\frac{3}{4} = 0.75$.

Final Answer:

The probability that a randomly selected child does not like to eat potato chips is 0.75.

This corresponds to option (C).

Given:

The number of students for each blood group is given in the table:

Blood group A: 10 students

Blood group AB: 13 students

Blood group B: 12 students

Blood group O: 15 students

To Find:

The probability that a randomly selected student has blood group B.

Solution:

First, we need to find the total number of students in the class.

Total number of students = Sum of the number of students in each blood group

Total students = $10 + 13 + 12 + 15$

Total students = $23 + 12 + 15$

Total students = $35 + 15$

Total students = $50$

The number of students with blood group B is given as 12.

The probability of an event is calculated as the ratio of the number of favorable outcomes to the total number of possible outcomes.

$P(\text{Event}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

In this case, the event is selecting a student with blood group B.

Number of favorable outcomes (students with blood group B) = 12

Total number of possible outcomes (total students) = 50

So, the probability of selecting a student with blood group B is:

$P(\text{Blood group B}) = \frac{12}{50}$

Simplifying the fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 2:

$P(\text{Blood group B}) = \frac{\cancel{12}^{6}}{\cancel{50}_{25}} = \frac{6}{25}$

Note: Based on the provided options, it appears there might be a discrepancy in the question's data or options, as $\frac{6}{25}$ is not listed. However, if the question were asking for the probability of blood group O, the calculation would be $\frac{15}{50} = \frac{3}{10}$, which is Option (C). Given the options, it is highly likely that option (C) is the intended correct answer, corresponding to the probability of blood group O based on the provided frequencies.

Final Answer:

Based on the given data, the probability of selecting a student with blood group B is $\frac{12}{50} = \frac{6}{25}$. However, this is not among the options. The probability of selecting a student with blood group O is $\frac{15}{50} = \frac{3}{10}$, which corresponds to option (C).

Assuming option (C) is the intended answer, it represents the probability of blood group O.

The final answer is $\boxed{\text{(C)}}$.

Given:

Total number of times two coins are tossed = 1000.

The outcomes and their recorded frequencies are:

• Number of Heads = 2, Frequency = 200
• Number of Heads = 1, Frequency = 550
• Number of Heads = 0, Frequency = 250

To Find:

The probability of getting at most one head.

Solution:

The event "at most one head" means getting either 0 heads or 1 head.

From the given data, the number of times this event occurred is the sum of the frequencies for 0 heads and 1 head.

Number of favorable outcomes (at most one head) = (Frequency of 0 Heads) + (Frequency of 1 Head)

Number of favorable outcomes = $250 + 550 = 800$

The total number of trials is the total number of times the coins were tossed, which is 1000.

Total number of trials = 1000

The empirical probability of the event is calculated as:

$P(\text{Event}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of trials}}$

Substituting the values:

$P(\text{at most one head}) = \frac{800}{1000}$

Now, we simplify the fraction:

$P(\text{at most one head}) = \frac{800}{1000} = \frac{80}{100} = \frac{8}{10} = \frac{4}{5}$

Final Answer:

The probability for at most one head is $\frac{4}{5}$.

This corresponds to option (C).

Given:

Total number of bulbs selected = 80.

The observed life times and their frequencies are given in the table:

To Find:

The probability that a randomly selected bulb has a life time of 1150 hours.

Solution:

The probability of an event based on observed frequencies is calculated using the empirical probability formula:

$P(\text{Event}) = \frac{\text{Number of times the event occurred}}{\text{Total number of trials}}$

The event we are interested in is that the life time of a randomly selected bulb is exactly 1150 hours.

From the given frequency table, the observed life times in the sample are 300, 500, 700, 900, and 1100 hours.

The value 1150 hours is not present in the list of observed life times in this sample of 80 bulbs.

This means that the number of times a bulb had a life time of exactly 1150 hours in this study is 0.

Number of favorable outcomes (life time is 1150 hours) = 0

Total number of possible outcomes (total number of bulbs selected) = 80

Using the probability formula:

$P(\text{Life time is 1150 hours}) = \frac{0}{80}$

$P(\text{Life time is 1150 hours}) = 0$

Based on this empirical data from the sample, the probability of selecting a bulb with a life time of 1150 hours is 0 because such an outcome did not occur in the experiment.

Final Answer:

The probability that the life time of a randomly selected bulb is 1150 hours is 0.

This corresponds to option (C).

Given:

Total number of bulbs selected = 80.

The observed life times and their frequencies from Question 29 are:

To Find:

The probability that a randomly selected bulb has a life less than 900 hours.

Solution:

The event "life less than 900 hours" includes the bulbs with recorded life times of 300 hours, 500 hours, and 700 hours.

To find the number of bulbs that satisfy this condition, we sum the frequencies for these life times:

Number of bulbs with life less than 900 hours = (Frequency for 300 hours) + (Frequency for 500 hours) + (Frequency for 700 hours)

Number of favorable outcomes = $10 + 12 + 23 = 45$

The total number of possible outcomes is the total number of bulbs selected from the lot, which is 80.

Total number of trials = 80

The empirical probability of the event is calculated as:

$P(\text{Life} < 900 \text{ hours}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of trials}}$

$P(\text{Life} < 900 \text{ hours}) = \frac{45}{80}$

Now, we simplify the fraction. Both the numerator (45) and the denominator (80) are divisible by 5.

$P(\text{Life} < 900 \text{ hours}) = \frac{\cancel{45}^{9}}{\cancel{80}_{16}} = \frac{9}{16}$

Final Answer:

The probability that bulbs selected randomly from the lot has life less than 900 hours is $\frac{9}{16}$.

This corresponds to option (D).

Given:

Data Set 1: 2, 8, 6, 5, 4, 5, 6, 3, 6, 4, 9, 1, 5, 6, 5

Mean of Data Set 1 is given as 5.

Data Set 2: 10, 12, 10, 2, 18, 8, 12, 6, 12, 10, 8, 10, 12, 16, 4

To Check and Reason:

Is it correct to say that the mean of Data Set 2 is 10, based on the given information? Provide a reason.

Solution:

First, let's verify the given mean of Data Set 1.

Data Set 1: 2, 8, 6, 5, 4, 5, 6, 3, 6, 4, 9, 1, 5, 6, 5

Number of observations in Data Set 1 ($n_1$) = 15

Sum of observations in Data Set 1 ($S_1$) = $2 + 8 + 6 + 5 + 4 + 5 + 6 + 3 + 6 + 4 + 9 + 1 + 5 + 6 + 5$

$S_1 = 75$

Mean of Data Set 1 ($\overline{x}_1$) = $\frac{S_1}{n_1} = \frac{75}{15} = 5$

The given mean of 5 for the first data set is correct.

Now, let's calculate the mean of Data Set 2.

Data Set 2: 10, 12, 10, 2, 18, 8, 12, 6, 12, 10, 8, 10, 12, 16, 4

Number of observations in Data Set 2 ($n_2$) = 15

Sum of observations in Data Set 2 ($S_2$) = $10 + 12 + 10 + 2 + 18 + 8 + 12 + 6 + 12 + 10 + 8 + 10 + 12 + 16 + 4$

$S_2 = 150$

Mean of Data Set 2 ($\overline{x}_2$) = $\frac{S_2}{n_2} = \frac{150}{15} = 10$

Conclusion and Reason:

Yes, it is correct to say that the mean of Data Set 2 is 10.

The reason is that the mean of Data Set 2, calculated directly from the given observations in Data Set 2 using the formula for the mean ($\text{Mean} = \frac{\text{Sum of observations}}{\text{Number of observations}}$), is found to be exactly 10.

The information about the first data set and its mean being 5 is also correct, but the assertion about the mean of the second data set is independently verifiable by calculating the mean of the second data set itself, which is provided in the question.

Given:

In a histogram, the area of each rectangle is proportional to the frequency of the corresponding class.

To Check and Reason:

Can we say that the lengths (heights) of the rectangles are also proportional to the frequencies?

Solution:

In a histogram, the horizontal axis represents the class intervals, and the vertical axis represents the frequency density.

The area of a rectangle in a histogram is given by:

$Area = Width \times Height$

We are given that the Area is proportional to the Frequency. Let $F$ be the frequency and $k$ be a constant of proportionality.

$Area = k \times F$

The width of the rectangle corresponds to the width of the class interval ($W$). The height of the rectangle corresponds to the frequency density ($H$).

In a histogram, the height (frequency density) is defined such that the area is proportional to the frequency. This is achieved by setting the height proportional to the frequency divided by the class width:

$H \propto \frac{F}{W}$

Often, the height is calculated as $H = \frac{F}{W} \times \text{Constant}$, where the constant is typically chosen such that the area is exactly equal to the frequency or proportional to it. A common method is to make the height proportional to the frequency density relative to a standard width, e.g., $H = \frac{F}{W} \times \text{Minimum Class Width}$.

Let's consider the relationship: $Area = W \times H$. Since $Area \propto F$, we have $W \times H \propto F$.

If the lengths (heights) of the rectangles ($H$) were proportional to the frequencies ($F$), this would mean $H \propto F$, or $H = c \times F$ for some constant $c$.

Substituting this into the area relationship: $W \times (c \times F) \propto F$

$(c \times W) \times F \propto F$

This implies that the factor $(c \times W)$ must be constant for the proportionality $Area \propto F$ to hold when $H \propto F$. Since $c$ is a constant, this requires the class width $W$ to be constant for all classes.

If the class intervals have equal widths, then $W$ is constant. In this case, $H \propto F$, and the heights are proportional to the frequencies.

If the class intervals have unequal widths, then $W$ is not constant. In this case, $H$ cannot be proportional to $F$ directly. The height is proportional to the frequency density ($\frac{F}{W}$), ensuring that the area ($W \times H$) is proportional to the frequency ($F$).

Therefore, the lengths (heights) of the rectangles are proportional to the frequencies only when the class intervals are of equal width.

Since histograms can have class intervals of unequal widths, we cannot generally say that the lengths (heights) of the rectangles are proportional to the frequencies.

Final Answer:

No, we cannot say that the lengths of the rectangles are always proportional to the frequencies.

Reason: The areas of the rectangles in a histogram are proportional to the frequencies (Area $\propto$ Frequency). The area of a rectangle is the product of its width (class width) and its height (length). So, Class Width $\times$ Height $\propto$ Frequency. If the class widths are unequal, the height must be proportional to the frequency density (Frequency / Class Width) to maintain the proportionality between area and frequency. Therefore, the height is proportional to the frequency only when all class widths are equal.

Given Data:

The data set is: 2, 3, 9, 16, 9, 3, 9.

The highest value in the observations is 16.

To Check and Reason:

Is it correct to say that 16 is the mode of the data because it is the highest value? Provide a reason.

Solution:

In statistics, the mode of a data set is the value that appears most frequently.

To find the mode, we need to count how many times each distinct value appears in the given data set.

The distinct values in the data set (2, 3, 9, 16, 9, 3, 9) are 2, 3, 9, and 16.

Let's find the frequency of each value:

• Value 2: Appears 1 time.
• Value 3: Appears 2 times.
• Value 9: Appears 3 times.
• Value 16: Appears 1 time.

The frequency of each value is:

Frequency of 2 = 1

Frequency of 3 = 2

Frequency of 9 = 3

Frequency of 16 = 1

The value with the highest frequency is 9, which appears 3 times.

Therefore, the mode of the data is 9.

The highest value in the data set is 16.

Comparing the mode (9) and the highest value (16), we see that they are different.

Conclusion and Reason:

No, it is not correct to say that 16 is the mode of the data just because it is the highest value.

Reason: The mode is defined as the observation that occurs with the highest frequency, not necessarily the observation with the largest value. In this data set, the value 9 has the highest frequency (3 times), whereas the value 16 has a frequency of only 1. Therefore, the mode of the data is 9, not 16.

Given:

The frequency distribution table is:

A graphical representation (histogram) of this distribution is provided (shown in the image).

To Check and Reason:

Is the graphical representation correct? Provide a reason.

Solution:

Let's examine the class intervals and their widths:

• Class 0 - 20: Width = $20 - 0 = 20$
• Class 20 - 40: Width = $40 - 20 = 20$
• Class 40 - 60: Width = $60 - 40 = 20$
• Class 60 - 100: Width = $100 - 60 = 40$

We observe that the class intervals have unequal widths. The first three classes have a width of 20, while the last class has a width of 40.

When constructing a histogram for a frequency distribution with unequal class intervals, the area of each rectangle must be proportional to the frequency of the corresponding class.

The area of a rectangle in a histogram is given by: Area = Width $\times$ Height.

Since Area $\propto$ Frequency, we have: Width $\times$ Height $\propto$ Frequency.

This implies that the Height of the rectangle must be proportional to $\frac{\text{Frequency}}{\text{Width}}$. The term $\frac{\text{Frequency}}{\text{Width}}$ is called the frequency density.

In a correctly drawn histogram with unequal class widths, the heights of the rectangles are proportional to the frequency densities, not the frequencies themselves.

Let's calculate the frequency density for each class:

• Class 0 - 20: Frequency Density = $\frac{10}{20} = 0.5$
• Class 20 - 40: Frequency Density = $\frac{15}{20} = 0.75$
• Class 40 - 60: Frequency Density = $\frac{20}{20} = 1.0$
• Class 60 - 100: Frequency Density = $\frac{25}{40} = 0.625$

The heights of the bars in a correct histogram should be proportional to these frequency densities (0.5, 0.75, 1.0, 0.625). This means the heights should be in the ratio 0.5 : 0.75 : 1.0 : 0.625. Multiplying by 40 to clear decimals, the ratio of heights should be 20 : 30 : 40 : 25.

If the graphical representation shows the heights of the rectangles directly proportional to the frequencies (10, 15, 20, 25), the ratio of heights would be 10 : 15 : 20 : 25, which simplifies to 2 : 3 : 4 : 5.

Comparing the required ratio of heights (20 : 30 : 40 : 25) with the ratio of direct frequencies (10 : 15 : 20 : 25), we see that they are different because of the unequal class widths.

