Chapter 13 Statistics

Given:

The marks obtained ($x_i$) by 30 students and their corresponding frequencies ($f_i$).

The data is presented in the table:

To Find:

The mean of the marks obtained by the students.

Solution:

We can find the mean using the direct method. The formula for the mean ($\overline{x}$) for ungrouped data with frequencies is given by:

$\overline{x} = \frac{\sum f_i x_i}{\sum f_i}$

We will create a table to calculate $\sum f_i$ and $\sum f_i x_i$.

So, $\sum f_i = 30$ and $\sum f_i x_i = 1779$.

Now, calculate the mean:

$\overline{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{1779}{30}$

$\overline{x} = \frac{177.9}{3}$

$\overline{x} = 59.3$

Answer:

The mean of the marks obtained by the students is $59.3$.

Given:

The percentage distribution of female teachers in primary schools of rural areas.

The data is presented in the table:

To Find:

The mean percentage of female teachers using all three methods (Direct Method, Assumed Mean Method, Step-deviation Method).

Solution:

First, we find the class mark ($x_i$) for each interval. The class mark is the midpoint of the class interval: $x_i = \frac{\text{Lower Limit} + \text{Upper Limit}}{2}$.

So, $\sum f_i = 35$.

Method 1: Direct Method

The formula for the mean ($\overline{x}$) using the direct method is $\overline{x} = \frac{\sum f_i x_i}{\sum f_i}$.

$\overline{x} = \frac{1390}{35}$

$\overline{x} = \frac{278}{7} \approx 39.714$

Method 2: Assumed Mean Method

The formula for the mean ($\overline{x}$) using the assumed mean method is $\overline{x} = a + \frac{\sum f_i d_i}{\sum f_i}$, where $a$ is the assumed mean and $d_i = x_i - a$.

Let's take the assumed mean $a$ as the class mark of the middle interval. Let $a = 50$.

$\overline{x} = a + \frac{\sum f_i d_i}{\sum f_i} = 50 + \frac{-360}{35}$

$\overline{x} = 50 - \frac{360}{35} = 50 - \frac{72}{7}$

$\overline{x} = \frac{50 \times 7 - 72}{7} = \frac{350 - 72}{7} = \frac{278}{7} \approx 39.714$

Method 3: Step-deviation Method

The formula for the mean ($\overline{x}$) using the step-deviation method is $\overline{x} = a + (\frac{\sum f_i u_i}{\sum f_i}) \times h$, where $a$ is the assumed mean, $u_i = \frac{x_i - a}{h}$, and $h$ is the class size.

The class size $h$ is the difference between the upper and lower limits of any class interval (assuming uniform class size), e.g., $25 - 15 = 10$. So, $h = 10$.

Let's again take the assumed mean $a = 50$.

$\overline{x} = a + (\frac{\sum f_i u_i}{\sum f_i}) \times h = 50 + (\frac{-36}{35}) \times 10$

$\overline{x} = 50 - \frac{36 \times 10}{35} = 50 - \frac{360}{35}$

$\overline{x} = 50 - \frac{72}{7} = \frac{350 - 72}{7} = \frac{278}{7} \approx 39.714$

Answer:

The mean percentage of female teachers by all three methods is $\frac{278}{7}$ or approximately $39.71\%$.

Given:

The distribution of the number of wickets taken by bowlers in one-day cricket matches.

The data is presented in the table:

To Find:

1. The mean number of wickets by choosing a suitable method.

2. The meaning of the calculated mean.

Solution:

First, we find the class mark ($x_i$) for each interval. The class mark is the midpoint of the class interval: $x_i = \frac{\text{Lower Limit} + \text{Upper Limit}}{2}$.

So, $\sum f_i = 45$.

The class intervals are of unequal width (40, 40, 50, 100, 100, 100). Therefore, the step-deviation method might not be the most suitable unless we make adjustments or use a large common factor, but the Assumed Mean Method is straightforward.

Let's use the Assumed Mean Method. The formula is $\overline{x} = a + \frac{\sum f_i d_i}{\sum f_i}$, where $a$ is the assumed mean and $d_i = x_i - a$.

Let's choose the assumed mean $a$ as the class mark of the interval with the highest frequency (100-150), so $a = 125$.

$\overline{x} = a + \frac{\sum f_i d_i}{\sum f_i} = 125 + \frac{1255}{45}$

Simplify the fraction $\frac{1255}{45}$ by dividing both by 5:

$\frac{1255}{45} = \frac{251}{9}$

$\overline{x} = 125 + \frac{251}{9}$

$\overline{x} = \frac{125 \times 9 + 251}{9} = \frac{1125 + 251}{9} = \frac{1376}{9}$

Performing the division:

$1376 \div 9 \approx 152.88...$

$\overline{x} \approx 152.89$

Meaning of the Mean:

The mean number of wickets ($\approx 152.89$) represents the average number of wickets taken by a bowler in this distribution of 45 bowlers in one-day cricket matches.

Answer:

The mean number of wickets is $\frac{1376}{9}$ or approximately $152.89$.

The mean signifies the average number of wickets taken per bowler in the given data set.

Given:

A survey regarding the number of plants in 20 houses.

The data is presented in the table:

To Find:

1. The mean number of plants per house.

2. The method used and the reason for choosing it.

Solution:

First, we find the class mark ($x_i$) for each interval: $x_i = \frac{\text{Lower Limit} + \text{Upper Limit}}{2}$.

So, $\sum f_i = 20$.

