# Selina Solutions Concise Maths Class 10 Chapter 24 Measures of Central Tendency Exercise 24(B)

Continuing from the previous exercise, this exercise contains problems on finding the mean for grouped data (both continuous and discontinuous) by three methods. Direct method, short-cut method and step-deviation method are the three methods discussed under this exercise. The Selina Solutions for Class 10 Maths can be used by students for solving problems. Further, the solutions of the Concise Selina Solutions for Class 10 Maths Chapter 24 Measures of Central Tendency Exercise 24(B) are available in PDF format, provided in the link below.

## Selina Solutions Concise Maths Class 10 Chapter 24 Measures of Central Tendency Exercise 24(B) Download PDF

### Access Selina Solutions Concise Maths Class 10 Chapter 24 Measures of Central Tendency Exercise 24(B)

1. The following table gives the ages of 50 students of a class. Find the arithmetic mean of their ages.

 Age – Years 16 – 18 18 – 20 20 – 22 22- 24 24-26 No. of Students 2 7 21 17 3

Solution:

 Age in years C.I. xi Number of students (fi) xifi 16 – 18 17 2 34 18 – 20 19 7 133 20 – 22 21 21 441 22 – 24 23 17 391 24 – 26 25 3 75 Total 50 1074

Mean = âˆ‘fi xi/ âˆ‘fi = 1074/50 = 21.48

2. The following table gives the weekly wages of workers in a factory.

 Weekly Wages (Rs) No. of Workers 50-55 5 55-60 20 60-65 10 65-70 10 70-75 9 75-80 6 80-85 12 85-90 8

Calculate the mean by using:

(i) Direct Method

(ii) Short – Cut Method

Solution:

(i) Direct Method

 Weekly Wages (Rs) Mid-Value xi No. of Workers (fi) fixi 50-55 52.5 5 262.5 55-60 57.5 20 1150.0 60-65 62.5 10 625.0 65-70 67.5 10 675.0 70-75 72.5 9 652.5 75-80 77.5 6 465.0 80-85 82.5 12 990.0 85-90 87.5 8 700.0 Total 80 5520.00

Mean = âˆ‘fi xi/ âˆ‘fi = 5520/80 = 69

(ii) Short – cut method

 Weekly wages (Rs) No. of workers (fi) Mid-value xi A = 72.5 di = x â€“ A fidi 50-55 5 52.5 -20 -100 55-60 20 57.5 -15 -300 60-65 10 62.5 -10 -100 65-70 10 67.5 -5 -50 70-75 9 A = 72.5 0 0 75-80 6 77.5 5 30 80-85 12 82.5 10 120 85-90 8 87.5 15 120 Total 80 -280

Here, A = 72.5

3. The following are the marks obtained by 70 boys in a class test:

 Marks No. of boys 30 – 40 10 40 – 50 12 50 – 60 14 60 – 70 12 70 – 80 9 80 – 90 7 90 – 100 6

Calculate the mean by:

(i) