# R.D. Sharma Solutions Class 10th: Ch 7 Statistics Exercise 7.3

## Chapter 7 Statistics R.D. Sharma Solutions for Class 10th Math Exercise 7.3

1. The following table gives the distribution of total household expenditure (in rupees) of manual workers in a city.
Find the average expenditure (in rupees) per household.

**Solution**

2. 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 number of plants per house .

Number of plants: 0-2 2-4 4-6 6-8 8-10 10-12 12-14

Number of houses: 1 2 1 5 6 2 3

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

**Solution**

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

Daily wages (in Rs). 100-120 120-140 140-160 160-180 180-200

Number of workers: 12 14 8 6 10

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

**Solution**

minute recorded and summarized as follows. Find the mean heart beats per minute for these

women, choosing a suitable method.

Number of heat beats per minute: 65 - 68 68 - 71 71 - 74 74 - 77 77 - 80 80 - 83 83 -86

Number of women: 2 4 3 8 7 4 2

**Solution**

Find the mean of each of the following frequency distributions: (5 − 14)

5. Class interval: 0-6 6-12 12-18 18-24 24-30

Frequency: 6 8 10 9 7

**Solution**

6. Class interval: 50 - 70 70 - 90 90 - 110 110 - 130 130 - 150 150 – 170

Frequency: 18 12 13 27 8 22

**Solution**

7. Class interval: 0-8 8- 16 16- 24 24-32 32-40

Frequency: 6 7 10 8 9

**Solution**

8. Class interval: 0-6 6-12 12-18 18-24 24-30

Frequency: 7 5 10 12 6

**Solution**

9. Class interval: 0- 10 10- 20 20-30 30-40 40-50

Frequency: 9 12 15 10 14

**Solution**

Frequency: 5 9 10 8 8

**Solution**

Frequency: 5 6 4 3 2

**Solution**

Frequency: 5 8 12 20 3 2

**Solution**

Frequency: 6 10 8 12 4

**Solution**

Frequency: 14 22 16 6 5 3 4

**Solution**

Size of item: 1-4 4-9 9- 16 16-27

Frequency: 6 12 26 20

**Solution**

16. The weekly observations on cost of living index in a certain city for the year 2004 - 2005 are given below. Compute the weekly cost of living index.

**Solution**

Marks: 0-10 10-20 20-30 30-40 40-50

Number of students: 20 24 40 36 20

Calculate the average marks by using all the three methods: direct method, assumed mean deviation and shortcut method.

**Solution**

_{1}and f

_{2}.

Class: 0-20 20-40 40-60 60-80 80-100 100–120

Frequency: 5 f

_{1}10 f

_{2}7 8

**Solution**

Class interval: 11-13 13-15 15-17 17-19 19-21 21-23 23-25

Frequency: 7 6 9 13 - 5 4

**Solution**

Class: 0-10 10-20 20-30 30-40 40-50

Frequency: 8 p 12 13 10

**Solution**

contained varying number of mangoes. The following was the distribution of mangoes

according to the number of boxes.

Number of mangoes: 50-52 53-55 56-58 59-61 62-64

Number of boxes: 15 110 135 115 25

Find the mean number of mangoes kept in a packing box. Which method of finding the

mean did you choose ?

**Solution**

Daily expenditure (in Rs): 100-150 150-200 200-250 250-300 300-350

Number of households: 4 5 1 2 2 2

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

**Solution**

23. To find out the concentration of SO

_{2}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 SO

_{2 }in the air.**Solution**

Number of days: 0-6 6-10 10-14 14-20 20-28 28-38 38-40

Number of students: 11 10 7 4 4 3 1

**Solution**

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

Literacy rate (in %): 45-55 55-65 65-75 75-85 85-95

Number of cities: 3 10 11 8 3

**Solution**

26 . The following is the cummlative frequency distribution (of less than type) of 100 persons each of age 20 years and above . Determine the mean age .

Age below (in years) : 30 40 50 60 70 80

Number of persons : 100 220 350 760 950 1000

**Solution**

We know that mean , x̅ = A + h(1/N Σf

Now, we have N = Σf_{i}u_{i})_{i}= 1000, h = 10, A = 55, Σf

_{i}u

_{i}= - 370

x̅ = 55 + 10 [1/100 × (-370)]

= 55 - 3.7

= 51.3 years

Class intervals : 11-13 13 -15 15 -17 17-19 19-21 21-23 23-25

Frequency : 3 6 9 13 f 5 4

**Solution**

We know that mean , x̅ = A + h(1/N Σf

Now, we have N = Σf

= 18 = 18 + 2 [1/(40+f) × (-8 + f)

⇒ 0 = -16+2f/40+f

⇒ -16 + 2f = 0

⇒ f = 8

28. Find the missing frequencies in the following distribution, if the sum of the frequencies is 120 and the mean is 50.

Class: 0-20 20-40 40-60 60-80 80-100

Frequency : 17 f

_{i}u_{i})Now, we have N = Σf

_{i}= 40 + f, h = 2, A = 18, Σf_{i}u_{i}= -8 + f= 18 = 18 + 2 [1/(40+f) × (-8 + f)

⇒ 0 = -16+2f/40+f

⇒ -16 + 2f = 0

⇒ f = 8

Class: 0-20 20-40 40-60 60-80 80-100

Frequency : 17 f

_{i}32 f_{2}19**Solution**

_{i}u

_{i})

Now, we have N = 68 + f

_{1}+ f

_{2}, h = 20, A = 50, Σf

_{i}u

_{i}= 4 - f

_{1}+ f

_{2}.

50 = 50 + 20 (4-f

_{1}+f

_{2}/68+f

_{1}+f

_{2})

⇒ f

_{1}- f_{2}= 4
⇒ f

_{1}= 4 + f_{2}.....(1)
Now, N = 68 + f

_{1}+ f_{2}= 120
f

_{1}+ f_{2}= 120 - 68 = 52
⇒ 4 + f

_{2}+ f_{2}= 52 [Using (1)]
⇒ f

_{2}= 24
so,

f

_{1}= 4 + 24 = 28
29. The daily incomes of a sample of 50 employees are tabulated as follows :

Income: 1-200 201-400 401-600 601-800

No.of employees : 14 15 14 7

Find the mean daily income of employees .

No.of employees : 14 15 14 7

Find the mean daily income of employees .

**Solution**

We know that mean , x̅ = A + h(1/N Σf

_{i}u_{i})
Now, we have N = Σf

_{i}= 50, h = 200, A = 500.5, Σf_{i}u_{i}= -36
x̅ = 500.5 + 200 [1/50 × (-36)]

= 500.5 - 144

= 356.5