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Statistics — Class 10 Mathematics

Mathematics · Chapter 13

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Chapter summary

Statistics deals with collecting, organising and interpreting numerical data. This chapter focuses on measures of central tendency for grouped data — the mean (using the direct, assumed-mean or step-deviation method), the median (using the median class and its formula) and the mode (using the modal class formula). It also explains the relationship . These measures summarise a large data set with a single representative value, which is essential for comparing and drawing conclusions from data.

Mean of grouped dataMedian of grouped dataMode of grouped dataRelationship between mean, median and mode

Key ideas & terms

Mean: The average value of the data.
Median: The middle value that divides ordered data into two halves.
Mode: The most frequently occurring value (or modal class).
Empirical relationship: .

Extra questions & answers

State the empirical relationship between mean, median and mode.

Name three methods to find the mean of grouped data.

Direct method, assumed-mean method, and step-deviation method.

Name the three measures of central tendency.

The mean, the median and the mode.

Name the methods of finding the mean of grouped data.

The direct method, the assumed-mean (deviation) method and the step-deviation method.

Write the formula for the median of grouped data.

, where is the lower limit of the median class, the cumulative frequency before it, its frequency and the class size.

Write the formula for the mode of grouped data.

, using the modal class and the frequencies around it.

State the empirical relationship between mean, median and mode.

The mean of 5 numbers is 18. If one number is removed, the mean becomes 16. Find the removed number.

Sum of 5 numbers ; sum of 4 numbers ; so the removed number .

What is a cumulative frequency?

The running total of the frequencies up to a given class, used to draw an ogive and to find the median.

What does an ogive help to find?

The median of grouped data graphically, as the x-coordinate of the point where the two cumulative frequency curves meet (or where ).

Activity 1 :

From the Table 13.3 find the mean by taking each of (i.e., and so on) as ‘’. What do you observe? You will find that the mean determined in each case is the same, i.e., . (Why ?)

Using the assumed mean formula, If a different class mark is chosen as , every deviation changes by the same fixed amount. This change is exactly cancelled when is added back in the formula. For example, taking , we get and . Therefore, Taking any of the other class marks as also gives . Thus, the calculated mean does not depend on the choice of the assumed mean.

Activity 2 :

1. Collect the marks obtained by all the students of your class in Mathematics in the latest examination conducted by your school. Form a grouped frequency distribution of the data obtained.

A model grouped frequency distribution is: marks , with respective frequencies . The total number of students is . Using class marks , . Hence, Thus, the mean mark in this illustrative data is .

2. Collect the daily maximum temperatures recorded for a period of days in your city. Present this data as a grouped frequency table.

A model grouped frequency table is: temperature in , , with respective frequencies . The total frequency is . The class marks are , so . Therefore, The mean daily maximum temperature in this illustrative data is approximately .

3. Measure the heights of all the students of your class (in cm) and form a grouped frequency distribution table of this data.

A model grouped frequency distribution is: height in , , with respective frequencies . The total number of students is . The class marks are , giving . Hence, Thus, the mean height in this illustrative data is .

EXERCISE 13.1

1. A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in houses in a locality. Find the mean number of plants per house. Number of plants: . Number of houses: . Which method did you use for finding the mean, and why?

The class marks are . Using the direct method, and . Therefore, The mean number of plants per house is . The direct method is convenient because the class marks and frequencies are small.

2. Consider the following distribution of daily wages of workers of a factory. Daily wages (in ₹): . Number of workers: . Find the mean daily wages of the workers of the factory by using an appropriate method.

Take the assumed mean and class size . The class marks are , so gives . Then, Since , The mean daily wage is ₹.

3. The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is ₹. Find the missing frequency . Daily pocket allowance (in ₹): . Number of children: .

The class marks are . The total frequency is , and Since the mean is , Therefore, Hence, the missing frequency is .

4. Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute were recorded and summarised as follows. Find the mean heartbeats per minute for these women, choosing a suitable method. Number of heartbeats per minute: . Number of women: .

The class marks are . Take and . The corresponding values are . Thus, Therefore, The mean number of heartbeats is per minute.

5. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes. Number of mangoes: . Number of boxes: . Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?

The class marks are . Take and . Then , and Also, . Therefore, The mean number of mangoes in a box is approximately . The step-deviation method is convenient because the class marks have a common difference of .

6. The table below shows the daily expenditure on food of households in a locality. Daily expenditure (in ₹): . Number of households: . Find the mean daily expenditure on food by a suitable method.

The class marks are . Take and . Then , and Hence, The mean daily expenditure on food is ₹.

7. To find out the concentration of in the air (in parts per million, i.e., ppm), the data was collected for localities in a certain city and is presented below: Concentration of (in ppm): . Frequency: . Find the mean concentration of in the air.

The class marks are . Therefore, Since , Thus, the mean concentration of is approximately .