The class marks ($x_i$) and frequencies ($f_i$) are relatively small numbers. Therefore, the Direct Method is suitable for calculating the mean as it involves simpler calculations without the need for assumed mean or deviations.

Method Used: Direct Method

The formula for the mean ($\overline{x}$) using the direct method is $\overline{x} = \frac{\sum f_i x_i}{\sum f_i}$.

$\overline{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{162}{20}$

$\overline{x} = \frac{16.2}{2} = 8.1$

Reason for using the Direct Method:

The values of the class marks ($x_i$) and frequencies ($f_i$) are small. Using the direct method avoids the calculation of deviations ($d_i$) or step-deviations ($u_i$), making the calculations simpler and less prone to errors compared to the other methods for this specific data set.

Answer:

The mean number of plants per house is $8.1$.

The method used is the Direct Method because the class marks and frequencies are small, making the calculation $\sum f_i x_i$ easy and efficient.

To find the mean daily wages of the workers, we can use the Step-Deviation Method because the class sizes are equal and the numerical values of the class marks ($x_i$) are large. This method simplifies the calculations.

First, we need to construct a table to calculate the necessary values.

Let the assumed mean ($a$) be 550. The class size ($h$) is $520 - 500 = 20$.

From the table above, we have:

The formula for the mean ($\bar{x}$) using the step-deviation method is:

Now, substituting the values in the formula:

$\bar{x} = 550 + 20 \left( \frac{-12}{50} \right)$

$\bar{x} = 550 - \frac{20 \times 12}{50}$

$\bar{x} = 550 - \frac{240}{50}$

$\bar{x} = 550 - \frac{24}{5}$

$\bar{x} = 550 - 4.8$

$\bar{x} = 545.2$

So, the mean daily wages of the workers of the factory is ₹ 545.20.

Alternate Solution (Using Direct Method)

We can also calculate the mean using the Direct Method. The formula for the direct method is $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$.

Let's create the calculation table for this method.

Using the formula for the direct method:

$\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{27260}{50}$

$\bar{x} = \frac{2726}{5}$

$\bar{x} = 545.2$

Both methods yield the same result. The mean daily wages of the workers is ₹ 545.20.

To find the missing frequency 'f', we are given that the mean pocket allowance is $\textsf{₹} 18$. We can use the Direct Method to find the mean, as it is straightforward when solving for an unknown frequency.

First, we create a table to organize our calculations. We need the class mark ($x_i$) for each interval and the product of the frequency and the class mark ($f_i x_i$).

From the table, we have:

Sum of frequencies, $\sum f_i = 7 + 6 + 9 + 13 + f + 5 + 4 = 44 + f$.

Sum of products, $\sum f_i x_i = 84 + 84 + 144 + 234 + 20f + 110 + 96 = 752 + 20f$.

The formula for the mean ($\overline{x}$) by the direct method is:

We are given $\overline{x} = 18$. Substituting the values into formula (i):

$18 = \frac{752 + 20f}{44 + f}$

Now, we solve for 'f':

$18(44 + f) = 752 + 20f$

$792 + 18f = 752 + 20f$

$792 - 752 = 20f - 18f$

$40 = 2f$

$f = \frac{40}{2} = 20$

Thus, the missing frequency f is 20.

We need to find the mean heartbeats per minute. Since the class size is uniform ($h=3$) and the class marks are decimal values, using the Step-Deviation Method will simplify the calculations. The Assumed Mean method is also a good choice.

Solution using Step-Deviation Method

Let's choose an assumed mean ($a$) from the middle of the data. The class mark for the interval 74-77 is $a=75.5$. The class size is $h = 71 - 68 = 3$.

We construct the following table:

From the table, we have $\sum f_i = 30$ and $\sum f_i u_i = (-6-8-3) + (7+8+6) = -17 + 21 = 4$.

The formula for the mean ($\overline{x}$) using the step-deviation method is:

$\overline{x} = a + h \left( \frac{\sum f_i u_i}{\sum f_i} \right)$

Substituting the values:

$\overline{x} = 75.5 + 3 \left( \frac{4}{30} \right)$

$\overline{x} = 75.5 + \frac{12}{30}$

$\overline{x} = 75.5 + 0.4$

$\overline{x} = 75.9$

The mean heartbeats per minute for these women is 75.9.

We need to find the mean number of mangoes kept in a packing box. The given class intervals are inclusive (e.g., 50-52, 53-55). For calculating the mean, this does not affect the class marks, so we can proceed directly. The class size is uniform ($h=3$, the difference between consecutive class marks like $54-51=3$). Given the large values of frequencies, the Step-Deviation Method is the most efficient choice.

Let's choose the assumed mean ($a$) as the class mark of the middle interval (56 - 58), so $a = 57$. The class size is $h=3$.

The calculation table is as follows:

From the table, we have $\sum f_i = 400$ and $\sum f_i u_i = (-30 - 110) + (115 + 50) = -140 + 165 = 25$.

The formula for the mean ($\overline{x}$) using the step-deviation method is:

$\overline{x} = a + h \left( \frac{\sum f_i u_i}{\sum f_i} \right)$

Substituting the values:

$\overline{x} = 57 + 3 \left( \frac{25}{400} \right)$

$\overline{x} = 57 + 3 \left( \frac{1}{16} \right)$

$\overline{x} = 57 + \frac{3}{16}$

$\overline{x} = 57 + 0.1875$

$\overline{x} = 57.1875$

The mean number of mangoes kept in a packing box is approximately 57.19.

