Chapter 14 Statistics

Here are five examples of data that can be collected from our day-to-day life:

1. The number of hours you sleep each night over a week.

2. The amount of money you spend each day for a month.

3. The temperature recorded at a specific time each day for a week.

4. The number of students present in your class each day.

5. The time taken to travel from home to school each morning.

Based on the definition of primary and secondary data:

Primary Data: Data collected by the investigator himself for a specific purpose.

Secondary Data: Data which has been collected by someone else for some other purpose.

All five examples of data given in Q.1 are collected by the person observing or experiencing them directly for their own use.

Therefore, all the data collected in Q.1 are examples of Primary Data.

Given

The marks obtained by 10 students in a mathematics test are:

55, 36, 95, 73, 60, 42, 25, 78, 75, 62

This form of data is known as raw data.

Solution

To analyze the given data, we first arrange the marks in ascending order. This helps in understanding the distribution and calculating various statistical measures.

The marks in ascending order are:

25, 36, 42, 55, 60, 62, 73, 75, 78, 95

Now, let's calculate some common statistical measures for this data.

1. Range of the Data

The range is the difference between the highest and the lowest values in the data set.

Highest marks = 95

Lowest marks = 25

Range = Highest marks - Lowest marks

Range = $95 - 25 = 70$

So, the range of the marks is 70.

2. Mean of the Data

The mean (or average) is the sum of all the observations divided by the total number of observations.

The formula for the mean is: Mean ($\bar{x}$) = $\frac{\text{Sum of all observations}}{\text{Number of observations}}$

Sum of marks = $25 + 36 + 42 + 55 + 60 + 62 + 73 + 75 + 78 + 95 \ $$ = 591$

Number of students ($n$) = 10

Mean = $\frac{591}{10} = 59.1$

So, the mean marks of the students is 59.1.

3. Median of the Data

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

The arranged data is: 25, 36, 42, 55, 60, 62, 73, 75, 78, 95.

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

For an even number of observations, the median is the average of the two middle terms, which are the $(\frac{n}{2})^{th}$ term and the $(\frac{n}{2} + 1)^{th}$ term.

$(\frac{10}{2})^{th}$ term = $5^{th}$ term

$(\frac{10}{2} + 1)^{th}$ term = $6^{th}$ term

From the arranged data, the $5^{th}$ term is 60 and the $6^{th}$ term is 62.

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

Median = $\frac{60 + 62}{2} = \frac{122}{2} = 61$

So, the median of the marks is 61.

4. Mode of the Data

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

Looking at the data: 25, 36, 42, 55, 60, 62, 73, 75, 78, 95.

Each mark appears only once. Since no value is repeated, there is no mode for this data set.

The given data represents the marks obtained by 30 students.

To prepare an ungrouped frequency distribution table, we list each unique mark and count how many times it appears in the data using tally marks.

Here is the ungrouped frequency distribution table:

Given

The number of plants that survived in 100 schools are given as raw data. The total number of observations is 100.

To Do

Prepare a grouped frequency distribution table for the given data using tally marks.

Solution

To create a grouped frequency distribution table, we first need to determine the range of the data and decide on suitable class intervals.

Step 1: Find the range of the data.

First, let's identify the minimum and maximum values in the dataset.

By observing the data:

Maximum value = 98

Minimum value = 23

Range = Maximum value - Minimum value = $98 - 23 = 75$

Step 2: Decide on the class intervals.

Since the data ranges from 23 to 98, we can create class intervals of size 10 to organize the data effectively. We can start the first interval from 20 to cover all the values. The class intervals will be 20-29, 30-39, 40-49, and so on, up to 90-99.

Step 3: Tally the data.

We will now go through the data and place a tally mark ($|$) against the corresponding class interval for each value. A group of five tally marks is represented as $\bcancel{||||}$.

After going through all 100 data points, we count the tally marks for each class to get the frequency.

The resulting grouped frequency distribution table is as follows:

This table presents the given data in a grouped format, showing how many schools fall into each category of surviving plants.

The given table is a grouped frequency distribution table.

This table has class intervals like 31-35, 36-40, etc.

In these intervals, the upper limit of one class (e.g., 35) does not match the lower limit of the next class (e.g., 36).

This type of frequency distribution is called a discontinuous or inclusive grouped frequency distribution.

Discontinuous Grouped Frequency Distribution Table:

To prepare a continuous grouped frequency distribution table, the upper limit of each class must coincide with the lower limit of the succeeding class.

We find the difference between the upper limit of any class and the lower limit of the next class.

Difference = Lower limit of next class - Upper limit of current class

Using the first two classes (31-35 and 36-40):

Difference = $36 - 35 = 1$

We divide this difference by 2 to get the adjustment factor.

Adjustment factor = Difference / 2 = $1 / 2 = 0.5$

To make the classes continuous, we subtract this adjustment factor from the lower limit of each class and add it to the upper limit of each class.

• New lower limit = Original lower limit - 0.5
• New upper limit = Original upper limit + 0.5

Continuous Grouped Frequency Distribution Table:

Given:

The blood groups of 30 students of Class VIII are recorded as:

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

To Find:

• Frequency distribution table of the blood groups.
• The most common blood group.
• The rarest blood group.

