PSEB 10th Class Maths Solutions Chapter 14 Statistics Ex 14.1

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.

PSEB 10th Class Maths Solutions Chapter 14 Statistics Ex 14.1 1

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

PSEB 10th Class Maths Solutions Chapter 14 Statistics Ex 14.1 2

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.

PSEB Solutions PSEB 10th Class Maths Solutions Chapter 14 Statistics Ex 14.2

Question 2.
Consider the following distribution of daily wages of 50 workers of a factory.

PSEB 10th Class Maths Solutions Chapter 14 Statistics Ex 14.1 3

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

PSEB 10th Class Maths Solutions Chapter 14 Statistics Ex 14.1 4

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.

PSEB Solutions PSEB 10th Class Maths Solutions Chapter 14 Statistics Ex 14.2

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.

PSEB 10th Class Maths Solutions Chapter 14 Statistics Ex 14.1 5

Solution;

PSEB 10th Class Maths Solutions Chapter 14 Statistics Ex 14.1 6

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.

PSEB Solutions PSEB 10th Class Maths Solutions Chapter 14 Statistics Ex 14.2

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.

PSEB 10th Class Maths Solutions Chapter 14 Statistics Ex 14.1 7

Solution:

PSEB 10th Class Maths Solutions Chapter 14 Statistics Ex 14.1 8

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.

PSEB Solutions PSEB 10th Class Maths Solutions Chapter 14 Statistics Ex 14.2

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.

PSEB 10th Class Maths Solutions Chapter 14 Statistics Ex 14.1 9

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.

PSEB 10th Class Maths Solutions Chapter 14 Statistics Ex 14.1 10

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.

PSEB Solutions PSEB 10th Class Maths Solutions Chapter 14 Statistics Ex 14.2

Question 6.
The table below shows the daily expenditure on food of 25 households in a locality.

PSEB 10th Class Maths Solutions Chapter 14 Statistics Ex 14.1 11

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

Solution:

PSEB 10th Class Maths Solutions Chapter 14 Statistics Ex 14.1 12

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.

PSEB Solutions PSEB 10th Class Maths Solutions Chapter 14 Statistics Ex 14.2

Question 7.
To find out the concentration of SO2 in the air (in parts per million, e., ppm), the data was collected for 30 localities in a certain city and is presented below:

PSEB 10th Class Maths Solutions Chapter 14 Statistics Ex 14.1 13

Find the mean concentration of SO2 in the air.
Solution:

PSEB 10th Class Maths Solutions Chapter 14 Statistics Ex 14.1 14

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 SO2 in the air is 0.0987 ppm.

PSEB Solutions PSEB 10th Class Maths Solutions Chapter 14 Statistics Ex 14.2

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.

PSEB 10th Class Maths Solutions Chapter 14 Statistics Ex 14.1 15

Solution:

PSEB 10th Class Maths Solutions Chapter 14 Statistics Ex 14.1 16

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.

PSEB Solutions PSEB 10th Class Maths Solutions Chapter 14 Statistics Ex 14.2

Question 9.
The following table gives the literacy rate (in percentage) of 35 citIes. Find the mean literacy rate.

PSEB 10th Class Maths Solutions Chapter 14 Statistics Ex 14.1 17

Solution:

PSEB 10th Class Maths Solutions Chapter 14 Statistics Ex 14.1 18

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%.

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