We chose the Step-Deviation Method because it simplifies the multiplication of frequencies with class marks, making the calculation easier and less prone to errors.

To find the mean daily expenditure on food, we can use a suitable method. Since the class size is uniform ($h = 150 - 100 = 50$) and the class marks are large, the Step-Deviation Method is a suitable and efficient choice.

Let's set up the calculation table. We choose the assumed mean ($a$) to be the class mark of the middle class interval (200 - 250), which is $a = 225$. The class size is $h=50$.

From the table, we get the following sums:

$\sum f_i = 25$

$\sum f_i u_i = (-8 - 5) + (2 + 4) = -13 + 6 = -7$

The formula for the mean ($\overline{x}$) using the step-deviation method is:

$\overline{x} = a + h \left( \frac{\sum f_i u_i}{\sum f_i} \right)$

Substituting the values we calculated:

$\overline{x} = 225 + 50 \left( \frac{-7}{25} \right)$

$\overline{x} = 225 - ( \frac{50 \times 7}{25} )$

$\overline{x} = 225 - (2 \times 7)$

$\overline{x} = 225 - 14$

$\overline{x} = 211$

Therefore, the mean daily expenditure on food is $\textsf{₹} 211$.

To find the mean concentration of $\text{SO}_2$, we can observe that the class marks will be small decimal numbers. The Step-Deviation Method is a good choice to simplify calculations, even with small numbers, as long as the class size is uniform.

The class size is $h = 0.04 - 0.00 = 0.04$. Let's choose an assumed mean ($a$) as the class mark of the interval 0.08 - 0.12, which is $a = 0.10$.

Now, we construct the table for our calculations:

From the table, we have the following sums:

$\sum f_i = 30$

$\sum f_i u_i = (-8 - 9) + (2 + 8 + 6) = -17 + 16 = -1$

The formula for the mean ($\overline{x}$) using the step-deviation method is:

$\overline{x} = a + h \left( \frac{\sum f_i u_i}{\sum f_i} \right)$

Substituting the values:

$\overline{x} = 0.10 + 0.04 \left( \frac{-1}{30} \right)$

$\overline{x} = 0.10 - \frac{0.04}{30}$

$\overline{x} = 0.10 - 0.00133...$

$\overline{x} \approx 0.09867$

Rounding to three decimal places, we get $\overline{x} \approx 0.099$.

Thus, the mean concentration of $\text{SO}_2$ in the air is approximately 0.099 ppm.

We need to find the mean number of days a student was absent. Looking at the data, the class intervals have varying widths (e.g., $6-0=6$, $10-6=4$, $20-14=6$). Because the class sizes are not uniform, we cannot use the Step-Deviation method. The Assumed Mean Method is the most appropriate choice here to keep the calculations with large numbers manageable. The Direct Method could also be used but would involve larger products.

Let's choose an assumed mean ($a$) from the middle of the data. The class mark for the interval 14-20 is $a=17$.

The calculation table is as follows:

From the table, we get the following sums:

$\sum f_i = 40$

$\sum f_i d_i = (-154 - 90 - 35) + (28 + 48 + 22) = -279 + 98 = -181$

The formula for the mean ($\overline{x}$) using the assumed mean method is:

$\overline{x} = a + \frac{\sum f_i d_i}{\sum f_i}$

Substituting the values:

$\overline{x} = 17 + \frac{-181}{40}$

$\overline{x} = 17 - 4.525$

$\overline{x} = 12.475$

Therefore, the mean number of days a student was absent is 12.475 days.

To find the mean literacy rate, we can use a suitable method. Since the class intervals have a uniform width ($h = 55 - 45 = 10$) and the frequencies are reasonably sized, the Step-Deviation Method is an excellent choice for simplifying the calculations.

Let's prepare the calculation table. We choose the assumed mean ($a$) as the class mark of the middle interval (65 - 75), so $a = 70$. The class size is $h=10$.

From the table, we get the following sums:

$\sum f_i = 35$

$\sum f_i u_i = (-6 - 10) + (8 + 6) = -16 + 14 = -2$

The formula for the mean ($\overline{x}$) using the step-deviation method is:

$\overline{x} = a + h \left( \frac{\sum f_i u_i}{\sum f_i} \right)$

Substituting the values:

$\overline{x} = 70 + 10 \left( \frac{-2}{35} \right)$

$\overline{x} = 70 - \frac{20}{35}$

$\overline{x} = 70 - \frac{4}{7}$

$\overline{x} \approx 70 - 0.5714$

$\overline{x} \approx 69.4286$

Rounding to two decimal places, we get $\overline{x} \approx 69.43$.

Thus, the mean literacy rate for the 35 cities is 69.43 %.

The mode is the value that appears most frequently in a set of data.

Given Data:

The number of wickets taken in 10 cricket matches are: 2, 6, 4, 5, 0, 2, 1, 3, 2, 3.

To Find:

The mode of the given data.

Solution:

To find the mode, let's first arrange the data in ascending order to easily count the frequency of each observation.

0, 1, 2, 2, 2, 3, 3, 4, 5, 6

Now, we can create a frequency table to count how many times each value appears:

From the table, we can see that the number of wickets '2' has the highest frequency of 3.

Therefore, the mode of the data is 2.

To find the mode for grouped data, we use a specific formula after identifying the modal class (the class with the highest frequency).

Step 1: Identify the Modal Class

By observing the frequency table, the highest frequency is 8. This corresponds to the class interval 3 - 5.

