RBSE Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.1

Rajasthan Board RBSE Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.1 Textbook Exercise Questions and Answers.

RBSE Class 10 Maths Solutions Chapter 14 Statistics Ex 14.1

Question 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 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 the number of plants and number of houses is numerically small, therefore we should use the direct method.

Hence, the mean number of plants per house is 8.1 plants.

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:

Let assumed mean (a) = 550 and Class Size (h) = 20
∴ \(\bar{u}=\frac{\sum f_{i} u_{i}}{\sum f_{i}}=\frac{-12}{50}\) = -0.24
∵ Mean \((\overline{\mathrm{X}})=a+h \bar{u}\)
= 550 + (20) (-0.24)
= 550 - 4.8
= 545.2
So, the mean daily wages of the workers of the factory is Rs. 545.20

Question 3.
The following distribution shows the daily pocket allowance of children of a locality. The-mean pocket allowance is Rs. 18. Find the missing frequency f.

Solution:

Hence, the missing frequency is 20.

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

Solution:
Calculation Table for a mean of heartbeat per minute
Let the assumed mean of beats a = 75.5

Here, Assumed Mean (a) = 75.5 and Class Size (h) = 3
∴ \(\bar{u}=\frac{\sum f_{i} u_{i}}{\sum f_{i}}\)
= \(\frac{4}{30}\)
= 0.13 (approx)
∵ Mean \((\overline{\mathbf{X}})=a+h \bar{u}\)
= 75.5 + 3(0.13)
= 75.5 + 0.39
\(\overline{\mathbf{X}}\) = 75.89
So, the mean heartbeats 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 numbers of 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 the number of mangoes and number of boxes is numerically large, therefore we shall use the step-deviation method.

Here Assumed Mean (a) = 57 and Class Size (h) = 3
∴ \(\bar{u}=\frac{\sum f_{i} u_{i}}{\sum f_{i}}\)
\(\bar{u}=\frac{25}{400}\) = 0.0625
∵ Mean \((\overline{\mathbf{X}})=a+h \bar{u}\)
\(\overline{\mathbf{X}}\) = 57 + 3(0.0625)
= 57 + 0.1875
= 57.1875
= 57.19 (approx.)
Hence the mean number of mangoes kept in a packing box is 57.19.

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

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

Here Assumed Mean (a) = 225 and Class Size (h) = 50
∴ \(\bar{u}=\frac{\sum f_{i} u_{i}}{\sum f_{i}}\)
= \(-\frac{7}{25}\)
= -0.28
Mean \((\overline{\mathrm{X}})=a+h \vec{u}\)
= 225 + 50(-0.28)
= 225 - 14
= 211
Hence, the mean daily expenditure on food is Rs. 211.
 
Question 7.
To find out the concentration of SO₂ in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:

Find the mean concentration of SO₂ in the air.
Solution:
Table for Calculating Concentration of SO₂ in air

Here, Assumed Mean (a) = 0.10 and Class Size (h) = 0.04
∴ \(\bar{u}=\frac{\sum f_{i} u_{i}}{\sum f_{i}}=\frac{-1}{30}\) = -0.03 (approx.)
∵ Mean \((\overline{\mathrm{X}})=a+h \bar{u}\)
= 0.10 + 0.04(-0.03)
= 0.10 - 0.0012
= 0.099 (approx.)
Hence, the mean concentration of SO₂ in the air is 0.099 ppm

Question 8.
A Class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.

Solution:
Mean Computation Table for Absence of Students

Hence the mean number of days a student was absent is 12.48 days.

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

Solution:
Computation Table for Mean Literacy Rate

Here Assumed Mean (a) = 70 and Class Size (h) = 10
∴ \(\bar{u}=\frac{\sum f_{i} u_{i}}{\sum f_{i}}=\frac{-2}{35}\) = -0.057
∵ Mean \((\overline{\mathrm{X}})=a+h \bar{u}\)
= 70 + 10(-0.057)
= 70 - 0.57
= 69.43
Hence mean literacy rate is 69.43%.