Solution:

We will count the frequency of each blood group from the given data:

• Blood group A: 9 students
• Blood group B: 6 students
• Blood group O: 12 students
• Blood group AB: 3 students

Total number of students = $9 + 6 + 12 + 3 = 30$.

Here is the frequency distribution table for the given data:

From the frequency distribution table:

The highest frequency is 12, which corresponds to Blood Group O.

The lowest frequency is 3, which corresponds to Blood Group AB.

Therefore:

The most common blood group among these students is O.

The rarest blood group among these students is AB.

Given:

The distance (in km) of 40 engineers from their residence to their place of work.

Data: 5, 3, 10, 20, 25, 11, 13, 7, 12, 31, 19, 10, 12, 17, 18, 11, 32, 17, 16, 2, 7, 9, 7, 8, 3, 5, 12, 15, 18, 3, 12, 14, 2, 9, 6, 15, 15, 7, 6, 12.

To Construct:

A grouped frequency distribution table with class size 5, taking the first interval as 0-5 (5 not included).

To Find:

Main features observed from the table.

Solution:

We need to group the data into class intervals of size 5. The first interval is given as 0-5, where 5 is not included. This implies the class intervals are $0 \leq d < 5$, $5 \leq d < 10$, $10 \leq d < 15$, and so on, until the maximum value is covered. The maximum value in the data is 32, so we need intervals up to 35.

The class intervals will be: 0-5, 5-10, 10-15, 15-20, 20-25, 25-30, 30-35.

Now, we count the number of engineers falling into each class interval.

• 0-5: Distances $\geq 0$ and $< 5$. The values are 3, 3, 2, 3, 2. Frequency = 5.
• 5-10: Distances $\geq 5$ and $< 10$. The values are 5, 7, 10 (no, 10 is not included), 7, 8, 5, 7, 6, 9, 7, 6, 9. The values are 5, 7, 7, 8, 5, 7, 9, 6, 7, 6, 9. Frequency = 11.
• 10-15: Distances $\geq 10$ and $< 15$. The values are 10, 11, 13, 12, 10, 12, 11, 12, 12, 14, 12. Frequency = 11.
• 15-20: Distances $\geq 15$ and $< 20$. The values are 19, 17, 18, 17, 16, 15, 18, 15, 15. Frequency = 9.
• 20-25: Distances $\geq 20$ and $< 25$. The value is 20. Frequency = 1.
• 25-30: Distances $\geq 25$ and $< 30$. The value is 25. Frequency = 1.
• 30-35: Distances $\geq 30$ and $< 35$. The values are 31, 32. Frequency = 2.

Total frequency = $5 + 11 + 11 + 9 + 1 + 1 + 2 = 40$. This matches the total number of engineers.

Here is the grouped frequency distribution table:

Main features observed from the tabular representation:

1. The data is concentrated in the smaller distance intervals. A large number of engineers live within 15 km of their workplace ($5 + 11 + 11 = 27$ engineers).

2. Most engineers (11 in 5-10 km and 11 in 10-15 km, totaling 22) live between 5 km and 15 km (excluding 15 km) from their workplace.

3. Only a few engineers (1 in 20-25 km, 1 in 25-30 km, 2 in 30-35 km, totaling 4) live 20 km or more away from their workplace.

Given:

Relative humidity (in %) of a certain city for a month of 30 days.

Data: 98.1, 98.6, 99.2, 90.3, 86.5, 95.3, 92.9, 96.3, 94.2, 95.1, 89.2, 92.3, 97.1, 93.5, 92.7, 95.1, 97.2, 93.3, 95.2, 97.3, 96.2, 92.1, 84.9, 90.2, 95.7, 98.3, 97.3, 96.1, 92.1, 89.

(i) Construct a grouped frequency distribution table with classes 84 - 86, 86 - 88, etc.

We need to group the data into class intervals of size 2, starting from 84-86. The class intervals are of the form [Lower limit, Upper limit), meaning the lower limit is included, and the upper limit is not included.

The classes will be 84-86, 86-88, 88-90, 90-92, 92-94, 94-96, 96-98, 98-100.

Now, we count the frequency for each class interval:

• 84 - 86: Values $\geq 84$ and $< 86$. Value: 84.9 (1)
• 86 - 88: Values $\geq 86$ and $< 88$. Value: 86.5 (1)
• 88 - 90: Values $\geq 88$ and $< 90$. Values: 89.2, 89 (2)
• 90 - 92: Values $\geq 90$ and $< 92$. Values: 90.3, 90.2 (2)
• 92 - 94: Values $\geq 92$ and $< 94$. Values: 92.9, 92.3, 93.5, 92.7, 93.3, 92.1, 92.1 (7)
• 94 - 96: Values $\geq 94$ and $< 96$. Values: 95.3, 94.2, 95.1, 95.1, 95.2, 95.7 (6)
• 96 - 98: Values $\geq 96$ and $< 98$. Values: 96.3, 97.1, 97.2, 97.3, 96.2, 97.3, 96.1 (7)
• 98 - 100: Values $\geq 98$ and $< 100$. Values: 98.1, 98.6, 99.2, 98.3 (4)

Total frequency = $1 + 1 + 2 + 2 + 7 + 6 + 7 + 4 = 30$.

Here is the grouped frequency distribution table:

(ii) Which month or season do you think this data is about?