Therefore, the modal class is 3 - 5.

Step 2: Identify Values for the Mode Formula

• Lower limit of the modal class ($l$) = 3
• Frequency of the modal class ($f_1$) = 8
• Frequency of the class preceding the modal class ($f_0$) = 7
• Frequency of the class succeeding the modal class ($f_2$) = 2
• Class size ($h$) = $5 - 3 = 2$

Step 3: Calculate the Mode

The formula for the mode of grouped data is:

$\text{Mode} = l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h$

Substituting the values into the formula:

$\text{Mode} = 3 + \left(\frac{8 - 7}{2(8) - 7 - 2}\right) \times 2$

$\text{Mode} = 3 + \left(\frac{1}{16 - 9}\right) \times 2$

$\text{Mode} = 3 + \left(\frac{1}{7}\right) \times 2$

$\text{Mode} = 3 + \frac{2}{7}$

$\text{Mode} \approx 3 + 0.286$

$\text{Mode} \approx 3.286$

Rounding to two decimal places, the mode of the data is approximately 3.29.

This problem requires us to calculate the mode and the mean from the given grouped data, and then compare and interpret these two measures of central tendency.

Part 1: Calculation of the Mode

First, we identify the modal class by finding the class interval with the highest frequency. From the table, the highest frequency is 7, which corresponds to the class interval 40 - 55.

Now, we gather the necessary values for the mode formula:

• Lower limit of the modal class ($l$) = 40
• Frequency of the modal class ($f_1$) = 7
• Frequency of the class preceding the modal class ($f_0$) = 3
• Frequency of the class succeeding the modal class ($f_2$) = 6
• Class size ($h$) = $55 - 40 = 15$

The formula for the mode is:

$\text{Mode} = l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h$

Substituting the values:

$\text{Mode} = 40 + \left(\frac{7 - 3}{2(7) - 3 - 6}\right) \times 15$

$\text{Mode} = 40 + \left(\frac{4}{14 - 9}\right) \times 15$

$\text{Mode} = 40 + \left(\frac{4}{5}\right) \times 15$

$\text{Mode} = 40 + (4 \times 3)$

$\text{Mode} = 40 + 12 = 52$

Part 2: Calculation of the Mean

The table provides the sum of frequencies ($\sum f_i = 30$) and the sum of the product of frequencies and class marks ($\sum f_i x_i = 1860$). We can use the direct method to find the mean.

The formula for the mean is:

$\overline{x} = \frac{\sum f_i x_i}{\sum f_i}$

Substituting the values from the table:

$\overline{x} = \frac{1860}{30}$

$\overline{x} = 62$

Part 3: Comparison and Interpretation

We have calculated:

• Mode = 52
• Mean = 62

Interpretation:

The mode of 52 indicates that the most frequently occurring marks for the students are around 52. This represents the most common score in the class.

The mean of 62 represents the average mark obtained by a student. It summarizes the overall performance of the entire class into a single value.

Comparison: The mean (62) is greater than the mode (52). This suggests that while the most common score is 52, there is a significant number of students who scored marks higher than 52, which pulls the overall average up. In statistical terms, the data distribution is slightly skewed to the right (positively skewed).

We need to find the mode and the mean for the given frequency distribution and then compare them.

Calculation of Mode

First, we identify the modal class, which is the class interval with the highest frequency.

From the table, the highest frequency is 23, which corresponds to the class interval 35 - 45.

So, the modal class is 35 - 45.

Now, we identify the values for the mode formula:

• Lower limit of the modal class ($l$) = 35
• Frequency of the modal class ($f_1$) = 23
• Frequency of the class preceding the modal class ($f_0$) = 21
• Frequency of the class succeeding the modal class ($f_2$) = 14
• Class size ($h$) = $45 - 35 = 10$

The formula for the mode is:

$\text{Mode} = l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h$

Substituting the values:

$\text{Mode} = 35 + \left(\frac{23 - 21}{2(23) - 21 - 14}\right) \times 10$

$\text{Mode} = 35 + \left(\frac{2}{46 - 35}\right) \times 10$

$\text{Mode} = 35 + \left(\frac{2}{11}\right) \times 10$

$\text{Mode} = 35 + \frac{20}{11}$

$\text{Mode} \approx 35 + 1.818$

$\text{Mode} \approx 36.818$

Calculation of Mean

We will use the Step-Deviation Method to find the mean. Let's choose an assumed mean ($a$) as the class mark of the 35-45 interval, so $a = 40$. The class size ($h$) is 10.

From the table, we have $\sum f_i = 80$ and $\sum f_i u_i = (-18 - 22 - 21) + (14 + 10) = -61 + 24 = -37$.

The formula for the mean is:

$\overline{x} = a + h \left(\frac{\sum f_i u_i}{\sum f_i}\right)$

Substituting the values:

$\overline{x} = 40 + 10 \left(\frac{-37}{80}\right)$

$\overline{x} = 40 - \frac{370}{80} = 40 - \frac{37}{8}$

$\overline{x} = 40 - 4.625$

$\overline{x} = 35.375$

Comparison and Interpretation

The calculated values are:

• Mode $\approx$ 36.8 years
• Mean $\approx$ 35.4 years

Interpretation: The mode indicates that the maximum number of patients admitted to the hospital are of the age 36.8 years (approximately). The mean indicates that the average age of a patient admitted to the hospital is 35.4 years (approximately).