For example, the class 40-60 has frequency 20 and width 20, while the class 60-100 has frequency 25 and width 40. The frequency density of 40-60 is $20/20 = 1.0$, and the frequency density of 60-100 is $25/40 = 0.625$. A correct histogram should show the height for 60-100 is lower than the height for 40-60, despite having a higher frequency, because its frequency density is lower.

Conclusion:

No, the graphical representation is not correct.

Reason: The class intervals in the given frequency distribution are of unequal widths. In a histogram with unequal class widths, the heights of the rectangles should be proportional to the frequency densities (frequency/class width) to ensure that the area of each rectangle is proportional to the frequency. The provided graph likely shows heights proportional directly to frequencies, which is incorrect for unequal class widths.

Given:

Marks obtained by students in a diagnostic test:

$46, 52, 48, 11, 41, 62, 54, 53, 96, 40, 98, 44$.

To Find:

Which ‘average’ is a good representative of the data and why.

Solution:

The given data represents the marks of $12$ students.

Let's arrange the data in ascending order:

$11, 40, 41, 44, 46, 48, 52, 53, 54, 62, 96, 98$.

We need to consider the different types of averages: Mean, Median, and Mode.

The Mean is calculated by summing all the data points and dividing by the number of data points. The mean is affected by extreme values or outliers.

In this data set, the values $11$, $96$, and $98$ appear to be significantly different from the majority of the other values (which are mostly clustered between $40$ and $62$). These extreme values can be considered as outliers.

Calculating the mean would give:

Mean $= \frac{46+52+48+11+41+62+54+53+96+40+98+44}{12}$

Mean $= \frac{645}{12} \approx 53.75$

The mean is $\approx 53.75$. However, this value is influenced by the high marks ($96, 98$) and the low mark ($11$), pulling it away from the central cluster of marks.

The Mode is the value that appears most frequently in the data set.

Looking at the ordered data ($11, 40, 41, 44, 46, 48, 52, 53, 54, 62, 96, 98$), no value is repeated. Therefore, there is no mode for this data set.

The Median is the middle value of the data when arranged in order. The median is less affected by outliers compared to the mean.

Since there are $12$ data points (an even number), the median is the average of the $6^{th}$ and $7^{th}$ values in the ordered list.

Ordered data: $11, 40, 41, 44, 46, \underline{48}, \underline{52}, 53, 54, 62, 96, 98$.

The $6^{th}$ value is $48$.

The $7^{th}$ value is $52$.

Median $= \frac{6^{th} \text{ value} + 7^{th} \text{ value}}{2}$

Median $= \frac{48 + 52}{2} = \frac{100}{2} = 50$.

The median is $50$. This value is closer to the main cluster of data points and is not significantly skewed by the extreme values.

Conclusion:

Given the presence of outliers (extreme values) in the data set, the Mean is not a good representative because it is heavily influenced by these outliers.

The Mode does not exist for this data set.

The Median is the least affected measure of central tendency by extreme values. Therefore, the Median is the most appropriate average to represent this data set.

The median mark is $50$.

Given:

A set of numbers: $3, 14, 18, 20, 5$.

The child states that the median of these numbers is $18$.

To Find:

What the child doesn’t understand about finding the median.

Solution:

The median is the middle value in a data set that is arranged in either ascending or descending order.

The steps to find the median are:

1. Arrange the data in ascending or descending order.

2. Find the middle value(s) of the ordered data.

Let's find the correct median for the given set of numbers: $3, 14, 18, 20, 5$.

Step 1: Arrange the numbers in ascending order.

The ordered set is: $3, 5, 14, 18, 20$.

Step 2: Find the middle value.

There are $5$ numbers in the ordered set. Since the number of data points is odd ($5$), the median is the single middle value.

In the ordered set $3, 5, \underline{14}, 18, 20$, the middle value is the $3^{rd}$ number, which is $14$.

The correct median of the numbers $3, 14, 18, 20, 5$ is $14$.

The child stated that the median is $18$. The value $18$ is the third number in the original list ($3, 14, \underline{18}, 20, 5$), before ordering.

Therefore, what the child doesn’t understand about finding the median is the crucial first step: that the data must be arranged in order (either ascending or descending) before identifying the middle value.

Given:

Number of goals scored in 10 matches: $1, 3, 2, 5, 8, 6, 1, 4, 7, 9$.

The provided calculation for the median is $7$.

To Determine:

Whether the provided answer for the median is correct and the reason.

Solution:

The median of a data set is the middle value when the data is arranged in order (ascending or descending).

When the number of observations ($n$) is even, the median is the average of the $\left(\frac{n}{2}\right)^{th}$ and $\left(\frac{n}{2} + 1\right)^{th}$ observations after the data has been arranged in order.

The given data set is: $1, 3, 2, 5, 8, 6, 1, 4, 7, 9$.

The number of observations is $10$, which is an even number.

Step 1: Arrange the data in ascending order.

Ordered data: $1, 1, 2, 3, 4, 5, 6, 7, 8, 9$.

Step 2: Identify the two middle observations.

Since $n=10$, the middle observations are the $\left(\frac{10}{2}\right)^{th} = 5^{th}$ and $\left(\frac{10}{2} + 1\right)^{th} = 6^{th}$ observations in the ordered list.

Looking at the ordered list ($1, 1, 2, 3, \underline{4}, \underline{5}, 6, 7, 8, 9$):

The $5^{th}$ observation is $4$.

The $6^{th}$ observation is $5$.

Step 3: Calculate the median.

Median $= \frac{5^{th} \text{ ordered observation} + 6^{th} \text{ ordered observation}}{2}$

Median $= \frac{4 + 5}{2} = \frac{9}{2} = 4.5$.

The provided calculation used the $5^{th}$ and $6^{th}$ observations from the original, unordered list ($1, 3, 2, 5, \underline{8}, \underline{6}, 1, 4, 7, 9$). The 5th observation in the original list is 8 and the 6th is 6.

Conclusion:

No, the provided answer ($7$) is not correct.

The reason it is incorrect is that the median is calculated using the observations from the data set after it has been arranged in order. The calculation used the $5^{th}$ and $6^{th}$ observations from the original, unordered list, which is the incorrect procedure for finding the median.

The correct median for the given data set is $4.5$.

Given:

The statement: "In a histogram, the area of each rectangle is proportional to the class size of the corresponding class interval."

To Determine:

Whether the given statement is correct, and if not, to provide the correct statement.

Solution:

In a histogram, the area of each rectangle represents the frequency of the corresponding class interval.

The height of each rectangle is proportional to the frequency only if the class intervals have equal width.

When the class intervals have unequal width, the heights of the rectangles are adjusted so that the area of the rectangle is proportional to the frequency.

The height is calculated as $\text{Height} = \frac{\text{Frequency}}{\text{Width of the class}}$. This height is sometimes called the frequency density.

The area of a rectangle in a histogram is given by:

Area $= \text{Width of class} \times \text{Height of rectangle}$

When constructing a histogram with unequal class widths, we ensure that:

Area of rectangle $\propto$ Frequency

Area of rectangle $= k \times \text{Frequency}$, where $k$ is a constant of proportionality (often $k=1$).

Therefore, the area of each rectangle in a histogram is proportional to the frequency of the corresponding class interval, not the class size (width).

Conclusion:

No, the statement is not correct.

Correct Statement:

In a histogram, the area of each rectangle is proportional to the frequency of the corresponding class interval.

Given:

The class marks of a continuous distribution are: $1.04, 1.14, 1.24, 1.34, 1.44, 1.54, 1.64$.

The statement that the last interval is $1.55 - 1.73$.

To Determine:

Whether the statement about the last interval is correct and provide justification.

Solution:

In a continuous distribution with equally spaced class marks, the class width is constant for all intervals.

The class width ($h$) can be found by taking the difference between any two consecutive class marks.

Let's calculate the class width using the first two class marks:

$h = 1.14 - 1.04 = 0.10$

Let's verify with another pair:

$h = 1.64 - 1.54 = 0.10$

The class width is $0.10$.

For a continuous distribution, if 'm' is the class mark and 'h' is the class width, the lower limit ($l$) and upper limit ($u$) of the class interval are given by:

$l = m - \frac{h}{2}$

$u = m + \frac{h}{2}$

We want to find the last interval, which corresponds to the last class mark, $m = 1.64$.

The class width is $h = 0.10$, so $\frac{h}{2} = \frac{0.10}{2} = 0.05$.

For the last class mark $1.64$:

Lower limit ($l$) $= 1.64 - 0.05 = 1.59$

Upper limit ($u$) $= 1.64 + 0.05 = 1.69$

The correct last class interval is $1.59 - 1.69$.

The statement claims that the last interval is $1.55 - 1.73$. Let's check the class width of this proposed interval:

Proposed class width $= \text{Upper limit} - \text{Lower limit} = 1.73 - 1.55 = 0.18$

The class width of the proposed interval ($0.18$) is different from the consistent class width ($0.10$) calculated from the given sequence of class marks.

In a continuous distribution with equally spaced class marks, the class width must be uniform throughout the distribution.

Conclusion:

No, it is not correct to say that the last interval will be $1.55 - 1.73$.

Justification:

The given class marks form an arithmetic progression with a common difference (class width) of $0.10$. For a continuous distribution with these class marks, all intervals must have a width of $0.10$. The proposed interval $1.55 - 1.73$ has a width of $0.18$, which is inconsistent with the determined class width. The correct last interval, based on the last class mark $1.64$ and a class width of $0.10$, is $1.59 - 1.69$.

Given:

Frequency distribution showing the number of hours children watched TV:

Total number of children = $8 + 16 + 4 + 2 = 30$.

The statement: The number of children who watched TV for 10 or more hours a week is 22.

To Determine:

Whether the given statement is correct and provide justification.

Solution:

We need to find the number of children who watched TV for 10 or more hours a week.

Looking at the frequency distribution table, the intervals that represent "10 or more hours" are:

1. The interval $10 - 15$ hours.

2. The interval $15 - 20$ hours.

The number of children in the $10 - 15$ hours interval is $4$.

The number of children in the $15 - 20$ hours interval is $2$.

To find the total number of children who watched TV for 10 or more hours, we sum the frequencies of these intervals:

Number of children who watched for 10 or more hours = (Frequency of 10 - 15) + (Frequency of 15 - 20)

Number of children who watched for 10 or more hours $= 4 + 2 = 6$.

The given statement says that the number of children who watched TV for 10 or more hours is $22$.

Our calculation shows that the actual number of children is $6$.

The value $22$ seems to be the sum of the frequencies of the first two intervals ($8 + 16 = 24$), which represents children who watched for less than 10 hours (specifically, less than 5 or between 5 and 10 hours).

Conclusion:

No, it is not correct to say that the number of children who watched TV for 10 or more hours a week is 22.

Justification:

The number of children who watched TV for 10 or more hours is the sum of the frequencies of the classes with a lower limit of 10 or more. These are the classes $10 - 15$ and $15 - 20$. The frequencies for these classes are 4 and 2, respectively. Therefore, the total number of children who watched for 10 or more hours is $4 + 2 = 6$. The statement that it is 22 is incorrect.

Given:

The question asks if the experimental probability of an event can be a negative number.

To Determine:

Whether experimental probability can be negative and provide justification.

Solution:

The experimental probability (or empirical probability) of an event is calculated based on the results of an experiment or observation. It is defined as:

Experimental Probability (E) $= \frac{\text{Number of times the event occurred}}{\text{Total number of trials}}$

Let's analyze the components of this formula:

1. Number of times the event occurred: This represents a count of how many times a specific outcome happened during the experiment. A count can be zero (if the event never happened) or a positive integer (if the event happened one or more times). It can never be a negative number.

2. Total number of trials: This represents the total number of times the experiment was conducted or the total number of observations made. This must be a positive integer (you cannot perform a negative number of trials or zero trials if calculating a probability based on outcomes). In a practical experiment, this number is always positive.

Since the number of times the event occurred is always non-negative ($\geq 0$), and the total number of trials is always positive ($> 0$), the ratio of these two numbers must be non-negative.

Experimental Probability $= \frac{\text{Non-negative number}}{\text{Positive number}} \geq 0$

Also, the number of times an event occurs can be at most equal to the total number of trials. Therefore, the ratio is always less than or equal to 1.

$0 \leq \text{Experimental Probability} \leq 1$

Conclusion:

No, the experimental probability of an event cannot be a negative number.

Reason:

Experimental probability is calculated as the ratio of the frequency of the event (which is always non-negative) to the total number of trials (which is always positive). The division of a non-negative number by a positive number will always result in a non-negative number. Probabilities must lie within the range $[0, 1]$.

Given:

The question asks if the experimental probability of an event can be greater than 1.

To Determine:

Whether experimental probability can be greater than 1 and provide justification.

Solution:

The experimental probability of an event is calculated using the formula:

Experimental Probability $= \frac{\text{Number of times the event occurred}}{\text{Total number of trials}}$

Let's consider the components of this formula:

1. Number of times the event occurred: This is a count of the favorable outcomes in the experiment. This number can be zero or a positive integer. It cannot be less than 0.

2. Total number of trials: This is the total count of repetitions of the experiment. This number must be a positive integer.

By definition, the number of times an event occurs during an experiment can never exceed the total number of times the experiment was performed (total trials).

Number of times the event occurred $\leq$ Total number of trials

Since the total number of trials is a positive value, we can divide both sides of the inequality by the "Total number of trials":

$\frac{\text{Number of times the event occurred}}{\text{Total number of trials}} \leq \frac{\text{Total number of trials}}{\text{Total number of trials}}$

Experimental Probability $\leq 1$

Combining this with the fact that experimental probability cannot be negative (as explained in the previous answer), we get:

$0 \leq \text{Experimental Probability} \leq 1$

This range indicates that the experimental probability must always be between 0 and 1, inclusive.

Conclusion:

No, the experimental probability of an event cannot be greater than 1.