8. A class teacher has the following absentee record of students of a class for the whole term. Find the mean number of days a student was absent. Number of days: . Number of students: .

The class marks are . Thus, Since , Therefore, the mean number of days a student was absent is , or approximately days.

9. The following table gives the literacy rate (in percentage) of cities. Find the mean literacy rate. Literacy rate (in %): . Number of cities: .

The class marks are . Therefore, Since , Hence, the mean literacy rate is approximately .

Activity 3 :

Continuing with the same groups as formed in Activity 2 and the situations assigned to the groups. Ask each group to find the mode of the data. They should also compare this with the mean, and interpret the meaning of both.

For the illustrative Mathematics-marks data in Activity 2, the modal class is . Using , , , , and , Its mean is . Thus, the average mark is , while marks near are the most typical by frequency. For the illustrative temperature data, the modal class is , giving Its mean is approximately . For the illustrative height data, the modal class is , giving Its mean is . In each case, the mean represents the arithmetic average, whereas the mode represents the most frequent or most typical value.

EXERCISE 13.2

1. The following table shows the ages of the patients admitted in a hospital during a year: Age (in years): . Number of patients: . Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

The modal class is . Here, , , , , and . Therefore, The class marks are , and Since , Thus, the mean age is approximately years, while the most frequently represented age is approximately years.

2. The following data gives the information on the observed lifetimes (in hours) of electrical components: Lifetimes (in hours): . Frequency: . Determine the modal lifetimes of the components.

The modal class is . Here, , , , , and . Thus, Therefore, the modal lifetime of the components is approximately hours.

3. The following data gives the distribution of total monthly household expenditure of families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure: Expenditure (in ₹): . Number of families: .

The modal class is ₹. Here, , , , , and . Therefore, For the mean, the class marks are . Their frequency-weighted sum is , so Hence, the modal monthly expenditure is approximately ₹, and the mean monthly expenditure is ₹.

4. The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures. Number of students per teacher: . Number of states / U.T.: .

The modal class is . Here, , , , , and . Thus, The class marks are . Their frequency-weighted sum is , and the total frequency is . Therefore, The mean indicates that there are about students per teacher on average, whereas the mode indicates that the most common ratio is about students per teacher.

5. The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches. Runs scored: . Number of batsmen: . Find the mode of the data.

The modal class is . Here, , , , , and . Therefore, Hence, the mode is approximately runs.

6. A student noted the number of cars passing through a spot on a road for periods each of minutes and summarised it in the table given below. Find the mode of the data: Number of cars: . Frequency: .

The modal class is . Here, , , , , and . Thus, Therefore, the modal number of cars passing during a -minute period is approximately .

EXERCISE 13.3

1. The following frequency distribution gives the monthly consumption of electricity of consumers of a locality. Find the median, mean and mode of the data and compare them. Monthly consumption (in units): . Number of consumers: .

The cumulative frequencies are . Since , , so the median class is . Thus, The class marks are , and . Therefore, The modal class is also , with , , , , and . Hence, Thus, The three values are very close, so they all describe a central electricity consumption of about units.

2. If the median of the distribution given below is , find the values of and . Class interval: . Frequency: . Total: .

Since the total frequency is , so The given median lies in the class . Therefore, , , , , and the cumulative frequency before the median class is . Using the median formula, Hence, Since , Therefore, and .

3. A life insurance agent found the following data for distribution of ages of policy holders. Calculate the median age, if policies are given only to persons having age years onwards but less than year. Age (in years): Below , Below , Below , Below , Below , Below , Below , Below , Below . Number of policy holders: .

Converting the cumulative frequencies into ordinary frequencies gives: , , , , , , , , and . Here, , so . The cumulative frequency just before the class is , making the median class. Thus, Therefore, the median age is approximately years.

4. The lengths of leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table: Length (in mm): . Number of leaves: . Find the median length of the leaves. (Hint : The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to .)

The continuous class intervals are . Their cumulative frequencies are . Since , , so the median class is . Here, , , , and . Therefore, Hence, the median length of the leaves is .

5. The following table gives the distribution of the life time of neon lamps: Life time (in hours): . Number of lamps: . Find the median life time of a lamp.

The cumulative frequencies are . Since , . The median class is . Here, , , , and . Thus, Therefore, the median lifetime of a neon lamp is approximately hours.

6. surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows: Number of letters: . Number of surnames: . Determine the median number of letters in the surnames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames.

The cumulative frequencies are . Since , , so the median class is . Thus, The class marks are . Therefore, and The modal class is . Here, , , , , and . Hence, Therefore, the median is letters, the mean is letters, and the modal size is approximately letters.

7. The distribution below gives the weights of students of a class. Find the median weight of the students. Weight (in kg): . Number of students: .

The cumulative frequencies are . Since , . The median class is . Here, , , , and . Therefore, Hence, the median weight of the students is approximately .