Comparison: The mode is slightly higher than the mean. This suggests that while the most common age group is 35-45, the overall distribution of patient ages is slightly skewed towards the younger side, pulling the average age down a little.

To determine the modal lifetimes, we need to find the mode of the given grouped frequency distribution.

Step 1: Identify the Modal Class

The modal class is the class interval with the highest frequency.

From the given table, the highest frequency is 61, which corresponds to the class interval 60 - 80.

Therefore, the modal class is 60 - 80.

Step 2: Identify Values for the Mode Formula

• Lower limit of the modal class ($l$) = 60
• Frequency of the modal class ($f_1$) = 61
• Frequency of the class preceding the modal class ($f_0$) = 52
• Frequency of the class succeeding the modal class ($f_2$) = 38
• Class size ($h$) = $80 - 60 = 20$

Step 3: Calculate the Mode

The formula for the mode of grouped data is:

$\text{Mode} = l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h$

Substituting the values into the formula:

$\text{Mode} = 60 + \left(\frac{61 - 52}{2(61) - 52 - 38}\right) \times 20$

$\text{Mode} = 60 + \left(\frac{9}{122 - 90}\right) \times 20$

$\text{Mode} = 60 + \left(\frac{9}{32}\right) \times 20$

$\text{Mode} = 60 + \frac{180}{32}$

$\text{Mode} = 60 + \frac{45}{8}$

$\text{Mode} = 60 + 5.625$

$\text{Mode} = 65.625$

The modal lifetime of the components is 65.625 hours.

We need to find the modal monthly expenditure and the mean monthly expenditure for the given data.

Calculation of Modal Monthly Expenditure

First, we identify the modal class, which is the class with the highest frequency.

From the table, the highest frequency is 40, which corresponds to the class interval 1500 - 2000.

The required values for the mode formula are:

• $l$ (Lower limit of modal class) = 1500
• $f_1$ (Frequency of modal class) = 40
• $f_0$ (Frequency of preceding class) = 24
• $f_2$ (Frequency of succeeding class) = 33
• $h$ (Class size) = 500

Using the mode formula:

$\text{Mode} = l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h$

$\text{Mode} = 1500 + \left(\frac{40 - 24}{2(40) - 24 - 33}\right) \times 500$

$\text{Mode} = 1500 + \left(\frac{16}{80 - 57}\right) \times 500$

$\text{Mode} = 1500 + \left(\frac{16}{23}\right) \times 500$

$\text{Mode} = 1500 + \frac{8000}{23}$

$\text{Mode} \approx 1500 + 347.826$

$\text{Mode} \approx 1847.826$

Calculation of Mean Monthly Expenditure

We use the Step-Deviation Method. Let the assumed mean ($a$) be 2750. The class size ($h$) is 500.

From the table, $\sum f_i = 200$ and $\sum f_i u_i = -35$.

Using the mean formula:

$\overline{x} = a + h \left(\frac{\sum f_i u_i}{\sum f_i}\right)$

$\overline{x} = 2750 + 500 \left(\frac{-35}{200}\right)$

$\overline{x} = 2750 - \frac{500 \times 35}{200}$

$\overline{x} = 2750 - (2.5 \times 35)$

$\overline{x} = 2750 - 87.5$

$\overline{x} = 2662.5$

The modal monthly expenditure is ₹ 1847.83 (approx).

The mean monthly expenditure is ₹ 2662.50.

We need to find the mode and the mean of the teacher-student ratio data and interpret the results.

Calculation of Mode

First, we identify the modal class by finding the interval with the highest frequency.

The highest frequency is 10, which corresponds to the class interval 30 - 35. This is our modal class.

The values for the formula are:

• $l$ (Lower limit of modal class) = 30
• $f_1$ (Frequency of modal class) = 10
• $f_0$ (Frequency of preceding class) = 9
• $f_2$ (Frequency of succeeding class) = 3
• $h$ (Class size) = 5

Using the mode formula:

$\text{Mode} = l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h$

$\text{Mode} = 30 + \left(\frac{10 - 9}{2(10) - 9 - 3}\right) \times 5$

$\text{Mode} = 30 + \left(\frac{1}{20 - 12}\right) \times 5$

$\text{Mode} = 30 + \frac{5}{8}$

$\text{Mode} = 30 + 0.625 = 30.625$

Calculation of Mean

We use the Step-Deviation Method. Let the assumed mean ($a$) be 32.5. The class size ($h$) is 5.

From the table, $\sum f_i = 35$ and $\sum f_i u_i = -23$.

Using the mean formula:

$\overline{x} = a + h \left(\frac{\sum f_i u_i}{\sum f_i}\right)$

$\overline{x} = 32.5 + 5 \left(\frac{-23}{35}\right)$

$\overline{x} = 32.5 - \frac{115}{35} = 32.5 - \frac{23}{7}$

$\overline{x} \approx 32.5 - 3.286$

$\overline{x} \approx 29.214$

Interpretation of Measures

• Mode $\approx$ 30.6. This implies that most of the states/U.T. have a student-teacher ratio of about 30.6.
• Mean $\approx$ 29.2. This implies that on average, the student-teacher ratio across all states/U.T. is about 29.2.

To find the mode of the given data, we will follow the standard procedure for grouped frequency distributions.

Step 1: Identify the Modal Class

The modal class is the class with the highest frequency. Looking at the table, the highest frequency is 18, which corresponds to the class interval 4000 - 5000.

Therefore, the modal class is 4000 - 5000.