Justification:

The experimental probability is calculated as the ratio of the number of times an event occurs to the total number of trials. Since the number of times an event occurs can never exceed the total number of trials, this ratio will always be less than or equal to 1. A probability of more than 1 would imply that the event occurred more times than the experiment was conducted, which is impossible.

Given:

The statement: "As the number of tosses of a coin increases, the ratio of the number of heads to the total number of tosses will be $\frac{1}{2}$."

To Determine:

Whether the given statement is correct, and if not, to write the correct statement.

Solution:

The statement relates to the concept of experimental probability and the Law of Large Numbers.

The theoretical probability of getting a head when tossing a fair coin is $0.5$ or $\frac{1}{2}$. This is because there are two equally likely outcomes (Heads and Tails), and one of them is the favorable outcome (Heads).

The experimental probability of getting a head after a certain number of tosses is calculated as:

Experimental Probability $= \frac{\text{Number of Heads observed}}{\text{Total number of tosses}}$

The Law of Large Numbers states that as the number of trials (tosses in this case) increases, the experimental probability tends to approach the theoretical probability.

This means that the ratio $\frac{\text{Number of Heads}}{\text{Total number of tosses}}$ gets closer and closer to $\frac{1}{2}$ as the total number of tosses becomes very large.

However, the statement says the ratio "will be $\frac{1}{2}$". This implies that for a sufficiently large number of tosses, the ratio will become exactly equal to $\frac{1}{2}$. This is not guaranteed by the Law of Large Numbers for any finite number of trials, no matter how large. The ratio approaches $\frac{1}{2}$ as the number of tosses tends towards infinity, but it may never be exactly $\frac{1}{2}$ for any finite number of tosses (unless the number of tosses is even and the number of heads is exactly half of the tosses, which is not a certainty even for a large number). For example, after 1000 tosses, you might get 498 heads, 500 heads, or 502 heads, making the ratio 0.498, 0.5, or 0.502, respectively. The Law of Large Numbers suggests that results like 498 or 502 become much more likely than getting something far away from 500, but it doesn't say you *must* get exactly 500.

Conclusion:

No, the statement is not entirely correct.

Correct Statement:

As the number of tosses of a coin increases, the ratio of the number of heads to the total number of tosses will approach $\frac{1}{2}$.

Or, more formally:

$\lim\limits_{\text{Total tosses} \to \infty} \left( \frac{\text{Number of Heads}}{\text{Total number of tosses}} \right) = \frac{1}{2}$

Given:

Heights (in cm) of 30 girls of Class IX:

$140, 140, 160, 139, 153, 153, 146, 150, 148, 150,$

$152, 146,$

$154, 150, 160, 148, 150, 148, 140, 148, 153, 138, 152, 150, 148, 138, 152, 140, 146, 148$.

To Prepare:

A frequency distribution table for the given data.

Solution:

To prepare a frequency distribution table, we first determine the range of the data and decide on appropriate class intervals.

Minimum height = 138 cm

Maximum height = 160 cm

Let's choose class intervals of width 5 cm. We can start the first interval from a value less than or equal to the minimum height, for example, 135 cm.

The class intervals will be 135-140, 140-145, 145-150, 150-155, 155-160, 160-165.

Since height is a continuous variable, the intervals should be interpreted as lower limit inclusive and upper limit exclusive, except possibly for the last interval to include the maximum value. So, the intervals are $135 \le h < 140$, $140 \le h < 145$, and so on, up to $160 \le h \le 165$ for the last interval.

Now, we count the number of girls whose height falls into each class interval by tallying the frequencies.

Frequency Distribution Table

To verify, the sum of the frequencies is $3 + 4 + 8 + 13 + 0 + 2 = 30$, which matches the total number of girls.

The table above is the frequency distribution table for the given heights.

Given:

The observations arranged in ascending order are: $26, 29, 42, 53, x, x+2, 70, 75, 82, 93$.

The median of the data is $65$.

To Find:

The value of $x$.

Solution:

The given data set has $10$ observations.

The number of observations ($n$) is $10$, which is an even number.

For a data set with an even number of observations arranged in ascending order, the median is the average of the $(\frac{n}{2})^{th}$ and $(\frac{n}{2} + 1)^{th}$ observations.

In this case, $n=10$, so the median is the average of the $(\frac{10}{2})^{th} = 5^{th}$ observation and the $(\frac{10}{2} + 1)^{th} = 6^{th}$ observation.

From the given ordered data:

The $5^{th}$ observation is $x$.

The $6^{th}$ observation is $x+2$.

The formula for the median is:

Median $= \frac{5^{th} \text{ observation} + 6^{th} \text{ observation}}{2}$

Substitute the values into the formula:

$65 = \frac{x + (x+2)}{2}$

Simplify and solve the equation for $x$:

$65 = \frac{2x + 2}{2}$

$65 = x + 1$

$x = 65 - 1$

$x = 64$

Let's verify if the data remains in ascending order when $x=64$. The 5th term is 64 and the 6th term is $64+2=66$. The terms around them are 53 and 70. Since $53 \le 64 \le 66 \le 70$, the ascending order is maintained.

The value of $x$ is $64$.

Given:

Mortality Table showing the number of persons surviving at different ages out of a sample of one million:

To Find:

(i) The probability of a person aged 60 dying within a year.

(ii) The probability that a person aged 61 will live for 4 years.

Solution:

Probabilities based on mortality tables are experimental probabilities calculated from observed data.

(i) Probability of a person aged 60 dying within a year:

A person aged 60 dying within a year means they die between their $60^{th}$ and $61^{st}$ birthdays.

Number of persons surviving at age 60 = $16090$.

Number of persons surviving at age 61 = $11490$.

The number of persons who died between age 60 and 61 is the difference between the number surviving at age 60 and the number surviving at age 61.

Number of deaths between 60 and 61 = (Number surviving at 60) - (Number surviving at 61)

Number of deaths between 60 and 61 = $16090 - 11490 = 4600$.

The probability of a person aged 60 dying within a year is the ratio of the number of deaths between 60 and 61 to the number of persons surviving at age 60.

Probability (Death within a year for aged 60) $= \frac{\text{Number of deaths between 60 and 61}}{\text{Number surviving at 60}}$

Probability $= \frac{4600}{16090}$

Probability $\approx 0.28589$ (approximately).

(ii) Probability that a person aged 61 will live for 4 years:

A person aged 61 living for 4 years means they survive from age 61 to age $61 + 4 = 65$.

Number of persons surviving at age 61 = $11490$.

Number of persons surviving at age 65 = $2320$.

The probability that a person aged 61 will live for 4 years is the ratio of the number of persons surviving at age 65 to the number of persons surviving at age 61 (assuming the initial sample of one million). This represents the proportion of those alive at 61 who are still alive at 65.

Probability (Living for 4 years for aged 61) $= \frac{\text{Number surviving at 65}}{\text{Number surviving at 61}}$

Probability $= \frac{2320}{11490}$

Probability $\approx 0.2019$ (approximately).

Given:

The blood groups of 30 students are:

A, B, O, A, AB, O, A, O, B, A,

O, B, A, AB, B, A, AB, B, A, A,

O, A, AB, B, A, O, B, A, B, A.

Total number of students = 30.

To Prepare:

A frequency distribution table for the given data.

Solution:

To prepare a frequency distribution table, we need to count the number of students for each blood group.

Let's count the frequency of each blood group:

• Blood Group A: Count the occurrences of 'A'.
• Blood Group B: Count the occurrences of 'B'.
• Blood Group O: Count the occurrences of 'O'.
• Blood Group AB: Count the occurrences of 'AB'.

Counting from the list:

• A: 10 times
• B: 9 times
• O: 6 times
• AB: 5 times

Total count = $10 + 9 + 6 + 5 = 30$. This matches the total number of students.

Now, we can construct the frequency distribution table.

Frequency Distribution Table of Blood Groups

Given:

The value of $\pi$ up to 35 decimal places: $3.14159265358979323846264338327950288$.

To Prepare:

A frequency distribution table for the digits 0 to 9 after the decimal point.

Solution:

The digits after the decimal point are: $1, 4, 1, 5, 9, 2, 6, 5, 3, 5, 8, 9, 7, 9, 3, 2, 3, 8, 4, 6, 2, 6, 4, 3, 3, 8, 3, 2, 7, 9, 5, 0, 2, 8, 8$.

We count the frequency of each digit from 0 to 9 in this list of 35 digits.

Frequency Distribution Table of Digits

Given:

The scores (out of 100) obtained by 33 students in a mathematics test:

69, 48, 84, 58, 48, 73, 83, 48, 66, 58,

84, 66, 64, 71, 64, 66, 69, 66, 83, 66,

69, 71, 81, 71, 73, 69, 66, 66, 64, 58,

64, 69, 69.

Total number of students = 33.

To Prepare:

A frequency distribution table for the given data.

Solution:

To prepare a frequency distribution table, we list each distinct score and count how many times it appears in the data.

The distinct scores are: 48, 58, 64, 66, 69, 71, 73, 81, 83, 84.

Let's count the frequency of each score:

• Score 48: Appears 3 times.
• Score 58: Appears 3 times.
• Score 64: Appears 4 times.
• Score 66: Appears 7 times.
• Score 69: Appears 6 times.
• Score 71: Appears 3 times.
• Score 73: Appears 2 times.
• Score 81: Appears 1 time.
• Score 83: Appears 2 times.
• Score 84: Appears 2 times.

Total frequency = $3 + 3 + 4 + 7 + 6 + 3 + 2 + 1 + 2 + 2 = 33$. This matches the total number of students.

Now, we construct the frequency distribution table.

Frequency Distribution Table of Mathematics Scores

Given:

Mid-points and corresponding frequencies of a distribution:

To Prepare:

A continuous grouped frequency distribution table.

Also, to find the size of the class intervals.

Solution:

The mid-points are given as $5, 15, 25, 35, 45$.

In a continuous grouped frequency distribution with equal class widths, the difference between consecutive mid-points is equal to the class width.

Let $h$ be the class width.

$h = 15 - 5 = 10$

$h = 25 - 15 = 10$

$h = 35 - 25 = 10$

$h = 45 - 35 = 10$

The class width is consistent and is $h = 10$.

Size of Class Intervals:

The size of the class intervals is the class width, which is $10$.

Now, let's determine the lower limit ($l$) and upper limit ($u$) for each class interval.

If 'm' is the mid-point and 'h' is the class width, then:

$l = m - \frac{h}{2}$

$u = m + \frac{h}{2}$

Here, $h = 10$, so $\frac{h}{2} = \frac{10}{2} = 5$.

For Mid-point = 5:

$l = 5 - 5 = 0$

$u = 5 + 5 = 10$

Class interval: $0 - 10$. Frequency: 4.

For Mid-point = 15:

$l = 15 - 5 = 10$

$u = 15 + 5 = 20$

Class interval: $10 - 20$. Frequency: 8.

For Mid-point = 25:

$l = 25 - 5 = 20$

$u = 25 + 5 = 30$

Class interval: $20 - 30$. Frequency: 13.

For Mid-point = 35:

$l = 35 - 5 = 30$

$u = 35 + 5 = 40$

Class interval: $30 - 40$. Frequency: 12.

For Mid-point = 45:

$l = 45 - 5 = 40$

$u = 45 + 5 = 50$

Class interval: $40 - 50$. Frequency: 6.

Now, we can prepare the continuous grouped frequency distribution table.

Continuous Grouped Frequency Distribution Table

Given:

A discontinuous grouped frequency distribution table:

To Convert and Find:

Convert the given distribution into a continuous grouped frequency distribution.

Determine which intervals would include 153.5 and 157.5 in the continuous distribution.

Solution:

The given frequency distribution is discontinuous because there are gaps between the upper limit of one class and the lower limit of the next class (e.g., the upper limit of the first class is 153 and the lower limit of the second class is 154).

To convert it into a continuous distribution, we find the difference between the lower limit of a class and the upper limit of the preceding class. We then subtract half of this difference from the lower limit of each class and add half of this difference to the upper limit of each class.

The difference between the upper limit of the first class (153) and the lower limit of the second class (154) is $154 - 153 = 1$.

Half of this difference is $\frac{1}{2} = 0.5$.

Now, we adjust the class limits:

• For the class 150 - 153: Lower limit = $150 - 0.5 = 149.5$. Upper limit = $153 + 0.5 = 153.5$. New interval: 149.5 - 153.5.
• For the class 154 - 157: Lower limit = $154 - 0.5 = 153.5$. Upper limit = $157 + 0.5 = 157.5$. New interval: 153.5 - 157.5.
• For the class 158 - 161: Lower limit = $158 - 0.5 = 157.5$. Upper limit = $161 + 0.5 = 161.5$. New interval: 157.5 - 161.5.
• For the class 162 - 165: Lower limit = $162 - 0.5 = 161.5$. Upper limit = $165 + 0.5 = 165.5$. New interval: 161.5 - 165.5.
• For the class 166 - 169: Lower limit = $166 - 0.5 = 165.5$. Upper limit = $169 + 0.5 = 169.5$. New interval: 165.5 - 169.5.
• For the class 170 - 173: Lower limit = $170 - 0.5 = 169.5$. Upper limit = $173 + 0.5 = 173.5$. New interval: 169.5 - 173.5.

The frequencies remain the same for the new intervals.

Continuous Grouped Frequency Distribution Table

Now, let's consider where 153.5 and 157.5 would be included in this continuous distribution.

In a continuous frequency distribution, the standard convention is that a class interval $(a - b)$ includes all values greater than or equal to the lower limit ($a$) and strictly less than the upper limit ($b$). That is, the interval is $[a, b)$.

Let's look at the new continuous intervals:

• 149.5 - 153.5 represents $[149.5, 153.5)$
• 153.5 - 157.5 represents $[153.5, 157.5)$
• 157.5 - 161.5 represents $[157.5, 161.5)$
• and so on.

For the value 153.5:

It is the upper limit of the interval 149.5 - 153.5.

It is the lower limit of the interval 153.5 - 157.5.

