Punjab State Board PSEB 10th Class Maths Book Solutions Chapter Statistics Ex 14.1 Textbook Exercise Questions and Answers.

## PSEB Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.1

Question 1.

A survey was conducted by a group of students as a part of their enviroment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

Which method did you use for finding the mean, and why?

Solution:

Since, number of plants and houses are small in their values; so we must use direct method

Mean X = \(\overline{\mathrm{X}}=\frac{\Sigma f_{i} x_{i}}{\Sigma f_{i}}\)

\(\frac{162}{20}\) = 8.1

Hence, mean number of plants per house is 8.1.

Question 2.

Consider the following distribution of daily wages of 50 workers of a factory.

Find the mean daily wages of the workers of the factory by using an appropriate method.

Solution:

From given data,

Assumed Mean (a) = 150

and width of the class (h) = 20

∴ \(\bar{u}=\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}\)

= \(\frac{-12}{50}\) = – 0.24

Using formula, Mean \((\overline{\mathrm{X}})=a+h \bar{u}\)

= 150 + (20) (- 0.24)

= 150 – 4.8 = 145.2

Hence,mean daily wages of the workers of factory is ₹ 145.20.

Question 3.

The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is 18. Find the missing frequency f.

Solution;

From the above data,

Assumed mean (a) = 18

Using formula, Mean \((\overline{\mathrm{X}})=a+\frac{\sum f_{i} d_{i}}{\Sigma f_{i}}\)

\(\bar{X}=18+\frac{2 f-40}{44+f}\)

But. Mean of data \((\bar{x})\) = 18 …(Given)

∴ 18 = 18 + \(\frac{2 f-40}{44+f}\)

or \(\frac{2 f-40}{44+f}\) = 18 – 18 = 0

or 2f – 40 = 0

or 2f = 40

or f = \(\frac{40}{2}\) = 20

Hence, missing frequency f is 20.

Question 4.

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

Solution:

From above data,

Assumed Mean (a) = 75.5

Width of Class (h) = 3

∴ \(\bar{u}=\frac{\sum f_{i} u_{i}}{\sum f_{i}}\)

= \(\frac{4}{30}\) (approx.)

Using formula,

Mean \((\overline{\mathrm{X}})=a+h \bar{u}\)

= 75.5 + 3 (0.13) = 75.5 + 0.39

\(\overline{\mathrm{X}}\) = 75.89

Hence, mean heart beats per minute for women is 75.89.

Question 5.

In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number mangoes. The following was the distribution of mangoes according to the number of boxes.

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?

Solution:

Since, values of number of mamgoes and number of boxes are large numerically. So, we must use step-deviation method.

From above data,

Assumed Mean (a) = 57

Width of class (h) = 3

∴ \(\bar{u}=\frac{\sum f_{i} u_{i}}{\sum f_{i}}\)

\(\bar{u}=\frac{25}{400}\) = 0.0625

Using formula, Mean \((\overline{\mathrm{X}})=a+h \bar{u}\)

\(\bar{X}\) = 57 + 3(0.0625)

= 57 + 0.1875 = 57.1875 = 57.19.

Hence, mean number of mangoes kept in a packing box is 57.19.

Question 6.

The table below shows the daily expenditure on food of 25 households in a locality.

Find the mean daily expenditure on food by a suitable method.

Solution:

From above data,

Assumed mean (a) = 225

Width of class (h) = 50

∴ \(\bar{u}=\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}\)

\(\bar{u}=-\frac{7}{25}\) = 0.28

Using formula,

Mean \((\overline{\mathrm{X}})=a+h \bar{u}\)

\(\bar{X}\) = 225 + 50 (- 0.28)

\(\bar{X}\) = 225 – 14

\(\bar{X}\) = 211

Hence, mean daily expenditure on food is ₹ 211.

Question 7.

To find out the concentration of SO_{2} in the air (in parts per million, e., ppm), the data was collected for 30 localities in a certain city and is presented below:

Find the mean concentration of SO_{2} in the air.

Solution:

From above data,

Assumed mean (a) = 0.10

Width of class (h) = 0.04

∴ \(\bar{u}=\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}\)

\(\bar{u}=-\frac{1}{30}\) = – 0.33(approx.)

Using formula, Mean \((\overline{\mathrm{X}})=a+h \bar{u}\)

\(\bar{X}\) = 0.10 + 0.04 (- 0.33)

= 0.10 – 0.0013 = 0.0987 (approx.)

Hence, mean concentration of SO_{2} in the air is 0.0987 ppm.

Question 8.

A class teacher has the folloing absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.

Solution:

From above data,

Assumed Mean (a) = 17

Using formula, Mean(\(\bar{X}\)) = a + \(\frac{\Sigma f_{i} d_{i}}{\Sigma f_{i}}\)

\(\bar{X}\) = 17 + \(-\frac{181}{40}\)

= 17 – 4.52 = 12.48.

Hence, mean 12.48 number of days a student was absent.

Question 9.

The following table gives the literacy rate (in percentage) of 35 citIes. Find the mean literacy rate.

Solution:

From the above data,

Assumed Mean (a) = 70

Width of class (h) = 10

∴ \(\bar{u}=\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}=\frac{-2}{35}\) = – 0.057

Using formula, Mean \((\overline{\mathrm{X}})=a+h \bar{u}\)

\(\bar{X}\) = 70 + 10 (- 0.057)

= 70 – 0.57 = 69.43

Hence, mean literacy rate is 69.43%.