Step 2: Identify Values for the Mode Formula

• Lower limit of the modal class ($l$) = 4000
• Frequency of the modal class ($f_1$) = 18
• Frequency of the class preceding the modal class ($f_0$) = 4
• Frequency of the class succeeding the modal class ($f_2$) = 9
• Class size ($h$) = $5000 - 4000 = 1000$

Step 3: Calculate the Mode

The formula for the mode is:

$\text{Mode} = l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h$

Substituting the identified values:

$\text{Mode} = 4000 + \left(\frac{18 - 4}{2(18) - 4 - 9}\right) \times 1000$

$\text{Mode} = 4000 + \left(\frac{14}{36 - 13}\right) \times 1000$

$\text{Mode} = 4000 + \left(\frac{14}{23}\right) \times 1000$

$\text{Mode} = 4000 + \frac{14000}{23}$

$\text{Mode} \approx 4000 + 608.696$

$\text{Mode} \approx 4608.696$

The mode of the runs scored is approximately 4608.7.

To find the mode of the data, we will use the formula for the mode of a grouped frequency distribution.

Step 1: Identify the Modal Class

The modal class is the class interval with the maximum frequency. From the table, the maximum frequency is 20, which corresponds to the class interval 40 - 50.

Thus, the modal class is 40 - 50.

Step 2: Identify Values for the Mode Formula

• Lower limit of the modal class ($l$) = 40
• Frequency of the modal class ($f_1$) = 20
• Frequency of the class preceding the modal class ($f_0$) = 12
• Frequency of the class succeeding the modal class ($f_2$) = 11
• Class size ($h$) = $50 - 40 = 10$

Step 3: Calculate the Mode

The formula for the mode is:

$\text{Mode} = l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h$

Substituting the values into the formula:

$\text{Mode} = 40 + \left(\frac{20 - 12}{2(20) - 12 - 11}\right) \times 10$

$\text{Mode} = 40 + \left(\frac{8}{40 - 23}\right) \times 10$

$\text{Mode} = 40 + \left(\frac{8}{17}\right) \times 10$

$\text{Mode} = 40 + \frac{80}{17}$

$\text{Mode} \approx 40 + 4.706$

$\text{Mode} \approx 44.706$

The mode of the data is approximately 44.7 cars.

To find the median height, we first need to convert the given cumulative frequency distribution of the 'less than' type into a continuous grouped frequency distribution.

Step 1: Constructing the Frequency Distribution Table

We can determine the frequency of each class interval by finding the difference between consecutive cumulative frequencies.

Step 2: Finding the Median Class

The total number of observations is $N = 51$.

We need to find the class corresponding to the $\frac{N}{2}^{th}$ observation.

$\frac{N}{2} = \frac{51}{2} = 25.5$

We look for the cumulative frequency which is just greater than or equal to 25.5. From the table, this value is 29, which corresponds to the class interval 145 - 150.

Therefore, the median class is 145 - 150.

Step 3: Calculating the Median

We use the following formula for the median of grouped data:

Here, we have:

• $l$ (Lower limit of the median class) = 145
• $N$ (Total frequency) = 51
• $cf$ (Cumulative frequency of the class preceding the median class) = 11
• $f$ (Frequency of the median class) = 18
• $h$ (Class size) = $150 - 145 = 5$

Substituting these values into formula (i):

$\text{Median} = 145 + \left(\frac{25.5 - 11}{18}\right) \times 5$

$\text{Median} = 145 + \left(\frac{14.5}{18}\right) \times 5$

$\text{Median} = 145 + \frac{72.5}{18}$

$\text{Median} \approx 145 + 4.028$

$\text{Median} \approx 149.028$

The median height is approximately 149.03 cm.

We are given the median and the total frequency, and we need to find two missing frequencies, $x$ and $y$. We will use this information to form two linear equations and solve them.

Step 1: Form an equation using the Total Frequency

The sum of all frequencies is given as 100.

$2 + 5 + x + 12 + 17 + 20 + y + 9 + 7 + 4 = 100$

$76 + x + y = 100$

$x + y = 100 - 76$

Step 2: Use the Median to form a second equation

First, we construct the cumulative frequency ($cf$) table.

The median is given as 525. Since 525 lies in the class interval 500 - 600, the median class is 500 - 600.

We have:

• $l$ (Lower limit of median class) = 500
• $N$ (Total frequency) = 100, so $\frac{N}{2} = 50$
• $cf$ (Cumulative frequency of the class preceding the median class) = $36 + x$
• $f$ (Frequency of the median class) = 20
• $h$ (Class size) = 100

Using the median formula: $\text{Median} = l + \left(\frac{\frac{N}{2} - cf}{f}\right) \times h$

$525 = 500 + \left(\frac{50 - (36 + x)}{20}\right) \times 100$

$525 - 500 = \left(\frac{50 - 36 - x}{20}\right) \times 100$

$25 = \left(\frac{14 - x}{\cancel{20}_1}\right) \times \cancel{100}^5$

$25 = (14 - x) \times 5$

$\frac{25}{5} = 14 - x$

$5 = 14 - x$

$x = 14 - 5 = 9$

Step 3: Solve for y

Substitute the value of $x=9$ into equation (i):

$9 + y = 24$

$y = 24 - 9 = 15$

Therefore, the values of the missing frequencies are x = 9 and y = 15.

We need to find the three measures of central tendency (mean, median, and mode) for the given data and then compare them.

Calculation of Median

First, we create a cumulative frequency ($cf$) table to find the median.