The relative humidity values are very high (mostly above 90%). This level of humidity is typically observed during the rainy season (monsoon).

(iii) What is the range of this data?

The range of the data is the difference between the maximum and minimum values in the dataset.

Maximum value = 99.2

Minimum value = 84.9

Range = Maximum value - Minimum value

Range = $99.2 - 84.9$

Range = $14.3$

Given:

The heights (in cm) of 50 students.

Data: 161, 150, 154, 165, 168, 161, 154, 162, 150, 151, 162, 164, 171, 165, 158, 154, 156, 172, 160, 170, 153, 159, 161, 170, 162, 165, 166, 168, 165, 164, 154, 152, 153, 156, 158, 162, 160, 161, 173, 166, 161, 159, 162, 167, 168, 159, 158, 153, 154, 159.

(i) Represent the data by a grouped frequency distribution table.

We need to construct a grouped frequency distribution table with class intervals 160 - 165, 165 - 170, etc. Since heights are measured to the nearest centimetre and the class boundaries meet (165, 165), these are continuous class intervals of the form [Lower limit, Upper limit), where the lower limit is included, and the upper limit is not included.

First, find the minimum and maximum values in the data.

Minimum height = 150 cm

Maximum height = 173 cm

With a class size of 5 and intervals starting around the minimum value to reach the maximum, the required class intervals are:

[150, 155), [155, 160), [160, 165), [165, 170), [170, 175)

Now, we count the frequency of heights falling in each interval:

• 150 - 155: Heights $\geq 150$ and $< 155$. Values are 150, 154, 150, 151, 154, 153, 154, 152, 153, 154, 153, 154. Count = 12.
• 155 - 160: Heights $\geq 155$ and $< 160$. Values are 158, 156, 158, 156, 158, 159, 159, 158, 159, 159. Count = 10.
• 160 - 165: Heights $\geq 160$ and $< 165$. Values are 161, 161, 162, 162, 164, 160, 162, 164, 160, 161, 162, 161, 162, 161. Count = 14.
• 165 - 170: Heights $\geq 165$ and $< 170$. Values are 165, 168, 165, 165, 166, 168, 165, 166, 167, 168. Count = 10.
• 170 - 175: Heights $\geq 170$ and $< 175$. Values are 171, 172, 170, 170, 173. Count = 5.

Total frequency = $12 + 10 + 14 + 10 + 5 = 51$. Oh, let's recount from the list. 150: 2, 151: 1, 152: 1, 153: 3, 154: 5. Total [150, 155) = 12. 156: 2, 158: 3, 159: 4. Total [155, 160) = 9. 160: 2, 161: 5, 162: 5, 164: 2. Total [160, 165) = 14. 165: 4, 166: 2, 167: 1, 168: 3. Total [165, 170) = 10. 170: 2, 171: 1, 172: 1, 173: 1. Total [170, 175) = 5. Total sum = $12 + 9 + 14 + 10 + 5 = 50$. The counts are correct now.

Here is the grouped frequency distribution table:

(ii) What can you conclude about their heights from the table?

From the grouped frequency distribution table, we can observe the following features about the students' heights:

1. The height of most students lies in the class interval 160 - 165 cm, as it has the maximum frequency of 14.

2. A large number of students have heights between 160 cm and 170 cm ($14 + 10 = 24$ students).

3. The number of students with heights below 160 cm is $12 + 9 = 21$.

4. The number of students with heights 165 cm or more is $10 + 5 = 15$.

5. The heights are distributed across the range from 150 cm to 175 cm.

(i) Grouped Frequency Distribution Table

To create the grouped frequency distribution table for the given data, we will use the specified class intervals: 0.00 - 0.04, 0.04 - 0.08, and so on. We assume that the upper limit of each class is not included in the interval (e.g., 0.04 is included in 0.04 - 0.08, not in 0.00 - 0.04).

The concentration of sulphur dioxide (in ppm) for 30 days is tallied as follows:

(ii) Number of days with SO₂ concentration more than 0.11 ppm

To find the number of days the concentration of sulphur dioxide was more than 0.11 ppm, we need to sum the frequencies of the class intervals that cover values greater than 0.11.

These intervals are:

• 0.12 - 0.16
• 0.16 - 0.20
• 0.20 - 0.24

From the frequency distribution table created in part (i), we have:

Frequency for the interval 0.12 - 0.16 = 2 days

Frequency for the interval 0.16 - 0.20 = 4 days

Frequency for the interval 0.20 - 0.24 = 2 days

Total number of days with concentration more than 0.11 ppm = Sum of frequencies of these intervals

Total days = $2 + 4 + 2 = 8$

Therefore, the concentration of sulphur dioxide was more than 0.11 parts per million for 8 days.

To prepare the frequency distribution table, we need to count how many times each possible outcome (number of heads) occurred. The possible outcomes when tossing three coins are 0, 1, 2, or 3 heads.

By counting the occurrences of each number from the given data:

• The number 0 appears 6 times.
• The number 1 appears 10 times.
• The number 2 appears 9 times.
• The number 3 appears 5 times.

We can now construct the frequency distribution table using this information.

Given:

The value of $\pi$ upto 50 decimal places is 3.14159265358979323846264338327950288419716939937510.

We are concerned with the 50 digits after the decimal point:

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, 4, 1, 9, 7, 1, 6, 9, 3, 9, 9, 3, 7, 5, 1, 0.