According to the convention $[a, b)$, the value 153.5 is included in the interval where it is the lower limit.

Thus, 153.5 would be included in the interval 153.5 - 157.5.

For the value 157.5:

It is the upper limit of the interval 153.5 - 157.5.

It is the lower limit of the interval 157.5 - 161.5.

According to the convention $[a, b)$, the value 157.5 is included in the interval where it is the lower limit.

Thus, 157.5 would be included in the interval 157.5 - 161.5.

Given:

The expenditure of a family on different heads in a month:

To Draw:

A bar graph to represent the given data.

Solution:

To draw a bar graph, we follow these steps:

1. Draw two perpendicular axes, the horizontal axis (x-axis) and the vertical axis (y-axis).

2. Represent the 'Heads of Expenditure' on the x-axis and the 'Expenditure (in $\textsf{₹}$)' on the y-axis.

3. Choose a suitable scale for the y-axis. The expenditures range from $\textsf{₹}1000$ to $\textsf{₹}4000$. A scale of 1 unit representing $\textsf{₹}500$ or $\textsf{₹}1000$ would be appropriate. Let's take 1 unit on the y-axis $= \textsf{₹}500$.

4. On the x-axis, draw bars of uniform width for each head of expenditure. The bars should be equally spaced from each other.

5. The height of each bar will correspond to the expenditure for that head, based on the chosen scale on the y-axis.

Based on the data and the scale (1 unit = $\textsf{₹}500$), the height of each bar will be:

• Food: $\frac{4000}{500} = 8$ units
• Education: $\frac{2500}{500} = 5$ units
• Clothing: $\frac{1000}{500} = 2$ units
• House Rent: $\frac{3500}{500} = 7$ units
• Others: $\frac{2500}{500} = 5$ units
• Savings: $\frac{1500}{500} = 3$ units

Construction of the Bar Graph:

• Draw the x-axis and label it "Heads of Expenditure".
• Draw the y-axis and label it "Expenditure (in $\textsf{₹}$)". Mark the scale along the y-axis, starting from 0 and increasing in steps of 500 (0, 500, 1000, 1500, ..., 4000, 4500).
• Draw the first bar for 'Food' with a height of 8 units (up to $\textsf{₹}4000$).
• Leave a uniform gap and draw the second bar for 'Education' with a height of 5 units (up to $\textsf{₹}2500$).
• Leave a uniform gap and draw the third bar for 'Clothing' with a height of 2 units (up to $\textsf{₹}1000$).
• Leave a uniform gap and draw the fourth bar for 'House Rent' with a height of 7 units (up to $\textsf{₹}3500$).
• Leave a uniform gap and draw the fifth bar for 'Others' with a height of 5 units (up to $\textsf{₹}2500$).
• Leave a uniform gap and draw the sixth bar for 'Savings' with a height of 3 units (up to $\textsf{₹}1500$).
• Ensure all bars have the same width and the gaps between them are equal.
• Label each bar clearly with the corresponding head of expenditure (Food, Education, Clothing, House Rent, Others, Savings).

(Note: The bar graph cannot be visually displayed in this text format, but the steps described above explain how to construct it.)

Given:

Expenditure on Education of a country during a five year period (2002-2006), in crores of rupees, for different heads:

To Draw:

A bar graph to represent the given data.

Solution:

To draw a bar graph for the given data, we follow these steps:

1. Draw two perpendicular axes: a horizontal axis (x-axis) and a vertical axis (y-axis).

2. Represent the 'Heads of Education Expenditure' on the x-axis and the 'Expenditure (in crores of $\textsf{₹}$)' on the y-axis.

3. Choose a suitable scale for the y-axis. The expenditure values range from 10 to 240. A scale of 1 unit representing 20 or 25 crores of rupees would be appropriate. Let's take 1 unit on the y-axis $= 25$ crores of $\textsf{₹}$.

4. On the x-axis, draw bars of uniform width for each head of expenditure. The bars should be equally spaced from each other.

5. The height of each bar will be proportional to the expenditure for that head, according to the chosen scale on the y-axis. We can calculate the height needed for each category:

• Elementary Education: $\frac{240}{25} = 9.6$ units
• Secondary Education: $\frac{120}{25} = 4.8$ units
• University Education: $\frac{190}{25} = 7.6$ units
• Teacher's Training: $\frac{20}{25} = 0.8$ units
• Social Education: $\frac{10}{25} = 0.4$ units
• Other Educational Programmes: $\frac{115}{25} = 4.6$ units
• Cultural Programmes: $\frac{25}{25} = 1$ unit
• Technical Education: $\frac{125}{25} = 5$ units

Construction of the Bar Graph:

• Draw the x-axis and label it "Heads of Education Expenditure".
• Draw the y-axis and label it "Expenditure (in crores of $\textsf{₹}$)". Mark the scale along the y-axis, starting from 0 and increasing in steps of 25 (0, 25, 50, 75, ..., 250).
• Draw the first bar for 'Elementary Education' with a height corresponding to 240 crores (9.6 units).
• Leave a uniform gap and draw the second bar for 'Secondary Education' with a height corresponding to 120 crores (4.8 units).
• Leave a uniform gap and draw the third bar for 'University Education' with a height corresponding to 190 crores (7.6 units).
• Leave a uniform gap and draw the fourth bar for 'Teacher's Training' with a height corresponding to 20 crores (0.8 units).
• Leave a uniform gap and draw the fifth bar for 'Social Education' with a height corresponding to 10 crores (0.4 units).
• Leave a uniform gap and draw the sixth bar for 'Other Educational Programmes' with a height corresponding to 115 crores (4.6 units).
• Leave a uniform gap and draw the seventh bar for 'Cultural Programmes' with a height corresponding to 25 crores (1 unit).
• Leave a uniform gap and draw the eighth bar for 'Technical Education' with a height corresponding to 125 crores (5 units).
• Ensure all bars have the same width and the gaps between them are equal.
• Label each bar clearly with the corresponding head of expenditure.

(Note: A visual representation of the bar graph is not possible in this text format, but the instructions provided describe how to construct it accurately.)

Given:

Frequencies of commonly used letters from a page of a book:

To Draw:

A bar graph to represent the given data.

Solution:

To draw a bar graph for the given data, we follow these steps:

1. Draw two perpendicular axes: a horizontal axis (x-axis) and a vertical axis (y-axis).

2. Represent the 'Letters' on the x-axis and the 'Frequency' on the y-axis.

3. Choose a suitable scale for the y-axis (Frequency). The frequencies range from 70 to 125. A scale where 1 unit represents 10 frequencies, or perhaps 5 frequencies, would be appropriate. Let's use a scale where 1 unit on the y-axis represents 10 frequencies.

4. On the x-axis, draw bars of uniform width for each letter. The bars should be equally spaced from each other.

5. The height of each bar will be proportional to the frequency of the corresponding letter, based on the chosen scale on the y-axis. We can calculate the height needed for each letter (in units of 10 frequencies):

• Letter 'a': $\frac{75}{10} = 7.5$ units
• Letter 'e': $\frac{125}{10} = 12.5$ units
• Letter 'i': $\frac{80}{10} = 8$ units
• Letter 'o': $\frac{70}{10} = 7$ units
• Letter 'r': $\frac{80}{10} = 8$ units
• Letter 't': $\frac{95}{10} = 9.5$ units
• Letter 'u': $\frac{75}{10} = 7.5$ units

Construction of the Bar Graph:

• Draw the x-axis and label it "Letters". Mark points along the axis for each letter (a, e, i, o, r, t, u), leaving equal space between them.
• Draw the y-axis and label it "Frequency". Mark the scale along the y-axis, starting from 0 and increasing in steps of 10 (0, 10, 20, ..., 130).
• Draw the first bar for 'a' with a height corresponding to a frequency of 75 (7.5 units).
• Leave a uniform gap and draw the second bar for 'e' with a height corresponding to a frequency of 125 (12.5 units).
• Leave a uniform gap and draw the third bar for 'i' with a height corresponding to a frequency of 80 (8 units).
• Leave a uniform gap and draw the fourth bar for 'o' with a height corresponding to a frequency of 70 (7 units).
• Leave a uniform gap and draw the fifth bar for 'r' with a height corresponding to a frequency of 80 (8 units).
• Leave a uniform gap and draw the sixth bar for 't' with a height corresponding to a frequency of 95 (9.5 units).
• Leave a uniform gap and draw the seventh bar for 'u' with a height corresponding to a frequency of 75 (7.5 units).
• Ensure all bars have the same width and the gaps between them are equal.
• Label each bar clearly with the corresponding letter (a, e, i, o, r, t, u).

(Note: A visual representation of the bar graph is not possible in this text format, but the instructions provided describe how to construct it accurately.)

Given:

A frequency distribution table:

The mean of the data is $\bar{x} = 20.2$.

To Find:

The value of p.

Solution:

The mean of a frequency distribution is given by the formula:

$\bar{x} = \frac{\sum (f_i \times x_i)}{\sum f_i}$

First, let's calculate the sum of frequencies ($\sum f_i$):

$\sum f_i = 6 + 8 + p + 10 + 6$

$\sum f_i = 30 + p$

Next, let's calculate the sum of the products of frequency and the corresponding value ($\sum (f_i \times x_i)$):

$\sum (f_i \times x_i) = (6 \times 10) + (8 \times 15) + (p \times 20) + (10 \times 25) + (6 \times 30)$

$\sum (f_i \times x_i) = 60 + 120 + 20p + 250 + 180$

$\sum (f_i \times x_i) = (60 + 120 + 250 + 180) + 20p$

$\sum (f_i \times x_i) = 610 + 20p$

Now, substitute the given mean and the calculated sums into the mean formula:

$20.2 = \frac{610 + 20p}{30 + p}$

Multiply both sides by $(30 + p)$ to eliminate the denominator:

$20.2 (30 + p) = 610 + 20p$

$20.2 \times 30 + 20.2 \times p = 610 + 20p$

$606 + 20.2p = 610 + 20p$

Rearrange the equation to solve for p. Subtract $20p$ from both sides:

$606 + 20.2p - 20p = 610$

$606 + 0.2p = 610$

Subtract 606 from both sides:

$0.2p = 610 - 606$

$0.2p = 4$

Divide by 0.2:

$p = \frac{4}{0.2}$

$p = \frac{40}{2}$

$p = 20$

Thus, the value of p is 20.

Given:

A frequency distribution table:

To Obtain:

The mean of the given distribution.

Solution:

The mean of a frequency distribution is calculated using the formula:

$\bar{x} = \frac{\sum (f_i \times x_i)}{\sum f_i}$

where $f_i$ is the frequency of the $i^{th}$ observation and $x_i$ is the $i^{th}$ observation (variable value).

First, we calculate the product of frequency and variable value ($f_i \times x_i$) for each row and sum them up ($\sum (f_i \times x_i)$).

Now, calculate the sum of the frequencies ($\sum f_i$) and the sum of the products ($\sum (f_i \times x_i)$).

$\sum f_i = 4 + 8 + 14 + 11 + 3 = 40$

$\sum (f_i \times x_i) = 16 + 48 + 112 + 110 + 36 = 322$

Substitute these sums into the mean formula:

$\bar{x} = \frac{322}{40}$

$\bar{x} = \frac{32.2}{4}$

$\bar{x} = 8.05$

The mean of the given distribution is $8.05$.

Given:

Total number of students in the class = 50.

Number of girls = 30.

Mean marks scored by girls = 73.

Mean marks scored by boys = 71.

To Determine:

The mean score of the whole class.

Solution:

Number of boys in the class = Total number of students - Number of girls

Number of boys = $50 - 30 = 20$.

The mean is calculated as $\text{Mean} = \frac{\text{Sum of observations}}{\text{Number of observations}}$.

So, the Sum of observations = Mean $\times$ Number of observations.

Total marks scored by girls = Mean marks of girls $\times$ Number of girls

Total marks scored by girls = $73 \times 30$

Total marks scored by girls = $2190$

Total marks scored by boys = Mean marks of boys $\times$ Number of boys

Total marks scored by boys = $71 \times 20$

Total marks scored by boys = $1420$

Total marks scored by the whole class = Total marks scored by girls + Total marks scored by boys

Total marks scored by the whole class = $2190 + 1420 = 3610$

The mean score of the whole class is the total marks scored by the class divided by the total number of students in the class.

Mean score of the whole class $= \frac{\text{Total marks scored by the whole class}}{\text{Total number of students}}$

Mean score of the whole class $= \frac{3610}{50}$

Mean score of the whole class $= \frac{361}{5}$

Mean score of the whole class $= 72.2$

The mean score of the whole class is $72.2$.

Given:

Number of observations ($n$) = 50.

Incorrect mean ($\bar{x}_{\text{incorrect}}$) = 80.4.

Value misread = 69 (incorrect value).

Correct value = 96.

To Find:

The correct mean of the 50 observations.

Solution:

The mean is calculated using the formula:

Mean $= \frac{\text{Sum of observations}}{\text{Number of observations}}$

From this, the sum of observations can be found as:

Sum of observations $=$ Mean $\times$ Number of observations.

Using the incorrect mean, we can find the incorrect sum of observations:

Incorrect sum of observations $= \text{Incorrect Mean} \times n$

Incorrect sum of observations $= 80.4 \times 50$

Incorrect sum of observations $= 4020$

The incorrect sum was obtained because one value (96) was misread as 69. To find the correct sum, we need to subtract the incorrectly read value and add the correct value to the incorrect sum.

Correct sum of observations $=$ Incorrect sum $-$ Incorrect value $+$ Correct value

Correct sum of observations $= 4020 - 69 + 96$

Correct sum of observations $= 4020 + (96 - 69)$

Correct sum of observations $= 4020 + 27$

Correct sum of observations $= 4047$

Now, we can calculate the correct mean using the correct sum and the total number of observations:

Correct Mean $= \frac{\text{Correct sum of observations}}{\text{Number of observations}}$

Correct Mean $= \frac{4047}{50}$

Correct Mean $= \frac{404.7}{5}$

Correct Mean $= 80.94$

The correct mean of the 50 observations is 80.94.