The total number of consumers is $N = 68$. We find the position of the median as $\frac{N}{2} = \frac{68}{2} = 34$.

The cumulative frequency just greater than 34 is 42, which corresponds to the class interval 125 - 145. This is the median class.

We have:

• $l$ (Lower limit of median class) = 125
• $\frac{N}{2}$ = 34
• $cf$ (Cumulative frequency of the class preceding the median class) = 22
• $f$ (Frequency of the median class) = 20
• $h$ (Class size) = $145 - 125 = 20$

Using the median formula: $\text{Median} = l + \left(\frac{\frac{N}{2} - cf}{f}\right) \times h$

$\text{Median} = 125 + \left(\frac{34 - 22}{20}\right) \times 20$

$\text{Median} = 125 + \frac{12}{20} \times 20$

$\text{Median} = 125 + 12 = 137$

Calculation of Mean

We use the Step-Deviation Method. Let the assumed mean ($a$) be 135. The class size ($h$) is 20.

Using the mean formula: $\overline{x} = a + h \left(\frac{\sum f_i u_i}{\sum f_i}\right)$

$\overline{x} = 135 + 20 \left(\frac{7}{68}\right)$

$\overline{x} = 135 + \frac{140}{68} = 135 + \frac{35}{17}$

$\overline{x} \approx 135 + 2.0588$

$\overline{x} \approx 137.06$

Calculation of Mode

The highest frequency is 20, so the modal class is 125 - 145.

We have:

• $l$ (Lower limit of modal class) = 125
• $f_1$ (Frequency of modal class) = 20
• $f_0$ (Frequency of preceding class) = 13
• $f_2$ (Frequency of succeeding class) = 14
• $h$ (Class size) = 20

Using the mode formula: $\text{Mode} = l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h$

$\text{Mode} = 125 + \left(\frac{20 - 13}{2(20) - 13 - 14}\right) \times 20$

$\text{Mode} = 125 + \left(\frac{7}{40 - 27}\right) \times 20$

$\text{Mode} = 125 + \frac{140}{13}$

$\text{Mode} \approx 125 + 10.769$

$\text{Mode} \approx 135.77$

Comparison of Results

The three measures of central tendency are:

• Mean = 137.06 units
• Median = 137 units
• Mode = 135.77 units

We can see that the mean, median, and mode are very close to each other. This indicates that the data distribution is approximately symmetric.

We are given the median and the total frequency. We need to find the two missing frequencies, $x$ and $y$. We will form two linear equations using the given information and solve them.

Step 1: Form an equation using the Total Frequency

The sum of all frequencies is given as 60.

$5 + x + 20 + 15 + y + 5 = 60$

$45 + x + y = 60$

$x + y = 60 - 45$

Step 2: Use the Median to form a second equation

First, we create the cumulative frequency ($cf$) table.

The median is given as 28.5, which lies in the class interval 20 - 30. So, this is the median class.

We have:

• $l$ (Lower limit of median class) = 20
• $N$ (Total frequency) = 60, so $\frac{N}{2} = 30$
• $cf$ (Cumulative frequency of the class preceding the median class) = $5 + x$
• $f$ (Frequency of the median class) = 20
• $h$ (Class size) = 10

Using the median formula: $\text{Median} = l + \left(\frac{\frac{N}{2} - cf}{f}\right) \times h$

$28.5 = 20 + \left(\frac{30 - (5 + x)}{20}\right) \times 10$

$28.5 - 20 = \left(\frac{30 - 5 - x}{20}\right) \times 10$

$8.5 = \left(\frac{25 - x}{\cancel{20}_2}\right) \times \cancel{10}^1$

$8.5 = \frac{25 - x}{2}$

$8.5 \times 2 = 25 - x$

$17 = 25 - x$

$x = 25 - 17 = 8$

Step 3: Solve for y

Now, substitute the value of $x=8$ into equation (i):

$8 + y = 15$

$y = 15 - 8 = 7$

Thus, the values of the missing frequencies are x = 8 and y = 7.

The given data is a cumulative frequency distribution of the 'below' type. To find the median age, we first need to convert this into a continuous grouped frequency distribution.

Step 1: Constructing the Frequency Distribution Table

We create class intervals and calculate the frequency for each class by finding the difference between consecutive cumulative frequencies.

Step 2: Finding the Median Class

The total number of policy holders is $N = 100$.

We find the position of the median as $\frac{N}{2} = \frac{100}{2} = 50$.

The cumulative frequency just greater than 50 is 78. This corresponds to the class interval 35 - 40.

Therefore, the median class is 35 - 40.

Step 3: Calculating the Median Age

We use the median formula with the following values:

• $l$ (Lower limit of median class) = 35
• $\frac{N}{2}$ = 50
• $cf$ (Cumulative frequency of the class preceding the median class) = 45
• $f$ (Frequency of the median class) = 33
• $h$ (Class size) = 5

The formula is: $\text{Median} = l + \left(\frac{\frac{N}{2} - cf}{f}\right) \times h$

$\text{Median} = 35 + \left(\frac{50 - 45}{33}\right) \times 5$

$\text{Median} = 35 + \left(\frac{5}{33}\right) \times 5$

$\text{Median} = 35 + \frac{25}{33}$

$\text{Median} \approx 35 + 0.7576$

$\text{Median} \approx 35.76$

The median age of the policy holders is approximately 35.76 years.

The given frequency distribution is not continuous. To find the median, we first need to make the class intervals continuous. We do this by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class.