(i) To Make:

A frequency distribution of the digits from 0 to 9 after the decimal point.

Solution (i):

We count the occurrence of each digit (0, 1, 2, 3, 4, 5, 6, 7, 8, 9) in the 50 decimal places.

• Digit 0: appears 2 times.
• Digit 1: appears 5 times.
• Digit 2: appears 5 times.
• Digit 3: appears 8 times.
• Digit 4: appears 4 times.
• Digit 5: appears 5 times.
• Digit 6: appears 4 times.
• Digit 7: appears 4 times.
• Digit 8: appears 5 times.
• Digit 9: appears 8 times.

Total frequency = $2 + 5 + 5 + 8 + 4 + 5 + 4 + 4 + 5 + 8 = 50$.

Here is the frequency distribution table:

(ii) To Find:

The most and the least frequently occurring digits.

Solution (ii):

From the frequency distribution table:

The highest frequency is 8, which corresponds to the digits 3 and 9.

The lowest frequency is 2, which corresponds to the digit 0.

Therefore:

The most frequently occurring digits are 3 and 9.

The least frequently occurring digit is 0.

(i) Grouped Frequency Distribution Table

To create the grouped frequency distribution table, we are given a class width of 5 and one of the class intervals as 5 - 10. Based on the range of the data (from 1 to 17), the required class intervals are 0 - 5, 5 - 10, 10 - 15, and 15 - 20. We will assume that the lower limit of each class is included, and the upper limit is excluded (e.g., 5 is included in the 5-10 interval, but 10 is included in the 10-15 interval).

The frequency distribution table for the given data is as follows:

(ii) Number of children who watched television for 15 or more hours a week

To find the number of children who watched TV for 15 or more hours, we look at the frequency of the class intervals that include 15 and above.

From the table, the class interval for "15 or more hours" is 15 - 20.

The number of children in this class interval is 2.

Therefore, 2 children watched television for 15 or more hours a week.

To construct the grouped frequency distribution table, we use the given class intervals with a size of 0.5, starting from 2.0 - 2.5. We tally the given data of the lives of 40 batteries into these classes. We assume the standard convention where the lower limit of a class is included, and the upper limit is excluded (e.g., a battery life of 2.5 years would be included in the 2.5 - 3.0 interval).

The grouped frequency distribution table is as follows:

(i) How many students were born in the month of November?

To answer this question, we observe the bar graph and locate the bar corresponding to the month of November on the horizontal axis (Months of Birth).

The height of the bar for November corresponds to the value on the vertical axis (Number of Students). Following the top of the bar for November to the left, we can see that it aligns with the number 4.

Therefore, 4 students were born in the month of November.

(ii) In which month were the maximum number of students born?

To find the month with the maximum number of births, we need to identify the tallest bar in the graph, as the height of the bar represents the number of students.

By observing the graph, the tallest bar corresponds to the month of August. The height of this bar reaches 6, which is the highest value among all the months.

Therefore, the maximum number of students were born in the month of August.

Given

Monthly expenditures of a family under various heads:

To Draw

A bar graph for the given data.

Steps for Construction

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

2. On the horizontal axis, represent the 'Heads' (categories of expenditure). Leave uniform space between the bars.

3. On the vertical axis, represent the 'Expenditure (in thousand $\textsf{₹}$)'. Since the maximum expenditure is 5 thousand rupees, we can choose a scale where 1 unit on the y-axis represents 1 thousand rupees. Mark the y-axis from 0 up to a value slightly greater than 5 (e.g., 6).

4. For each 'Head', draw a bar of uniform width. The height of each bar should correspond to the expenditure value for that head on the vertical axis.

• For 'Grocery', draw a bar of height 4 units.
• For 'Rent', draw a bar of height 5 units.
• For 'Education of children', draw a bar of height 5 units.
• For 'Medicine', draw a bar of height 2 units.
• For 'Fuel', draw a bar of height 2 units.
• For 'Entertainment', draw a bar of height 1 unit.
• For 'Miscellaneous', draw a bar of height 1 unit.

5. Label the axes: 'Heads of Expenditure' on the x-axis and 'Expenditure (in thousand $\textsf{₹}$)' on the y-axis.

6. Give a suitable title to the bar graph, e.g., "Monthly Expenditure of a Family".

The resulting bar graph is shown below:

Given

A grouped frequency distribution table showing the marks obtained by 90 students in a mathematics test, with unequal class widths.

To Draw

A histogram for the given data.

Solution

Since the class widths in the given frequency distribution table are unequal, the heights of the rectangles in the histogram must be adjusted to be proportional to the frequency.

The formula for the adjusted height of the rectangle is:

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

First, we find the width of each class interval:

The class widths are 20, 10, 10, 10, 10, 10, and 30. The minimum class width is 10.

Now, we calculate the adjusted frequency for each class:

Steps for Drawing the Histogram

1. Represent the 'Marks' on the x-axis and the 'Adjusted Frequency' on the y-axis.

2. Draw adjacent rectangles for each class interval. The width of each rectangle is its class width, and the height is its adjusted frequency.

The resulting histogram is shown below:

Given

A frequency distribution table showing the marks obtained by 51 students.

To Draw

A frequency polygon for the given data.