Given:

The observations in ascending order are: $6, 14, 15, 17, x+1, 2x-13, 30, 32, 34, 43$.

The median of the data is $24$.

To Find:

The value of $x$.

Solution:

The given data set consists of $10$ observations.

The number of observations ($n$) is $10$, which is an even number.

For a data set with an even number of observations arranged in ascending order, the median is the average of the $(\frac{n}{2})^{th}$ and $(\frac{n}{2} + 1)^{th}$ observations.

In this case, $n=10$, so the median is the average of the $(\frac{10}{2})^{th} = 5^{th}$ observation and the $(\frac{10}{2} + 1)^{th} = 6^{th}$ observation.

From the given ordered data:

The $5^{th}$ observation is $x+1$.

The $6^{th}$ observation is $2x-13$.

The formula for the median is:

Median $= \frac{5^{th} \text{ observation} + 6^{th} \text{ observation}}{2}$

Substitute the given median value and the expressions for the observations into the formula:

$24 = \frac{(x+1) + (2x-13)}{2}$

Simplify and solve the equation for $x$.

Multiply both sides by 2:

$24 \times 2 = (x+1) + (2x-13)$

$48 = x + 1 + 2x - 13$

Combine like terms on the right side:

$48 = (x + 2x) + (1 - 13)$

$48 = 3x - 12$

Add 12 to both sides:

$48 + 12 = 3x$

$60 = 3x$

Divide by 3:

$x = \frac{60}{3}$

$x = 20$

Let's check if the data remains in ascending order when $x=20$.

The $5^{th}$ observation is $x+1 = 20+1 = 21$.

The $6^{th}$ observation is $2x-13 = 2(20)-13 = 40-13 = 27$.

The ordered data becomes: $6, 14, 15, 17, 21, 27, 30, 32, 34, 43$.

Since $17 \le 21 \le 27 \le 30$, the ascending order is maintained.

The value of $x$ is $20$.

Given:

The points scored by a basketball team in a series of matches:

$17, 2, 7, 27, 25, 5, 14, 18, 10, 24, 48, 10, 8, 7, 10, 28$.

To Find:

The median and mode for the given data.

Solution:

First, let's arrange the data in ascending order to find the median:

$2, 5, 7, 7, 8, 10, 10, 10, 14, 17, 18, 24, 25, 27, 28, 48$.

Count the number of observations ($n$). There are 16 observations.

Since $n=16$ (an even number), the median is the average of the $(\frac{n}{2})^{th}$ and $(\frac{n}{2}+1)^{th}$ observations in the ordered list.

The $(\frac{16}{2})^{th} = 8^{th}$ observation is $10$.

The $(\frac{16}{2}+1)^{th} = 9^{th}$ observation is $14$.

Median $= \frac{8^{th} \text{ observation} + 9^{th} \text{ observation}}{2}$

Median $= \frac{10 + 14}{2} = \frac{24}{2} = 12$.

Next, let's find the mode. The mode is the observation that occurs most frequently in the data set.

Let's count the frequency of each distinct observation in the ordered list:

• 2: 1 time
• 5: 1 time
• 7: 2 times
• 8: 1 time
• 10: 3 times
• 14: 1 time
• 17: 1 time
• 18: 1 time
• 24: 1 time
• 25: 1 time
• 27: 1 time
• 28: 1 time
• 48: 1 time

The observation with the highest frequency is $10$, which appears 3 times.

The mode of the data is $10$.

The median of the data is $12$ and the mode is $10$.

Given:

A histogram (Fig. 14.2) depicting the daily wages of workers in a factory.

(The image shows Daily Wages (in $\textsf{₹}$) on the horizontal axis and Number of Workers (Frequency) on the vertical axis.)

To Construct:

The frequency distribution table from the given histogram.

Solution:

In a histogram, the horizontal axis represents the class intervals, and the height of each rectangle (bar) represents the frequency of the corresponding class interval, as shown on the vertical axis.

We will read the class intervals from the horizontal axis and the frequencies (heights of the bars) from the vertical axis.

From the histogram, we observe the following class intervals and their corresponding frequencies:

• The first bar is for the interval 150 - 200. Its height corresponds to 5 on the frequency axis.
• The second bar is for the interval 200 - 250. Its height corresponds to 10 on the frequency axis.
• The third bar is for the interval 250 - 300. Its height corresponds to 20 on the frequency axis.
• The fourth bar is for the interval 300 - 350. Its height corresponds to 9 on the frequency axis.
• The fifth bar is for the interval 350 - 400. Its height corresponds to 6 on the frequency axis.
• The sixth bar is for the interval 400 - 450. Its height corresponds to 3 on the frequency axis.

Now, we can construct the frequency distribution table using these values.

Frequency Distribution Table of Daily Wages

(Note: The sum of frequencies is $5 + 10 + 20 + 9 + 6 + 3 = 53$).

Given:

Data from a survey of 4000 households relating income level and number of television sets:

Total number of households surveyed = 4000.

To Find:

The probability of the following events:

(i) A household earning $\textsf{₹}10000 – \textsf{₹}14999$ per year and having exactly one television.

(ii) A household earning $\textsf{₹}25000$ and more per year and owning 2 televisions.

(iii) A household not having any television.

Solution:

The probability of an event is calculated as:

Probability $= \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$

Here, the total number of outcomes is the total number of households surveyed, which is 4000.

(i) Probability of a household earning $\textsf{₹}10000 – \textsf{₹}14999$ and having exactly one television:

From the table, locate the row for "10000 - 14999" income level and the column for "1" television.

The number of households in this category is 240.

Number of favorable outcomes = 240.

Total number of outcomes = 4000.

Probability (Income 10000-14999 and 1 TV) $= \frac{240}{4000}$

Probability $= \frac{24}{400} = \frac{6}{100} = 0.06$

(ii) Probability of a household earning $\textsf{₹}25000$ and more and owning 2 televisions:

From the table, locate the row for "25000 and above" income level and the column for "2" televisions.

The number of households in this category is 760.

Number of favorable outcomes = 760.

Total number of outcomes = 4000.

Probability (Income $\ge$ 25000 and 2 TVs) $= \frac{760}{4000}$

Probability $= \frac{76}{400} = \frac{19}{100} = 0.19$

(iii) Probability of a household not having any television:

This means the household has 0 televisions. We need to sum the frequencies in the column for "0" televisions across all income levels.

Number of households with 0 TVs $= 20 + 10 + 0 + 0 + 0 = 30$.

Number of favorable outcomes = 30.

Total number of outcomes = 4000.

Probability (0 TVs) $= \frac{30}{4000}$

Probability $= \frac{3}{400} = 0.0075$

Given:

Frequency distribution of the sum of two dice thrown 500 times:

Total number of trials = 500.

To Find:

The experimental probability of getting the sums:

(i) 3

(ii) more than 10

(iii) less than or equal to 5

(iv) between 8 and 12

Solution:

The experimental probability of an event is given by:

Probability (Event) $= \frac{\text{Frequency of the Event}}{\text{Total number of trials}}$

(i) Probability of getting a sum 3:

From the table, the frequency of getting a sum of 3 is 30.

Probability (Sum = 3) $= \frac{\text{Frequency of sum 3}}{\text{Total number of trials}}$

Probability (Sum = 3) $= \frac{30}{500} = \frac{3}{50} = 0.06$

(ii) Probability of getting a sum more than 10:

A sum more than 10 includes sums 11 and 12.

Frequency of sum 11 = 28

Frequency of sum 12 = 15

Frequency of sum more than 10 = Frequency of sum 11 + Frequency of sum 12

Frequency of sum more than 10 = $28 + 15 = 43$

Probability (Sum > 10) $= \frac{\text{Frequency of sum > 10}}{\text{Total number of trials}}$

Probability (Sum > 10) $= \frac{43}{500} = 0.086$

(iii) Probability of getting a sum less than or equal to 5:

A sum less than or equal to 5 includes sums 2, 3, 4, and 5.

Frequency of sum 2 = 14

Frequency of sum 3 = 30

Frequency of sum 4 = 42

Frequency of sum 5 = 55

Frequency of sum $\le$ 5 = Frequency(2) + Frequency(3) + Frequency(4) + Frequency(5)

Frequency of sum $\le$ 5 = $14 + 30 + 42 + 55 = 141$

Probability (Sum $\le$ 5) $= \frac{\text{Frequency of sum } \le \text{ 5}}{\text{Total number of trials}}$

Probability (Sum $\le$ 5) $= \frac{141}{500} = 0.282$

(iv) Probability of getting a sum between 8 and 12:

A sum between 8 and 12 (exclusive of 8 and 12) includes sums 9, 10, and 11.

Frequency of sum 9 = 53

Frequency of sum 10 = 46

Frequency of sum 11 = 28

Frequency of sum between 8 and 12 = Frequency(9) + Frequency(10) + Frequency(11)

Frequency of sum between 8 and 12 = $53 + 46 + 28 = 127$

Probability (Sum between 8 and 12) $= \frac{\text{Frequency of sum between 8 and 12}}{\text{Total number of trials}}$

Probability (Sum between 8 and 12) $= \frac{127}{500} = 0.254$

Given:

Results of examining 700 cartons for defective bulbs:

Total number of cartons examined = 700.

To Find:

The probability that a randomly selected carton has:

(i) no defective bulb.

(ii) defective bulbs from 2 to 6.

(iii) defective bulbs less than 4.

Solution:

The experimental probability of an event is given by:

Probability (Event) $= \frac{\text{Number of favorable outcomes}}{\text{Total number of trials}}$

Here, the total number of trials is the total number of cartons examined, which is 700.

(i) Probability of a carton having no defective bulb:

This corresponds to the category "Number of defective bulbs = 0".

From the table, the frequency for 0 defective bulbs is 400.

Number of favorable outcomes = 400.

Total number of outcomes = 700.

Probability (0 defective bulbs) $= \frac{400}{700} = \frac{4}{7}$

(ii) Probability of a carton having defective bulbs from 2 to 6:

This includes the categories where the number of defective bulbs is 2, 3, 4, 5, or 6.

We need to sum the frequencies for these categories.

Frequency (2 defective) = 48

Frequency (3 defective) = 41

Frequency (4 defective) = 18

Frequency (5 defective) = 8

Frequency (6 defective) = 3

Number of favorable outcomes = $48 + 41 + 18 + 8 + 3 = 118$

Total number of outcomes = 700.

Probability (2 to 6 defective bulbs) $= \frac{118}{700} = \frac{59}{350}$

(iii) Probability of a carton having defective bulbs less than 4:

This includes the categories where the number of defective bulbs is 0, 1, 2, or 3.

We need to sum the frequencies for these categories.

Frequency (0 defective) = 400

Frequency (1 defective) = 180

Frequency (2 defective) = 48

Frequency (3 defective) = 41

Number of favorable outcomes = $400 + 180 + 48 + 41 = 669$

Total number of outcomes = 700.

Probability (less than 4 defective bulbs) $= \frac{669}{700}$

Given:

Frequency distribution of the number of defective parts produced by a machine over 200 working days:

Total number of working days (trials) = 200.

To Determine:

The experimental probability for tomorrow's output having:

(i) no defective part.

(ii) atleast one defective part.

(iii) not more than 5 defective parts.

(iv) more than 13 defective parts.

Solution:

The experimental probability of an event is given by:

Probability (Event) $= \frac{\text{Frequency of the Event}}{\text{Total number of trials}}$

Here, the total number of trials is the total number of days observed, which is 200.

(i) Probability of no defective part:

This corresponds to the number of defective parts = 0.

From the table, the frequency for 0 defective parts is 50 days.

Number of favorable outcomes = 50.

Total number of outcomes = 200.

Probability (0 defective parts) $= \frac{50}{200} = \frac{1}{4} = 0.25$

(ii) Probability of at least one defective part:

At least one defective part means the number of defective parts is 1 or more (1, 2, 3, ..., 13). This is the complement of having no defective parts.

Probability (at least one defective part) = 1 - Probability (no defective part)

Probability (at least one defective part) $= 1 - \frac{50}{200} = 1 - \frac{1}{4} = \frac{3}{4} = 0.75$

Alternatively, we can sum the frequencies for 1, 2, 3, ..., 13 defective parts:

Sum of frequencies for 1 to 13 defective parts = $32 + 22 + 18 + 12 + 12 + 10 + 10 + 10 + 8 + 6 + 6 + 2 + 2 = 150$.

Probability (at least one defective part) $= \frac{150}{200} = \frac{15}{20} = \frac{3}{4} = 0.75$

(iii) Probability of not more than 5 defective parts:

Not more than 5 defective parts means the number of defective parts is less than or equal to 5 (0, 1, 2, 3, 4, or 5).

We need to sum the frequencies for these categories.

Frequency (0 defective) = 50

Frequency (1 defective) = 32

Frequency (2 defective) = 22

Frequency (3 defective) = 18

Frequency (4 defective) = 12

Frequency (5 defective) = 12

Number of favorable outcomes = $50 + 32 + 22 + 18 + 12 + 12 = 146$

Total number of outcomes = 200.

Probability (not more than 5 defective parts) $= \frac{146}{200} = \frac{73}{100} = 0.73$

(iv) Probability of more than 13 defective parts:

From the table, the maximum number of defective parts observed in the 200 days is 13. The category "more than 13" has a frequency of 0 days.

Number of favorable outcomes = 0.

Total number of outcomes = 200.

Probability (more than 13 defective parts) $= \frac{0}{200} = 0$

Given:

The age distribution of workers in a factory:

Total number of workers surveyed = $38 + 27 + 86 + 46 + 3 = 200$.

To Find:

The probability that a randomly selected person is:

(i) 40 years or more.

(ii) under 40 years.

(iii) having age from 30 to 39 years.

(iv) under 60 but over 39 years.

Solution:

The experimental probability of an event is calculated as:

Probability (Event) $= \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$

Here, the total number of outcomes is the total number of workers, which is 200.

