GSEB Solutions Class 10 Maths Chapter 14 Statistics Ex 14.1

   

Gujarat Board GSEB Solutions Class 10 Maths Chapter 14 Statistics Ex 14.1 Textbook Questions and Answers.

Gujarat Board Textbook Solutions 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 environment awareness program, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean of plants per house.
GSEB Solutions Class 10 Maths Chapter 14 Statistics Ex 14.1
Which method did you use for finding the mean and why?
Solution:
GSEB Solutions Class 10 Maths Chapter 14 Statistics Ex 14.1
\(\bar{x}\) = \(\frac{\Sigma f_{i} x_{i}}{\Sigma f_{i}}\) = \(\frac{162}{20}\) = 8.1 plants
Direct method for finding the mean as numerical values of x. and f are small.

Question 2.
Consider the following distribution of daily wages of 50 workers of a factory.
GSEB Solutions Class 10 Maths Chapter 14 Statistics Ex 14.1
Find the mean daily wages of the workers of the factory by using an appropriate method.
Solution:
GSEB Solutions Class 10 Maths Chapter 14 Statistics Ex 14.1
Using the step-deviation method
\(\bar{x}\) = a + \(\left(\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}\right)\) x h= 150 + \(\left(\frac{-12}{50}\right)\) x 20 = 150 – 4.8 = 145.20
Hence, the mean daily wages of the workers of the factory is 145.20.

GSEB Solutions Class 10 Maths Chapter 14 Statistics Ex 14.1

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.
GSEB Solutions Class 10 Maths Chapter 14 Statistics Ex 14.1
Solution:

Using the direct method,
\(\bar{x}\) = \(\frac{\Sigma f_{i} x_{i}}{\Sigma f_{i}}\)
⇒ 18 = \(\frac{20 f+752}{f+44}\)
⇒ 20f + 752 = 18(f + 44)
⇒ 20f + 752 = 18f + 792
⇒ 20f – 18f = 792 – 752
2f = 40
f = \(\frac {4}{10}\) = 20
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 were recorded and summarized as follows. Find the mean heartbeat per minute for these women, choosing a suitable method.
GSEB Solutions Class 10 Maths Chapter 14 Statistics Ex 14.1
Solution:
GSEB Solutions Class 10 Maths Chapter 14 Statistics Ex 14.1
Using the step-deviation method
\(\bar{x}\) = a + \(\left(\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}\right)\) x h = 75.5 + (\(\frac {4}{30}\)) x 3 = 75.5 + 0.4 = 75.9
Hence, the mean heart beat per minute is 75.9.

GSEB Solutions Class 10 Maths Chapter 14 Statistics Ex 14.1

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.
GSEB Solutions Class 10 Maths Chapter 14 Statistics Ex 14.1
Find the mean number of mangoes kept in a packing box.
Which method of finding the mean did you choose?
Solution:
GSEB Solutions Class 10 Maths Chapter 14 Statistics Ex 14.1
Using the step-deviation method,
\(\bar{x}\) = a + \(\left(\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}\right)\) x h = 57 + \(\left(\frac{25}{400}\right)\) x 3 = 57 + 0.19 = 57.19

Question 6.
The table below shows the daily expenditure on the food of 25 households in a locality.
GSEB Solutions Class 10 Maths Chapter 14 Statistics Ex 14.1
Find the mean daily expenditure on food by a suitable method.
Solution:
GSEB Solutions Class 10 Maths Chapter 14 Statistics Ex 14.1
Using the step-deviation method
\(\bar{x}\) = a + \(\left(\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}\right)\) x h = 225 + (\(\frac {-7}{25}\)) x 50 = 225 – 14 = ₹ 211
Hence, the daily expeiiditure on food is 211.

GSEB Solutions Class 10 Maths Chapter 14 Statistics Ex 14.1

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 n city and is presented below. Find the mean concentration of SO2 in the air.
GSEB Solutions Class 10 Maths Chapter 14 Statistics Ex 14.1
Solution:
GSEB Solutions Class 10 Maths Chapter 14 Statistics Ex 14.1
Using the step-deviation method,
\(\bar{x}\) = a + \(\left(\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}\right)\) x h = 0.14 + (\(\frac {-31}{30}\)) x 0.04 = 0.14 – 0.041 = 0.099 ppm
Therefore, the mean concentration of SO2 in the air is 0.099 ppm.

Question 8.
A class teacher has the following absent record of 40 students of a class for the whole term. Find the mean number of days a student was absent.
GSEB Solutions Class 10 Maths Chapter 14 Statistics Ex 14.1
Solution:
GSEB Solutions Class 10 Maths Chapter 14 Statistics Ex 14.1
Using direct method
\(\bar{x}\) = \(\left(\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}\right)\) = \(\frac {499}{40}\) = 12.47
Hence, the mean number of days a student was absent is 12.47.

GSEB Solutions Class 10 Maths Chapter 14 Statistics Ex 14.1

Question 9.
The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.
GSEB Solutions Class 10 Maths Chapter 14 Statistics Ex 14.1
Solution:
GSEB Solutions Class 10 Maths Chapter 14 Statistics Ex 14.1
Using the step-deviation method
\(\bar{x}\) = a + \(\left(\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}\right)\) x h = 70 + \(\frac {-2}{35}\) x 10 = 70 – \(\frac {4}{7}\) = 70 – 0.57 = 69.43%
Hence, the literacy rate is 69.43%.

GSEB Solutions Class 10 Maths Chapter 14 Statistics Ex 14.1

Leave a Comment

Your email address will not be published. Required fields are marked *