Solution

To draw a frequency polygon, we first need to find the mid-point (class mark) of each class interval. The mid-point is calculated as $\frac{\text{Lower Limit} + \text{Upper Limit}}{2}$.

We then plot the points (Class Mark, Frequency) and connect them.

Let's prepare the table with class marks:

To close the polygon, we consider two imaginary classes with zero frequency: one before the first class (-10 to 0) and one after the last class (100 to 110). Their class marks are -5 and 105, respectively.

The points to be plotted are: (-5, 0), (5, 5), (15, 10), (25, 4), (35, 6), (45, 7), (55, 3), (65, 2), (75, 2), (85, 3), (95, 9), and (105, 0).

Steps for Drawing the Frequency Polygon

1. Represent the 'Marks' (Class Marks) on the x-axis and the 'Number of Students' (Frequency) on the y-axis.

2. Plot the points calculated above.

3. Join these points with straight line segments to form the frequency polygon.

The resulting frequency polygon is shown below:

Given

A frequency distribution table showing the cost of living index and the number of weeks.

To Draw

A frequency polygon for the given data.

Solution

To draw a frequency polygon, we first find the mid-point (class mark) of each class interval using the formula: Mid-point = $\frac{\text{Lower Limit} + \text{Upper Limit}}{2}$.

Let's calculate the class marks and prepare the table for plotting:

To complete the polygon, we consider imaginary classes with zero frequency before the first class (130-140) and after the last class (200-210). Their class marks are 135 and 205, respectively.

The points to be plotted on the graph are (Class Mark, Frequency): (135, 0), (145, 5), (155, 10), (165, 20), (175, 9), (185, 6), (195, 2), and (205, 0).

Steps for Drawing the Frequency Polygon

1. Represent the 'Cost of living index' (Class Marks) on the x-axis and the 'Number of weeks' (Frequency) on the y-axis.

2. Plot the points listed above.

3. Join these points with straight line segments to get the required frequency polygon.

The resulting frequency polygon is shown below:

Given

The percentage of female fatality rates for various causes among women aged 15-44 worldwide.

(i) Represent the information given above graphically.

A bar graph is a suitable way to represent this data. We represent the 'Causes' on the horizontal axis and the 'Female fatality rate (%)' on the vertical axis.

(ii) Which condition is the major cause of women’s ill health and death worldwide?

From the given table and the bar graph, the highest female fatality rate is 31.8%, which corresponds to 'Reproductive health conditions'.

Therefore, the major cause of women’s ill health and death worldwide (in the age group 15-44) is Reproductive health conditions.

(iii) Try to find out, with the help of your teacher, any two factors which play a major role in the cause in (ii) above being the major cause.

Two factors that play a major role in 'Reproductive health conditions' being the major cause of women's ill health and death worldwide in the age group 15-44 could be:

1. Lack of access to adequate healthcare facilities and services: Many women, especially in developing regions, lack access to timely and quality healthcare, including antenatal care, safe delivery services, and postnatal care. This leads to preventable complications during pregnancy, childbirth, and postpartum periods becoming fatal.

2. Lack of awareness and education on reproductive health: Limited access to information about sexual and reproductive health, family planning, and hygiene practices contributes to a higher incidence of reproductive tract infections, complications from unsafe abortions, and difficulties in managing pregnancy-related issues effectively.

Given

Number of girls per thousand boys in different sections of Indian society.

(i) Represent the information above by a bar graph.

We use a bar graph to display the data, with 'Section' on the horizontal axis and 'Number of girls per thousand boys' on the vertical axis.

(ii) In the classroom discuss what conclusions can be arrived at from the graph.

Based on the bar graph, several conclusions can be discussed:

1. Sex Ratio Imbalance: In all sections represented, the number of girls per thousand boys is less than 1000. This indicates a skewed sex ratio across the board, with fewer females than males.

2. Comparison Across Sections:

• The sex ratio is highest among Scheduled Tribes (ST) (970), suggesting a relatively better status for girls in this section compared to others.
• The sex ratio is lowest in Urban areas (910). This might be counter-intuitive as one might expect better conditions in urban settings, leading to discussions about the prevalence of practices like sex-selective abortions.
• The sex ratio in Backward districts (950) is higher than in Non-backward districts (920). Similarly, Rural areas (930) have a better sex ratio than Urban areas (910).

3. Areas of Concern: The low sex ratio in Urban areas and among Non SC/ST populations highlights these as key areas for intervention and awareness campaigns to improve the value and survival of the girl child.

Given

The number of seats won by different political parties in a state assembly election.

(i) Draw a bar graph to represent the polling results.

A bar graph is used to represent the polling results, with the political parties on the x-axis and the number of seats won on the y-axis.

(ii) Which political party won the maximum number of seats?

By observing the data in the table or the heights of the bars in the graph, we can find the party with the maximum number of seats.

The highest number of seats won is 75.

Therefore, Political Party A won the maximum number of seats.

Given

The length of 40 leaves (in mm) and their frequency distribution.

(i) Draw a histogram to represent the given data.

The class intervals are discontinuous. To draw a histogram, we must first make them continuous. The gap between consecutive classes is $127 - 126 = 1$. We add and subtract half of this gap (0.5) to the upper and lower limits, respectively.