(i) Probability that the person is 40 years or more:

This includes workers in the age groups 40-49, 50-59, and 60 and above.

Number of workers aged 40 or more = (Number in 40-49) + (Number in 50-59) + (Number in 60 and above)

Number of favorable outcomes = $86 + 46 + 3 = 135$.

Total number of outcomes = 200.

Probability (Age $\ge$ 40) $= \frac{135}{200} = \frac{27}{40} = 0.675$

(ii) Probability that the person is under 40 years:

This includes workers in the age groups 20-29 and 30-39.

Number of workers aged under 40 = (Number in 20-29) + (Number in 30-39)

Number of favorable outcomes = $38 + 27 = 65$.

Total number of outcomes = 200.

Probability (Age < 40) $= \frac{65}{200} = \frac{13}{40} = 0.325$

(Note: This is also $1 - P(\text{Age} \ge 40) = 1 - 0.675 = 0.325$).

(iii) Probability of having age from 30 to 39 years:

This corresponds directly to the age group 30-39.

From the table, the number of workers in this age group is 27.

Number of favorable outcomes = 27.

Total number of outcomes = 200.

Probability (Age 30-39) $= \frac{27}{200} = 0.135$

(iv) Probability of being under 60 but over 39 years:

This includes workers whose age is greater than 39 and less than 60. Based on the given age groups, this corresponds to the age groups 40-49 and 50-59.

Number of workers aged under 60 but over 39 = (Number in 40-49) + (Number in 50-59)

Number of favorable outcomes = $86 + 46 = 132$.

Total number of outcomes = 200.

Probability (39 < Age < 60) $= \frac{132}{200} = \frac{33}{50} = 0.66$

Given:

Frequency distribution of marks obtained by students of Class VIII:

To Draw:

A histogram for the given distribution.

Solution:

To draw a histogram from a frequency distribution, the class intervals are represented on the horizontal axis and the frequencies on the vertical axis. The area of each rectangle in a histogram is proportional to the frequency of the corresponding class.

In the given table, the class intervals have unequal widths (e.g., 150-100=50, 200-150=50, 300-200=100, 500-300=200, 800-500=300). When class widths are unequal, we need to adjust the frequencies so that the heights of the rectangles are proportional to the frequencies.

The height of each rectangle in a histogram with unequal class widths is proportional to the frequency density.

Frequency Density $= \frac{\text{Frequency}}{\text{Width of the class}}$

To make the areas proportional to frequency, we adjust the height (which represents frequency density) relative to the minimum class width. We calculate the 'Length of the rectangle' (which is proportional to the frequency density) using the formula:

Length of the rectangle $= \frac{\text{Frequency}}{\text{Width of the class}} \times \text{Minimum class width}$

First, let's find the width of each class interval:

• 100 - 150: Width = $150 - 100 = 50$
• 150 - 200: Width = $200 - 150 = 50$
• 200 - 300: Width = $300 - 200 = 100$
• 300 - 500: Width = $500 - 300 = 200$
• 500 - 800: Width = $800 - 500 = 300$

The minimum class width is $50$.

Now, let's calculate the adjusted height (Length of the rectangle) for each class, using the minimum width (50) as the reference:

• Class 100 - 150: Height $\propto \frac{60}{50} \times 50 = 60$
• Class 150 - 200: Height $\propto \frac{100}{50} \times 50 = 100$
• Class 200 - 300: Height $\propto \frac{100}{100} \times 50 = 50$
• Class 300 - 500: Height $\propto \frac{80}{200} \times 50 = \frac{80}{4} = 20$
• Class 500 - 800: Height $\propto \frac{180}{300} \times 50 = \frac{180}{6} = 30$

Let's organize these calculations in a table:

Construction of the Histogram:

• Draw the horizontal axis (x-axis) and label it "Marks". Mark the class boundaries: 100, 150, 200, 300, 500, 800.
• Draw the vertical axis (y-axis) and label it "Length of the rectangle" or "Height proportional to Frequency Density". Choose a suitable scale for the y-axis to accommodate the calculated lengths (from 20 to 100). A scale where 1 unit represents 10 would be appropriate.
• Draw the first rectangle with the base from 100 to 150 on the x-axis and height 60 on the y-axis.
• Draw the second rectangle with the base from 150 to 200 on the x-axis and height 100 on the y-axis. There is no gap between consecutive bars in a histogram.
• Draw the third rectangle with the base from 200 to 300 on the x-axis and height 50 on the y-axis.
• Draw the fourth rectangle with the base from 300 to 500 on the x-axis and height 20 on the y-axis.
• Draw the fifth rectangle with the base from 500 to 800 on the x-axis and height 30 on the y-axis.

(Note: The visual bar graph cannot be displayed directly in this text format, but the steps described above, using the calculated 'Length of the rectangle' values for the heights and the given class intervals for the bases, explain how to construct the correct histogram.)

Given:

Marks obtained by 60 students in a mathematics olympiad:

46, 31, 74, 68, 42, 54, 14, 61, 83, 48,

37, 26, 8, 64, 57, 93, 72, 53, 59, 38,

16, 88, 75, 56, 46, 66, 45, 61, 54, 27,

27, 44, 63, 58, 43, 81, 64, 67, 36, 49,

50, 76, 38, 47, 55, 77, 62, 53, 40, 71,

60, 58, 45, 42, 34, 46, 40, 59, 42, 29.

Total number of students = 60.

To Prepare and Find:

Prepare a grouped frequency distribution table using the class intervals 0-9, 10-19, etc.

Find the number of students who secured more than 49 marks.

Solution:

We need to create class intervals with a width of 10, starting from 0-9.

The class intervals will be 0-9, 10-19, 20-29, 30-39, 40-49, 50-59, 60-69, 70-79, 80-89, 90-99.

We will now tally the marks into these intervals.

Grouped Frequency Distribution Table

Let me re-tally the data to ensure the total is 60.

Data: 46, 31, 74, 68, 42, 54, 14, 61, 83, 48, 37, 26, 8, 64, 57, 93, 72, 53, 59, 38, 16, 88, 75, 56, 46, 66, 45, 61, 54, 27, 27, 44, 63, 58, 43, 81, 64, 67, 36, 49, 50, 76, 38, 47, 55, 77, 62, 53, 40, 71, 60, 58, 45, 42, 34, 46, 40, 59, 42, 29.

Tallying again:

• 0-9: 8 ($|$) -> 1
• 10-19: 14, 16 ($||$) -> 2
• 20-29: 26, 27, 27, 29 ($||||$) -> 4
• 30-39: 31, 37, 38, 38, 36, 34 ($\bcancel{||||} \ |$) -> 6
• 40-49: 46, 42, 48, 44, 43, 49, 47, 40, 45, 40, 45, 42, 46, 42, 46 ($\bcancel{||||} \ \bcancel{||||} \ \bcancel{||||}$) -> 15 (Previous count 12 was wrong)
• 50-59: 54, 57, 53, 59, 56, 54, 58, 50, 55, 53, 58, 59 ($\bcancel{||||} \ \bcancel{||||} \ ||$) -> 12 (Previous count 10 was wrong)
• 60-69: 68, 61, 64, 66, 61, 63, 64, 67, 60 ($\bcancel{||||} \ ||||$) -> 9
• 70-79: 74, 72, 75, 76, 77, 71 ($\bcancel{||||} \ |$) -> 6
• 80-89: 83, 88, 81 ($\||$) -> 3
• 90-99: 93 ($|$) -> 1

Total frequency = $1 + 2 + 4 + 6 + 15 + 12 + 9 + 6 + 3 + 1 = 59$. Still not 60. Let me re-read the problem statement. "Two sections of Class IX having 30 students each". This means there are $30+30=60$ students. My table from the problem description has 6 rows * 10 columns = 60 entries. The data seems correct. Let me recount one last time.

Sorted data might help with tallying:

8, 14, 16, 26, 27, 27, 29, 31, 34, 36, 37, 38, 38, 40, 40, 42, 42, 42, 43, 44, 45, 45, 46, 46, 46, 47, 48, 49, 50, 53, 53, 54, 54, 55, 56, 57, 58, 58, 59, 59, 60, 61, 61, 62, 63, 64, 64, 66, 67, 68, 71, 72, 74, 75, 76, 77, 81, 83, 88, 93.

Let's count from the sorted list:

• 0-9: (8) -> 1
• 10-19: (14, 16) -> 2
• 20-29: (26, 27, 27, 29) -> 4
• 30-39: (31, 34, 36, 37, 38, 38) -> 6
• 40-49: (40, 40, 42, 42, 42, 43, 44, 45, 45, 46, 46, 46, 47, 48, 49) -> 15
• 50-59: (50, 53, 53, 54, 54, 55, 56, 57, 58, 58, 59, 59) -> 12
• 60-69: (60, 61, 61, 62, 63, 64, 64, 66, 67, 68) -> 10 (Previous count 9 was wrong)
• 70-79: (71, 72, 74, 75, 76, 77) -> 6
• 80-89: (81, 83, 88) -> 3
• 90-99: (93) -> 1

Total frequency = $1 + 2 + 4 + 6 + 15 + 12 + 10 + 6 + 3 + 1 = 60$. This matches the total number of students.

Grouped Frequency Distribution Table (Corrected Frequencies)

Now, we need to find the number of students who secured more than 49 marks.

"More than 49 marks" means marks starting from 50 up to the maximum score.

In our grouped frequency distribution, this includes the students in the following class intervals:

• 50 - 59
• 60 - 69
• 70 - 79
• 80 - 89
• 90 - 99

The frequencies for these intervals are:

• 50 - 59: 12 students
• 60 - 69: 10 students
• 70 - 79: 6 students
• 80 - 89: 3 students
• 90 - 99: 1 student

Number of students who secured more than 49 marks = Frequency (50-59) + Frequency (60-69) + Frequency (70-79) + Frequency (80-89) + Frequency (90-99)

Number of students who secured more than 49 marks $= 12 + 10 + 6 + 3 + 1 = 32$.

The number of students who secured more than 49 marks is 32.

Given:

The marks (out of 100) of 60 students in mathematics:

16, 13, 5, 80, 86, 7, 51, 48, 24, 56, 70, 19, 61, 17, 16, 36, 34, 42, 34, 35,

72, 55, 75, 31, 52, 28, 72, 97, 74, 45, 62, 68, 86, 35, 85, 36, 81, 75, 55, 26,

95, 31, 7, 78, 92, 62, 52, 56, 15, 63, 25, 36, 54, 44, 47, 27, 72, 17, 4, 30.

Total number of students = 60.

To Construct:

A grouped frequency distribution table with class width 10, starting from the interval 0-9.

Solution:

The class intervals will be 0-9, 10-19, 20-29, 30-39, 40-49, 50-59, 60-69, 70-79, 80-89, 90-99. We need to check the highest mark to see if we need an interval for 100. The highest mark is 97, so up to 90-99 is sufficient.

We will now tally the marks into these intervals.

Grouped Frequency Distribution Table of Mathematics Marks

Let me tally the data again to ensure the total is 60.

Data: 16, 13, 5, 80, 86, 7, 51, 48, 24, 56, 70, 19, 61, 17, 16, 36, 34, 42, 34, 35, 72, 55, 75, 31, 52, 28, 72, 97, 74, 45, 62, 68, 86, 35, 85, 36, 81, 75, 55, 26, 95, 31, 7, 78, 92, 62, 52, 56, 15, 63, 25, 36, 54, 44, 47, 27, 72, 17, 4, 30.

Tallying again:

• 0-9: 5, 7, 7, 4 ($||||$) -> 4
• 10-19: 16, 13, 19, 17, 16, 15, 17 ($\bcancel{||||} \ ||$) -> 7
• 20-29: 24, 28, 26, 25, 27 ($||\ |||$) -> 5
• 30-39: 36, 34, 34, 35, 31, 35, 36, 31, 36, 30 ($\bcancel{||||} \ \bcancel{||||}$) -> 10
• 40-49: 48, 42, 45, 44, 47 ($\bcancel{||||}$) -> 5
• 50-59: 51, 56, 55, 52, 55, 52, 56, 54 ($\bcancel{||||} \ |||$ ) -> 8
• 60-69: 61, 62, 68, 62, 63 ($\bcancel{||||}$) -> 5
• 70-79: 70, 72, 75, 72, 74, 75, 72, 78 ($\bcancel{||||} \ |||$ ) -> 8 (Previous count 7 was wrong)
• 80-89: 80, 86, 86, 85, 81 ($\bcancel{||||}$) -> 5
• 90-99: 97, 95, 92 ($|||$ ) -> 3 (Previous count 2 was wrong)

Total frequency = $4 + 7 + 5 + 10 + 5 + 8 + 5 + 8 + 5 + 3 = 60$. This matches the total number of students.

Grouped Frequency Distribution Table of Mathematics Marks (Corrected Frequencies)

Given:

The marks (out of 100) of 60 students in mathematics from Question 1:

16, 13, 5, 80, 86, 7, 51, 48, 24, 56, 70, 19, 61, 17, 16, 36, 34, 42, 34, 35, 72, 55, 75, 31, 52, 28, 72, 97, 74, 45, 62, 68, 86, 35, 85, 36, 81, 75, 55, 26, 95, 31, 7, 78, 92, 62, 52, 56, 15, 63, 25, 36, 54, 44, 47, 27, 72, 17, 4, 30.

Total number of students = 60.

To Construct:

A grouped frequency distribution table with class width 10, such that one of the classes is 10 - 20 (20 not included).

Solution:

The condition "10 - 20 (20 not included)" defines the class interval as $[10, 20)$. Since the class width is 10, the continuous class intervals will be:

$[0, 10), [10, 20), [20, 30), [30, 40), [40, 50), [50, 60), [60, 70), [70, 80), [80, 90), [90, 100)$.

We will now tally the given marks into these continuous intervals. Remember that a mark equal to the upper limit of an interval is included in the next interval.