Step 1: Create a Continuous Frequency Distribution Table

We adjust the class intervals and calculate the cumulative frequency ($cf$).

Step 2: Finding the Median Class

The total number of leaves is $N = 40$.

The position of the median is $\frac{N}{2} = \frac{40}{2} = 20$.

The cumulative frequency just greater than 20 is 29, which corresponds to the class interval 144.5 - 153.5. This is our median class.

Step 3: Calculating the Median Length

We use the median formula with the following values:

• $l$ (Lower limit of median class) = 144.5
• $\frac{N}{2}$ = 20
• $cf$ (Cumulative frequency of the class preceding the median class) = 17
• $f$ (Frequency of the median class) = 12
• $h$ (Class size) = $153.5 - 144.5 = 9$

The formula is: $\text{Median} = l + \left(\frac{\frac{N}{2} - cf}{f}\right) \times h$

$\text{Median} = 144.5 + \left(\frac{20 - 17}{12}\right) \times 9$

$\text{Median} = 144.5 + \left(\frac{3}{12}\right) \times 9$

$\text{Median} = 144.5 + \left(\frac{1}{4}\right) \times 9$

$\text{Median} = 144.5 + \frac{9}{4}$

$\text{Median} = 144.5 + 2.25$

$\text{Median} = 146.75$

The median length of the leaves is 146.75 mm.

To find the median life time of a lamp, we will calculate the cumulative frequency and then apply the median formula for grouped data.

Step 1: Construct the Cumulative Frequency Table

Step 2: Finding the Median Class

The total number of lamps is $N = 400$.

The position of the median is $\frac{N}{2} = \frac{400}{2} = 200$.

The cumulative frequency just greater than 200 is 216, which corresponds to the class interval 3000 - 3500. This is the median class.

Step 3: Calculating the Median Life Time

We use the median formula with the following values:

• $l$ (Lower limit of median class) = 3000
• $\frac{N}{2}$ = 200
• $cf$ (Cumulative frequency of the class preceding the median class) = 130
• $f$ (Frequency of the median class) = 86
• $h$ (Class size) = $3500 - 3000 = 500$

The formula is: $\text{Median} = l + \left(\frac{\frac{N}{2} - cf}{f}\right) \times h$

$\text{Median} = 3000 + \left(\frac{200 - 130}{86}\right) \times 500$

$\text{Median} = 3000 + \left(\frac{70}{86}\right) \times 500$

$\text{Median} = 3000 + \frac{35000}{86}$

$\text{Median} \approx 3000 + 406.977$

$\text{Median} \approx 3406.977$

The median life time of a lamp is approximately 3406.98 hours.

We need to calculate the median, mean, and mode for the given distribution of the number of letters in surnames.

Calculation of Median

First, we create the cumulative frequency ($cf$) table.

Total surnames $N = 100$, so $\frac{N}{2} = 50$. The cumulative frequency just greater than 50 is 76, so the median class is 7 - 10.

• $l = 7$, $\frac{N}{2} = 50$, $cf = 36$, $f = 40$, $h = 3$

$\text{Median} = 7 + \left(\frac{50 - 36}{40}\right) \times 3 = 7 + \frac{14 \times 3}{40} = 7 + \frac{42}{40} = 7 + 1.05 = 8.05$

Calculation of Mean

We use the Step-Deviation Method. Let the assumed mean ($a$) be 8.5. The class size ($h$) is 3.

$\overline{x} = a + h \left(\frac{\sum f_i u_i}{\sum f_i}\right) = 8.5 + 3 \left(\frac{-6}{100}\right) = 8.5 - 0.18 = 8.32$

Calculation of Mode

The highest frequency is 40, so the modal class is 7 - 10.

• $l = 7$, $f_1 = 40$, $f_0 = 30$, $f_2 = 16$, $h = 3$

$\text{Mode} = 7 + \left(\frac{40 - 30}{2(40) - 30 - 16}\right) \times 3 $$ = 7 + \left(\frac{10}{80 - 46}\right) \times 3 $$ = 7 + \frac{30}{34} \approx 7 + 0.88 = 7.88$

The calculated values are:

• Median number of letters = 8.05
• Mean number of letters = 8.32
• Modal size of surnames = 7.88

To find the median weight of the students, we first need to prepare a cumulative frequency table.

Step 1: Construct the Cumulative Frequency Table

Step 2: Finding the Median Class

The total number of students is $N = 30$.

The position of the median is $\frac{N}{2} = \frac{30}{2} = 15$.

The cumulative frequency just greater than 15 is 19, which corresponds to the class interval 55 - 60. This is the median class.

Step 3: Calculating the Median Weight

We use the median formula with the following values:

• $l$ (Lower limit of median class) = 55
• $\frac{N}{2}$ = 15
• $cf$ (Cumulative frequency of the class preceding the median class) = 13
• $f$ (Frequency of the median class) = 6
• $h$ (Class size) = $60 - 55 = 5$

The formula is: $\text{Median} = l + \left(\frac{\frac{N}{2} - cf}{f}\right) \times h$

$\text{Median} = 55 + \left(\frac{15 - 13}{6}\right) \times 5$

$\text{Median} = 55 + \left(\frac{2}{6}\right) \times 5$

$\text{Median} = 55 + \frac{10}{6}$

$\text{Median} = 55 + \frac{5}{3}$

$\text{Median} \approx 55 + 1.667$

$\text{Median} \approx 56.67$

The median weight of the students is approximately 56.67 kg.