The new continuous class intervals are:

Now we can draw the histogram using these continuous intervals on the x-axis and the frequencies on the y-axis.

(ii) Is there any other suitable graphical representation for the same data?

Yes, a Frequency Polygon is another suitable graphical representation for this data. It can be constructed by joining the mid-points of the tops of the bars of the histogram.

(iii) Is it correct to conclude that the maximum number of leaves are 153 mm long? Why?

No, this conclusion is incorrect.

Reason: The data is presented in groups. The class interval 145 - 153 (or 144.5 - 153.5) has the highest frequency of 12. This only means that the 12 leaves with the most common lengths have lengths that fall *somewhere* within this range. We do not know the exact individual lengths of these 12 leaves from the given table. Therefore, we cannot conclude that their length is exactly 153 mm.

Given

The life times (in hours) of 400 neon lamps.

(i) Represent the given information with the help of a histogram.

The class intervals are continuous and have a uniform width. We can draw the histogram with 'Life time' on the x-axis and 'Number of lamps' on the y-axis.

(ii) How many lamps have a life time of more than 700 hours?

To find the number of lamps with a lifetime of more than 700 hours, we need to sum the frequencies of the class intervals from 700-800 onwards.

Number of lamps = (Frequency of 700-800) + (Frequency of 800-900) + (Frequency of 900-1000)

Number of lamps = $74 + 62 + 48 = 184$

Therefore, 184 lamps have a life time of more than 700 hours.

Given

Frequency distribution of marks for two sections (A and B).

Solution

To draw the frequency polygons, we first need to find the mid-points (class marks) of the class intervals.

We plot the class marks on the x-axis and the frequencies on the y-axis for both sections and join the points with line segments. To close the polygons, we use class marks of preceding (-5) and succeeding (55) intervals with zero frequency.

Comparison of Performance

By observing the two frequency polygons on the graph:

1. The peak of Section B's polygon is higher in the 10-20 marks range, indicating a large number of students scored in this lower range.

2. The peak of Section A's polygon is in the 20-30 marks range. Overall, the polygon for Section A is shifted more towards the right (higher marks) compared to Section B.

3. Section A has a significantly higher number of students scoring more than 20 marks compared to Section B.

Conclusion: The performance of Section A is better than that of Section B, as more students in Section A have scored higher marks.

Given

Runs scored by two teams A and B in intervals of 6 balls.

Solution

First, we make the class intervals continuous by subtracting 0.5 from lower limits and adding 0.5 to upper limits. Then, we find the mid-point (class mark) of each continuous interval.

We plot these class marks on the x-axis and the runs on the y-axis for both teams on the same graph. To close the polygons, we use preceding (-2.5) and succeeding (63.5) class marks with zero frequency.

Given

Frequency distribution of the number of children in various age groups.

Solution

The class intervals have unequal widths. To draw an accurate histogram, we must adjust the height of the rectangles. The height (or length) of each rectangle is calculated as:

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

The minimum class width is 1 (for 1-2 and 2-3).

We now draw the histogram using the class intervals for the width of the bars and the adjusted frequencies for the height of the bars.

Given

Frequency distribution of the number of letters in 100 surnames.

(i) Draw a histogram to depict the given information.

The class intervals have unequal widths, so we must calculate the adjusted frequencies to draw the histogram correctly. The minimum class width is 2 (for 4-6 and 6-8).

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

We draw the histogram using the class intervals for the width and the adjusted frequencies for the height.

(ii) Write the class interval in which the maximum number of surnames lie.

To answer this, we look at the original frequency column in the given table, not the adjusted frequency.

The highest frequency is 44.

Therefore, the class interval in which the maximum number of surnames lie is 6 - 8.

Given:

The time (in hours) spent by 5 people in a week doing social work: 10, 7, 13, 20, 15.

Number of observations, $n = 5$.

To Find:

The mean (average) time spent per week.

Solution:

The mean of ungrouped data is calculated by the formula:

$ \text{Mean} (\overline{x}) = \frac{\text{Sum of all observations}}{\text{Number of observations}} $

Sum of all observations = $10 + 7 + 13 + 20 + 15$

Sum of all observations = $65$

Number of observations, $n = 5$

Mean $(\overline{x}) = \frac{65}{5}$

Mean $(\overline{x}) = 13$

Alternatively, we can show the addition:

Sum of observations = $10 + 7 + 13 + 20 + 15$

Sum of observations = 65

Mean $(\overline{x}) = \frac{65}{5} = 13$

The mean time in a week devoted by them for social work is 13 hours.

Given

The marks obtained by 30 students of Class IX, as presented in the frequency distribution table from Example 2 of the textbook.

To Find

The mean of the marks obtained by the 30 students.

Solution

To find the mean of the given data, we use the formula for the mean of data presented with frequencies:

We first need to calculate the product of each mark ($x_i$) and its corresponding frequency ($f_i$), which is $f_i x_i$. Then, we find the sum of these products ($\sum\limits f_i x_i$).

Let's create a table to organize the calculations:

From the table, we have:

Total number of students, $\sum\limits f_i = 30$

Sum of the products, $\sum\limits f_i x_i = 1779$

Now, we can calculate the mean:

Mean ($\overline{x}$) = $\frac{\sum\limits f_i x_i}{\sum\limits f_i} = \frac{1779}{30}$

Mean ($\overline{x}$) = 59.3

Thus, the mean of the marks obtained by the 30 students is 59.3.