Grouped Frequency Distribution Table of Mathematics Marks

(Note: The sum of frequencies $4 + 7 + 5 + 10 + 5 + 8 + 5 + 8 + 5 + 3 = 60$, which matches the total number of students.)

Given:

A continuous frequency distribution table showing Height (in cm) and Number of students (Frequency):

To Draw:

A histogram for the given distribution.

Solution:

To draw a histogram from a continuous frequency distribution, we use the class intervals on the horizontal axis and the frequencies on the vertical axis. Since the class intervals are continuous and have equal width (153-150 = 3, 156-153 = 3, etc.), the heights of the bars will be directly proportional to the frequencies.

Construction of the Histogram:

• Draw two perpendicular axes: a horizontal axis (x-axis) and a vertical axis (y-axis).
• Represent the 'Height (in cm)' on the x-axis. Mark the class boundaries along this axis: 150, 153, 156, 159, 162, 165, 168.
• Represent the 'Number of students (Frequency)' on the y-axis. Choose a suitable scale. The frequencies range from 5 to 14. A scale where 1 unit on the y-axis represents 2 students would be appropriate (or use a scale marking every unit). Let's mark the scale 0, 2, 4, 6, 8, 10, 12, 14.
• Draw rectangles (bars) for each class interval. The base of each rectangle lies on the x-axis and corresponds to the class interval (e.g., the first bar spans from 150 to 153).
• The height of each rectangle is equal to the frequency of the corresponding class.
  • For the interval 150 - 153, draw a bar with height 7.
  • For the interval 153 - 156, draw a bar with height 8.
  • For the interval 156 - 159, draw a bar with height 14.
  • For the interval 159 - 162, draw a bar with height 10.
  • For the interval 162 - 165, draw a bar with height 6.
  • For the interval 165 - 168, draw a bar with height 5.
• Since the class intervals are continuous, the bars should be adjacent to each other with no gaps in between.
• Label the axes clearly.

(Note: A visual representation of the histogram is not possible in this text format, but the instructions provided describe how to construct it using the class intervals as bases on the x-axis and the frequencies as heights on the y-axis.)

Given:

A grouped frequency distribution table showing Age (in years) and Number of teachers (Frequency):

To Draw:

A histogram for the given distribution.

Solution:

The given frequency distribution has discontinuous class intervals (e.g., 20-24, 25-29). To draw a histogram, the classes must be continuous.

The difference between the upper limit of one class and the lower limit of the next is $25 - 24 = 1$.

To make the classes continuous, we subtract half of this difference ($1/2 = 0.5$) from the lower limit of each class and add 0.5 to the upper limit of each class.

The new continuous class intervals are:

• 20 - 24 becomes $20 - 0.5$ to $24 + 0.5$, i.e., 19.5 - 24.5
• 25 - 29 becomes $25 - 0.5$ to $29 + 0.5$, i.e., 24.5 - 29.5
• 30 - 34 becomes $30 - 0.5$ to $34 + 0.5$, i.e., 29.5 - 34.5
• 35 - 39 becomes $35 - 0.5$ to $39 + 0.5$, i.e., 34.5 - 39.5
• 40 - 44 becomes $40 - 0.5$ to $44 + 0.5$, i.e., 39.5 - 44.5
• 45 - 49 becomes $45 - 0.5$ to $49 + 0.5$, i.e., 44.5 - 49.5
• 50 - 54 becomes $50 - 0.5$ to $54 + 0.5$, i.e., 49.5 - 54.5

The frequencies remain the same for these new intervals. The width of each continuous class is $24.5 - 19.5 = 5$, $29.5 - 24.5 = 5$, etc. The class width is constant (5) in the continuous distribution.

Since the class width is constant in the continuous distribution, the heights of the bars in the histogram will be directly proportional to the frequencies.

Construction of the Histogram:

• Draw two perpendicular axes: a horizontal axis (x-axis) and a vertical axis (y-axis).
• Represent the 'Age (in years)' on the x-axis. Mark the boundaries of the continuous class intervals along this axis: 19.5, 24.5, 29.5, 34.5, 39.5, 44.5, 49.5, 54.5.
• Represent the 'Number of teachers (Frequency)' on the y-axis. Choose a suitable scale. The frequencies range from 10 to 50. A scale where 1 unit on the y-axis represents 5 teachers would be appropriate. Mark the scale 0, 5, 10, 15, ..., 50.
• Draw rectangles (bars) for each continuous class interval. The base of each rectangle lies on the x-axis, spanning the continuous interval.
• The height of each rectangle is equal to the frequency of the corresponding class.
  • For the interval 19.5 - 24.5, draw a bar with height 10.
  • For the interval 24.5 - 29.5, draw a bar with height 28.
  • For the interval 29.5 - 34.5, draw a bar with height 32.
  • For the interval 34.5 - 39.5, draw a bar with height 48.
  • For the interval 39.5 - 44.5, draw a bar with height 50.
  • For the interval 44.5 - 49.5, draw a bar with height 35.
  • For the interval 49.5 - 54.5, draw a bar with height 12.
• Since the class intervals are continuous, the bars must be adjacent to each other with no gaps.
• Label the axes clearly.

(Note: A visual representation of the histogram is not possible in this text format, but the instructions provided describe how to construct it using the continuous class intervals as bases on the x-axis and the frequencies as heights on the y-axis.)

Given:

A discontinuous frequency distribution table showing the lengths of 62 leaves:

Let's check the total frequency: $7+10+12+17+7+5+3 = 61$. The problem states 62 leaves, but the table sums to 61. Assuming the table is correct as presented for the 61 leaves counted.

To Draw:

A histogram to represent the given data.

Solution:

The given frequency distribution has discontinuous class intervals (e.g., 118-126, 127-135). To draw a histogram, the classes must be continuous.

The difference between the upper limit of one class and the lower limit of the next is $127 - 126 = 1$.

To make the classes continuous, we subtract half of this difference ($1/2 = 0.5$) from the lower limit of each class and add 0.5 to the upper limit of each class.

The new continuous class intervals are:

• 118 - 126 becomes $118 - 0.5$ to $126 + 0.5$, i.e., 117.5 - 126.5
• 127 - 135 becomes $127 - 0.5$ to $135 + 0.5$, i.e., 126.5 - 135.5
• 136 - 144 becomes $136 - 0.5$ to $144 + 0.5$, i.e., 135.5 - 144.5
• 145 - 153 becomes $145 - 0.5$ to $153 + 0.5$, i.e., 144.5 - 153.5
• 154 - 162 becomes $154 - 0.5$ to $162 + 0.5$, i.e., 153.5 - 162.5
• 163 - 171 becomes $163 - 0.5$ to $171 + 0.5$, i.e., 162.5 - 171.5
• 172 - 180 becomes $172 - 0.5$ to $180 + 0.5$, i.e., 171.5 - 180.5

The frequencies remain the same for these new intervals. The width of each continuous class is $126.5 - 117.5 = 9$, $135.5 - 126.5 = 9$, etc. The class width is constant (9) in the continuous distribution.

Since the class width is constant in the continuous distribution, the heights of the bars in the histogram will be directly proportional to the frequencies.

Construction of the Histogram:

• Draw two perpendicular axes: a horizontal axis (x-axis) and a vertical axis (y-axis).
• Represent the 'Length (in mm)' on the x-axis. Mark the boundaries of the continuous class intervals along this axis: 117.5, 126.5, 135.5, 144.5, 153.5, 162.5, 171.5, 180.5.
• Represent the 'Number of leaves (Frequency)' on the y-axis. Choose a suitable scale. The frequencies range from 3 to 17. A scale where 1 unit on the y-axis represents 2 leaves would be appropriate (or use a scale marking every unit). Let's mark the scale 0, 2, 4, ..., 18.
• Draw rectangles (bars) for each continuous class interval. The base of each rectangle lies on the x-axis, spanning the continuous interval.
• The height of each rectangle is equal to the frequency of the corresponding class.
  • For the interval 117.5 - 126.5, draw a bar with height 7.
  • For the interval 126.5 - 135.5, draw a bar with height 10.
  • For the interval 135.5 - 144.5, draw a bar with height 12.
  • For the interval 144.5 - 153.5, draw a bar with height 17.
  • For the interval 153.5 - 162.5, draw a bar with height 7.
  • For the interval 162.5 - 171.5, draw a bar with height 5.
  • For the interval 171.5 - 180.5, draw a bar with height 3.
• Since the class intervals are continuous, the bars must be adjacent to each other with no gaps.
• Label the axes clearly.

(Note: A visual representation of the histogram is not possible in this text format, but the instructions provided describe how to construct it using the continuous class intervals as bases on the x-axis and the frequencies as heights on the y-axis.)

Given:

Frequency distribution of marks obtained by 80 students:

To Construct:

A histogram for the given distribution.

Solution:

The given frequency distribution has unequal class widths. To draw a histogram where the area of each rectangle is proportional to the frequency, we need to calculate the adjusted height for each bar using the frequency density and a reference class width.

First, calculate the width of each class interval:

• 10 - 20: Width = $20 - 10 = 10$
• 20 - 30: Width = $30 - 20 = 10$
• 30 - 50: Width = $50 - 30 = 20$
• 50 - 70: Width = $70 - 50 = 20$
• 70 - 100: Width = $100 - 70 = 30$

The minimum class width is $10$.

Now, calculate the 'Length of the rectangle' (height proportional to frequency density) for each class using the formula:

Length of the rectangle $= \frac{\text{Frequency}}{\text{Width of the class}} \times \text{Minimum class width}$

• Class 10 - 20: Length $= \frac{6}{10} \times 10 = 6$
• Class 20 - 30: Length $= \frac{17}{10} \times 10 = 17$
• Class 30 - 50: Length $= \frac{15}{20} \times 10 = \frac{15}{2} = 7.5$
• Class 50 - 70: Length $= \frac{16}{20} \times 10 = \frac{16}{2} = 8$
• Class 70 - 100: Length $= \frac{26}{30} \times 10 = \frac{26}{3} \approx 8.67$

Let's summarize the data for constructing the histogram:

Construction of the Histogram:

• Draw the horizontal axis (x-axis) and label it "Marks". Mark the class boundaries along this axis: 10, 20, 30, 50, 70, 100.
• Draw the vertical axis (y-axis) and label it "Length of the rectangle" or "Height proportional to Frequency Density". Choose a suitable scale for the y-axis to accommodate the calculated lengths (from 6 to 17, including decimals). A scale marking every unit or every 2 units would work.
• Draw rectangles (bars) for each class interval. The base of each rectangle lies on the x-axis, spanning the interval width.
• The height of each rectangle corresponds to the calculated 'Length of the rectangle'.
  • For the interval 10 - 20, draw a bar with height 6.
  • For the interval 20 - 30, draw a bar with height 17.
  • For the interval 30 - 50, draw a bar with height 7.5.
  • For the interval 50 - 70, draw a bar with height 8.
  • For the interval 70 - 100, draw a bar with height $\approx 8.67$.
• Since the class intervals are continuous, the bars must be adjacent to each other.
• Label the axes clearly.

(Note: A visual representation of the histogram is not possible in this text format. The base of each rectangle will correspond to the class interval width on the x-axis, and the height will be the calculated 'Length of the rectangle' on the y-axis.)

Given:

Frequency distribution table for the speed of cars:

To Draw:

A histogram and a frequency polygon for the given data.

Solution:

Part 1: Drawing the Histogram

The given frequency distribution has continuous class intervals with equal width ($40-30=10$, $50-40=10$, etc.). Therefore, the heights of the bars in the histogram will be directly proportional to the frequencies.

Construction of the Histogram:

• Draw two perpendicular axes: a horizontal axis (x-axis) and a vertical axis (y-axis).
• Represent the 'Speed (in km/h)' on the x-axis. Mark the class boundaries along this axis: 30, 40, 50, 60, 70, 80, 90, 100.
• Represent the 'Frequency' on the y-axis. Choose a suitable scale. The frequencies range from 3 to 65. A scale where 1 unit represents 5 or 10 frequencies would be appropriate. Let's use 1 unit = 5 frequencies. Mark the scale 0, 5, 10, ..., 65, 70.
• Draw rectangles (bars) for each class interval. The base of each rectangle lies on the x-axis and corresponds to the class interval.
• The height of each rectangle is equal to the frequency of the corresponding class.
  • For 30-40, height = 3.
  • For 40-50, height = 6.
  • For 50-60, height = 25.
  • For 60-70, height = 65.
  • For 70-80, height = 50.
  • For 80-90, height = 28.
  • For 90-100, height = 14.
• Since the class intervals are continuous, the bars are adjacent with no gaps.
• Label the axes clearly.

Part 2: Drawing the Frequency Polygon

A frequency polygon can be drawn by plotting points at the mid-points of the top of each rectangle of the histogram and joining them with line segments. Alternatively, we can plot points with coordinates (class mark, frequency) and join them.

First, calculate the mid-point (class mark) of each class interval:

Mid-point $= \frac{\text{Lower limit} + \text{Upper limit}}{2}$

• 30 - 40: Mid-point = $\frac{30+40}{2} = 35$. Frequency = 3. Point: (35, 3).
• 40 - 50: Mid-point = $\frac{40+50}{2} = 45$. Frequency = 6. Point: (45, 6).
• 50 - 60: Mid-point = $\frac{50+60}{2} = 55$. Frequency = 25. Point: (55, 25).
• 60 - 70: Mid-point = $\frac{60+70}{2} = 65$. Frequency = 65. Point: (65, 65).
• 70 - 80: Mid-point = $\frac{70+80}{2} = 75$. Frequency = 50. Point: (75, 50).
• 80 - 90: Mid-point = $\frac{80+90}{2} = 85$. Frequency = 28. Point: (85, 28).
• 90 - 100: Mid-point = $\frac{90+100}{2} = 95$. Frequency = 14. Point: (95, 14).

To make the frequency polygon start and end on the horizontal axis, we include two additional class intervals with zero frequency: one before the first interval and one after the last interval. The width of these intervals is also 10.