Given

The heights (in cm) of 9 students of a class are:

155, 160, 145, 149, 150, 147, 152, 144, 148.

To Find

The median of the given data.

Solution

First, we arrange the given data in ascending order to find the median.

Arranging the heights in ascending order:

144, 145, 147, 148, 149, 150, 152, 155, 160

The number of observations is $n = 9$, which is an odd number.

When the number of observations is odd, the median is the middlemost value, which is the $(\frac{n+1}{2})^{th}$ observation.

Median position = $(\frac{9+1}{2})^{th}$ observation = $(\frac{10}{2})^{th}$ observation = $5^{th}$ observation.

Now, we find the 5th value in the arranged data:

144, 145, 147, 148, 149, 150, 152, 155, 160

The 5th observation is 149.

Therefore, the median height of the students is 149 cm.

Given

The points scored by a Kabaddi team in a series of matches:

17, 2, 7, 27, 15, 5, 14, 8, 10, 24, 48, 10, 8, 7, 18, 28.

To Find

The median of the points scored.

Solution

To find the median, we first arrange the data in ascending order.

Arranged data:

2, 5, 7, 7, 8, 8, 10, 10, 14, 15, 17, 18, 24, 27, 28, 48

The number of observations is $n = 16$, which is an even number.

When the number of observations is even, the median is the average of the two middle observations, which are the $(\frac{n}{2})^{th}$ and $(\frac{n}{2} + 1)^{th}$ observations.

$(\frac{16}{2})^{th}$ observation = $8^{th}$ observation.

$(\frac{16}{2} + 1)^{th}$ observation = $9^{th}$ observation.

From the arranged data:

The 8th observation is 10.

The 9th observation is 14.

Median = $\frac{8^{th} \text{ observation} + 9^{th} \text{ observation}}{2}$

Median = $\frac{10 + 14}{2} = \frac{24}{2} = 12$

The median of the points scored by the team is 12.

Given

The marks obtained by 20 students (out of 10):

4, 6, 5, 9, 3, 2, 7, 7, 6, 5, 4, 9, 10, 10, 3, 4, 7, 6, 9, 9.

To Find

The mode of the marks.

Solution

The mode is the value that appears most frequently in a data set. To find the mode, let's create a frequency table for the marks.

By observing the frequency table, we can see that the mark '9' has the highest frequency of 4.

Therefore, the mode of the marks is 9.

Given

Salaries of 5 employees in a factory unit:

• 4 labourers at $\textsf{₹}$ 5,000 per month each.
• 1 supervisor at $\textsf{₹}$ 15,000 per month.

The list of salaries is: $\textsf{₹}$ 5,000, $\textsf{₹}$ 5,000, $\textsf{₹}$ 5,000, $\textsf{₹}$ 5,000, $\textsf{₹}$ 15,000.

To Calculate

The mean, median, and mode of the salaries.

Solution

1. Calculation of Mean

Mean = $\frac{\text{Sum of all salaries}}{\text{Number of employees}}$

Sum of salaries = $(4 \times 5000) + 15000 = 20000 + 15000 = \textsf{₹}\ 35,000$

Number of employees = 5

Mean = $\frac{35000}{5} = \textsf{₹}\ 7,000$

The mean salary is $\textsf{₹}$ 7,000.

2. Calculation of Median

First, arrange the salaries in ascending order:

5000, 5000, 5000, 5000, 15000

The number of observations is $n = 5$ (odd). The median is the middle value, i.e., the $(\frac{n+1}{2})^{th}$ term.

Median position = $(\frac{5+1}{2}) = 3^{rd}$ term.

The 3rd term in the ordered list is 5000.

The median salary is $\textsf{₹}$ 5,000.

3. Calculation of Mode

The mode is the most frequently occurring value in the data.

In the list of salaries, the value $\textsf{₹}$ 5,000 appears 4 times, while $\textsf{₹}$ 15,000 appears only once.

The mode salary is $\textsf{₹}$ 5,000.

In this case, the mean ($\textsf{₹}$ 7,000) is influenced by the single high salary of the supervisor. The median and mode ($\textsf{₹}$ 5,000) better represent the typical salary of an employee in this unit.

Given

The number of goals scored by a team in 10 matches are:

2, 3, 4, 5, 0, 1, 3, 3, 4, 3.

Total number of matches (observations), $n = 10$.

To Find

The mean, median, and mode of these scores.

Solution

1. Mean

The mean is calculated using the formula:

Mean = $\frac{\text{Sum of all scores}}{\text{Total number of matches}}$

Sum of scores = $2 + 3 + 4 + 5 + 0 + 1 + 3 + 3 + 4 + 3 = 28$

Mean = $\frac{28}{10} = 2.8$

Therefore, the mean score is 2.8.

2. Median

To find the median, we first arrange the scores in ascending order:

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

Wait, there are only 9 scores here. Let's re-check the original data: 2, 3, 4, 5, 0, 1, 3, 3, 4, 3. There are four 3s. Let's reorder.

Ascending order: 0, 1, 2, 3, 3, 3, 3, 4, 4, 5

The number of observations is $n = 10$, which is an even number. The median is the average of the two middle values, which are the $(\frac{n}{2})^{th}$ and $(\frac{n}{2}+1)^{th}$ terms.

The 5th term is 3.

The 6th term is 3.

Median = $\frac{5^{th} \text{ term} + 6^{th} \text{ term}}{2} = \frac{3 + 3}{2} = \frac{6}{2} = 3$

Therefore, the median score is 3.

3. Mode

The mode is the score that appears most frequently.

From the arranged data (0, 1, 2, 3, 3, 3, 3, 4, 4, 5), we can see that the score '3' occurs 4 times, which is more than any other score.

Therefore, the mode of the scores is 3.

Given

Marks obtained by 15 students:

41, 39, 48, 52, 46, 62, 54, 40, 96, 52, 98, 40, 42, 52, 60.

Total number of students (observations), $n = 15$.

To Find

The mean, median, and mode of this data.

Solution

1. Mean

Mean = $\frac{\text{Sum of all marks}}{\text{Total number of students}}$

Sum of marks = $41 + 39 + 48 + 52 + 46 + 62 + 54 + 40 + 96 + 52 \ $$ + 98 + 40 + 42 + 52 + 60 = 822$

Mean = $\frac{822}{15} = 54.8$

Therefore, the mean mark is 54.8.

2. Median

First, we arrange the marks in ascending order:

39, 40, 40, 41, 42, 46, 48, 52, 52, 52, 54, 60, 62, 96, 98

The number of observations is $n = 15$, which is an odd number. The median is the middle value, i.e., the $(\frac{n+1}{2})^{th}$ term.

Median position = $(\frac{15+1}{2}) = 8^{th}$ term.

The 8th term in the ordered list is 52.

Therefore, the median mark is 52.

3. Mode

The mode is the mark that appears most frequently.

From the arranged data, we can count the frequency of each mark. The mark '52' appears 3 times, which is more than any other mark.

Therefore, the mode of the marks is 52.

Given

The observations arranged in ascending order are:

29, 32, 48, 50, $x$, $x + 2$, 72, 78, 84, 95.

The total number of observations, $n = 10$.

The median of the data is 63.

To Find

The value of $x$.

Solution

Since the number of observations $n = 10$ is even, the median is the average of the two middle terms: the $(\frac{n}{2})^{th}$ term and the $(\frac{n}{2}+1)^{th}$ term.

5th term = $x$

6th term = $x + 2$

The formula for the median is:

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

Substituting the given values:

$63 = \frac{x + (x + 2)}{2}$

$63 = \frac{2x + 2}{2}$

$63 = x + 1$

$x = 63 - 1$

$x = 62$

Therefore, the value of $x$ is 62.

Given

The data set is: 14, 25, 14, 28, 18, 17, 18, 14, 23, 22, 14, 18.

To Find

The mode of the data.

Solution

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

Let's count the frequency of each number in the given data:

• 14 appears 4 times.
• 25 appears 1 time.
• 28 appears 1 time.
• 18 appears 3 times.
• 17 appears 1 time.
• 23 appears 1 time.
• 22 appears 1 time.

The number 14 has the highest frequency (it appears 4 times).

Therefore, the mode of the given data is 14.

Given

A frequency distribution table of the salaries of 60 workers.

To Find

The mean salary of the 60 workers.

Solution

To find the mean salary, we use the formula $\overline{x} = \frac{\sum f_i x_i}{\sum f_i}$, where $x_i$ is the salary and $f_i$ is the number of workers.

We will create a calculation table:

From the table, we have:

Total number of workers, $\sum f_i = 60$.

Sum of products, $\sum f_i x_i = 305000$.

Now, we calculate the mean:

Mean Salary = $\frac{305000}{60} = \frac{30500}{6} \approx 5083.33$

The mean salary of the 60 workers is approximately $\textsf{₹}$ 5083.33.

(i) Situation where the mean is an appropriate measure of central tendency.

The mean is a good measure of central tendency when the data is symmetrically distributed and does not have extreme values (outliers).

Example: Consider the marks obtained by a student in 5 different subjects: 75, 80, 82, 85, 88. The data values are close to each other without any extreme scores.

Mean = $\frac{75+80+82+85+88}{5} = \frac{410}{5} = 82$.

Here, the mean mark of 82 provides a fair and accurate summary of the student's overall performance.

(ii) Situation where the mean is not appropriate, but the median is.

The mean is not a good measure when the data is skewed or contains extreme values (outliers), because the mean is heavily influenced by these outliers. In such cases, the median provides a better representation of the central value.

Example: Consider the monthly income of 5 employees in a small office: $\textsf{₹}$ 15,000, $\textsf{₹}$ 16,000, $\textsf{₹}$ 17,000, $\textsf{₹}$ 18,000, and $\textsf{₹}$ 1,00,000 (the owner's income).

Mean Income = $\frac{15000+16000+17000+18000+100000}{5} = \frac{166000}{5} = \textsf{₹}\ 33,200$.

This mean value of $\textsf{₹}$ 33,200 is not representative of what a typical employee earns, as it is much higher than the salary of 4 out of the 5 people.

Now let's find the median. The data is already in order. The median is the middle value (3rd term), which is $\textsf{₹}$ 17,000.

The median income of $\textsf{₹}$ 17,000 is a much better and more appropriate measure of the central tendency for this data, as it is not affected by the outlier salary of the owner.