• Class before 30-40: 20 - 30. Mid-point = $\frac{20+30}{2} = 25$. Frequency = 0. Point: (25, 0).
• Class after 90-100: 100 - 110. Mid-point = $\frac{100+110}{2} = 105$. Frequency = 0. Point: (105, 0).

Construction of the Frequency Polygon:

• Using the same axes and scales as the histogram:
• Plot the points corresponding to (class mark, frequency): (25, 0), (35, 3), (45, 6), (55, 25), (65, 65), (75, 50), (85, 28), (95, 14), (105, 0).
• Join these points consecutively with line segments.
• The resulting closed figure is the frequency polygon.

(Note: A visual representation of the graphs is not possible in this text format. However, the steps outline how to construct both the histogram and the frequency polygon.)

Given:

Frequency distribution table for the speed of cars (from Question 7):

To Draw:

A frequency polygon representing the given data without drawing the histogram.

Solution:

To draw a frequency polygon without constructing a histogram, we plot points using the class marks of the intervals and their corresponding frequencies. These points are then joined by line segments. To make the polygon a closed figure and touch the horizontal axis, we include two additional class intervals with zero frequency, one preceding the first class and one succeeding the last class, each having the same width as the given classes.

First, calculate the class mark (mid-point) for each given class interval:

Class Mark $= \frac{\text{Lower limit} + \text{Upper limit}}{2}$

• 30 - 40: Class Mark = $\frac{30+40}{2} = 35$. Frequency = 3.
• 40 - 50: Class Mark = $\frac{40+50}{2} = 45$. Frequency = 6.
• 50 - 60: Class Mark = $\frac{50+60}{2} = 55$. Frequency = 25.
• 60 - 70: Class Mark = $\frac{60+70}{2} = 65$. Frequency = 65.
• 70 - 80: Class Mark = $\frac{70+80}{2} = 75$. Frequency = 50.
• 80 - 90: Class Mark = $\frac{80+90}{2} = 85$. Frequency = 28.
• 90 - 100: Class Mark = $\frac{90+100}{2} = 95$. Frequency = 14.

The width of each class interval is $40 - 30 = 10$. We need to include two intervals of width 10 with zero frequency at the beginning and end.

• Preceding class: The class mark preceding 35 with a width of 10 would be $35 - 10 = 25$. The interval is 20 - 30. Frequency = 0. Point: (25, 0).
• Succeeding class: The class mark succeeding 95 with a width of 10 would be $95 + 10 = 105$. The interval is 100 - 110. Frequency = 0. Point: (105, 0).

We now have the points to plot:

Construction of the Frequency Polygon:

• Draw two perpendicular axes: a horizontal axis (x-axis) and a vertical axis (y-axis).
• Represent the 'Speed (in km/h)' using the class marks on the x-axis. Mark the points 25, 35, 45, 55, 65, 75, 85, 95, 105 along the x-axis with appropriate spacing.
• Represent the 'Frequency' on the y-axis. Choose a suitable scale to accommodate frequencies up to 65. Mark the scale 0, 5, 10, ..., 65, 70 on the y-axis.
• Plot the points calculated above: (25, 0), (35, 3), (45, 6), (55, 25), (65, 65), (75, 50), (85, 28), (95, 14), and (105, 0).
• Join these points consecutively with straight line segments.
• The resulting closed figure is the frequency polygon.

(Note: A visual representation of the frequency polygon is not possible in this text format. However, follow these steps to draw the polygon by plotting the points (class mark, frequency) and connecting them.)

Given:

Frequency distribution of marks obtained by students of Section A and Section B:

To Draw:

Two frequency polygons representing the marks of both sections on the same graph.

To make an observation from the graph.

Solution:

To draw frequency polygons, we need the class marks and frequencies. The class intervals are continuous (0-15, 15-30, etc.) and have equal width ($15-0=15$, $30-15=15$, etc.), which is 15.

Calculate the class mark for each interval:

Class Mark $= \frac{\text{Lower limit} + \text{Upper limit}}{2}$

• 0 - 15: Class Mark = $\frac{0+15}{2} = 7.5$
• 15 - 30: Class Mark = $\frac{15+30}{2} = 22.5$
• 30 - 45: Class Mark = $\frac{30+45}{2} = 37.5$
• 45 - 60: Class Mark = $\frac{45+60}{2} = 52.5$
• 60 - 75: Class Mark = $\frac{60+75}{2} = 67.5$
• 75 - 90: Class Mark = $\frac{75+90}{2} = 82.5$

To make the polygons touch the x-axis, we include imaginary classes with zero frequency at the beginning and end. The class width is 15.

• Class before 0-15: -15 - 0. Class Mark = $\frac{-15+0}{2} = -7.5$. Frequency = 0. Point: (-7.5, 0).
• Class after 75-90: 90 - 105. Class Mark = $\frac{90+105}{2} = 97.5$. Frequency = 0. Point: (97.5, 0).

Points for Section A: (Class Mark, Frequency)

$(-7.5, 0), (7.5, 5), (22.5, 12), (37.5, 28), (52.5, 30), (67.5, 35), (82.5, 13), (97.5, 0)$.

Points for Section B: (Class Mark, Frequency)

$(-7.5, 0), (7.5, 3), (22.5, 16), (37.5, 25), (52.5, 27), (67.5, 40), (82.5, 10), (97.5, 0)$.

Construction of the Frequency Polygons:

• Draw two perpendicular axes: x-axis (Marks, using class marks) and y-axis (Frequency).
• Mark the class marks on the x-axis: -7.5, 7.5, 22.5, 37.5, 52.5, 67.5, 82.5, 97.5.
• Choose a suitable scale for the y-axis to accommodate frequencies up to 40. Mark the scale 0, 5, 10, ..., 40.
• For Section A: Plot the points $(-7.5, 0), (7.5, 5), (22.5, 12), (37.5, 28), (52.5, 30), (67.5, 35), (82.5, 13), (97.5, 0)$. Join these points with straight line segments.
• For Section B: Plot the points $(-7.5, 0), (7.5, 3), (22.5, 16), (37.5, 25), (52.5, 27), (67.5, 40), (82.5, 10), (97.5, 0)$. Join these points with straight line segments.
• Use different colors or line styles (e.g., solid line for A, dashed line for B) for the two polygons and provide a legend to distinguish them.
• Label the axes clearly.

Observation:

By observing the two frequency polygons drawn on the same graph, we can compare the performance of students in Section A and Section B.

Looking at the frequencies:

• Section B has more students in the lower marks (0-15, 15-30) compared to Section A.
• Section A has more students in the middle marks (30-45, 45-60) compared to Section B.
• Section B has more students in the higher marks (60-75) compared to Section A.
• Section A has slightly more students in the highest marks (75-90) compared to Section B.

The peak of the frequency polygon for Section A is around the 45-60 marks interval, while the peak for Section B is around the 60-75 marks interval.

Overall, Section B seems to have performed slightly better in terms of securing higher marks, as indicated by the higher frequency in the 60-75 interval and a shape that is shifted slightly towards the right compared to Section A.

Given:

A frequency distribution table:

The mean of the distribution is $\bar{x} = 50$.

To Find:

The value of $a$, and the frequencies of 30 and 70.

Solution:

The mean of a frequency distribution is given by the formula:

$\bar{x} = \frac{\sum (f_i \times x_i)}{\sum f_i}$

First, let's calculate the sum of frequencies ($\sum f_i$):

$\sum f_i = 17 + (5a + 3) + 32 + (7a - 11) + 19$

$\sum f_i = (17 + 3 + 32 - 11 + 19) + (5a + 7a)$

$\sum f_i = (20 + 32 - 11 + 19) + 12a$

$\sum f_i = (52 - 11 + 19) + 12a$

$\sum f_i = (41 + 19) + 12a$

$\sum f_i = 60 + 12a$

Next, let's calculate the sum of the products of frequency and the corresponding value ($\sum (f_i \times x_i)$):

$\sum (f_i \times x_i) = (17 \times 10) + ((5a + 3) \times 30) + (32 \times 50) + ((7a - 11) \times 70) + (19 \times 90)$

$\sum (f_i \times x_i) = 170 + (150a + 90) + 1600 + (490a - 770) + 1710$

$\sum (f_i \times x_i) = (170 + 90 + 1600 - 770 + 1710) + (150a + 490a)$

$\sum (f_i \times x_i) = (260 + 1600 - 770 + 1710) + 640a$

$\sum (f_i \times x_i) = (1860 - 770 + 1710) + 640a$

$\sum (f_i \times x_i) = (1090 + 1710) + 640a$

$\sum (f_i \times x_i) = 2800 + 640a$

Now, substitute the given mean ($\bar{x} = 50$) and the calculated sums into the mean formula:

$50 = \frac{2800 + 640a}{60 + 12a}$

Multiply both sides by $(60 + 12a)$:

$50 (60 + 12a) = 2800 + 640a$

$3000 + 600a = 2800 + 640a$

Rearrange the equation to solve for $a$. Subtract $600a$ from both sides:

$3000 = 2800 + 640a - 600a$

$3000 = 2800 + 40a$

Subtract 2800 from both sides:

$3000 - 2800 = 40a$

$200 = 40a$

Divide by 40:

$a = \frac{200}{40}$

$a = 5$

Now that we have the value of $a$, we can find the frequencies of 30 and 70.

Frequency of 30 $= 5a + 3$

Substitute $a=5$:

Frequency of 30 $= 5(5) + 3 = 25 + 3 = 28$

Frequency of 70 $= 7a - 11$

Substitute $a=5$:

Frequency of 70 $= 7(5) - 11 = 35 - 11 = 24$

The value of $a$ is 5, the frequency of 30 is 28, and the frequency of 70 is 24.

Let's check if these frequencies are non-negative. 28 $\ge$ 0 and 24 $\ge$ 0. The frequencies are valid.

Given:

Mean marks of boys ($\bar{x}_b$) = 70.

Mean marks of girls ($\bar{x}_g$) = 73.

Mean marks of all students ($\bar{x}$) = 71.

To Find:

The ratio of the number of boys to the number of girls.

Solution:

Let the number of boys be $n_b$ and the number of girls be $n_g$.

The mean of a group is calculated as: $\text{Mean} = \frac{\text{Sum of marks}}{\text{Number of students}}$.

So, the Sum of marks = Mean $\times$ Number of students.

Sum of marks obtained by boys = $\bar{x}_b \times n_b = 70 \times n_b = 70n_b$.

Sum of marks obtained by girls = $\bar{x}_g \times n_g = 73 \times n_g = 73n_g$.

The total number of students in the examination is $n_b + n_g$.

The total marks obtained by all students = Sum of marks by boys + Sum of marks by girls

Total marks $= 70n_b + 73n_g$.

The mean marks of all students is given by:

Mean of all students $= \frac{\text{Total marks}}{\text{Total number of students}}$

$\bar{x} = \frac{70n_b + 73n_g}{n_b + n_g}$

Substitute the given mean of all students ($\bar{x} = 71$) into the equation:

$71 = \frac{70n_b + 73n_g}{n_b + n_g}$

Multiply both sides by $(n_b + n_g)$:

$71(n_b + n_g) = 70n_b + 73n_g$

$71n_b + 71n_g = 70n_b + 73n_g$

Rearrange the terms to group $n_b$ and $n_g$ terms:

$71n_b - 70n_b = 73n_g - 71n_g$

$1n_b = 2n_g$

$n_b = 2n_g$

We need to find the ratio of the number of boys to the number of girls, which is $\frac{n_b}{n_g}$.

Divide both sides of the equation $n_b = 2n_g$ by $n_g$ (assuming $n_g \ne 0$, which must be true as the mean of girls' marks is given):

$\frac{n_b}{n_g} = \frac{2n_g}{n_g}$

$\frac{n_b}{n_g} = 2$

The ratio of the number of boys to the number of girls is $2:1$.

The ratio of the number of boys to the number of girls is 2:1.

Given:

Blood sugar levels (in mg/dl) of 25 patients:

87, 71, 83, 67, 85, 77, 69, 76, 65, 85, 85, 54, 70, 68, 80, 73, 78, 68, 85, 73, 81, 78, 81, 77, 75.

Total number of patients (observations) = 25.

To Find:

The mean, median, and mode of the given data.

Solution:

We will calculate the mean, median, and mode separately.

Calculation of Mean:

The mean is the sum of all observations divided by the number of observations.

Sum of observations = $87+71+83+67+85+77+69+76+65+85+85+54+70+68+80+73+78+68+85+73+81+78+81+77+75$

Sum of observations = $1914$

Number of observations ($n$) = $25$

Mean $= \frac{\text{Sum of observations}}{n}$

Mean $= \frac{1914}{25}$

Mean $= 76.56$

Calculation of Median:

To find the median, we first arrange the data in ascending order.

Ordered data: 54, 65, 67, 68, 68, 69, 70, 71, 73, 73, 75, 76, 77, 77, 78, 78, 80, 81, 81, 83, 85, 85, 85, 85, 87.

The number of observations is $n = 25$, which is an odd number.

The median is the $\left(\frac{n+1}{2}\right)^{th}$ observation in the ordered data.

Median $= \left(\frac{25+1}{2}\right)^{th}$ observation $= \left(\frac{26}{2}\right)^{th}$ observation $= 13^{th}$ observation.

Looking at the ordered data, the 13th observation is 77.

Median = 77.

Calculation of Mode:

The mode is the observation that occurs most frequently in the data set.

Let's count the frequency of each distinct observation from the ordered data:

• 54: 1
• 65: 1
• 67: 1
• 68: 2
• 69: 1
• 70: 1
• 71: 1
• 73: 2
• 75: 1
• 76: 1
• 77: 2
• 78: 2
• 80: 1
• 81: 2
• 85: 4
• 87: 1

The observation 85 appears most frequently (4 times).

Mode = 85.

The mean blood sugar level is 76.56 mg/dl.

The median blood sugar level is 77 mg/dl.

The mode blood sugar level is 85 mg/dl.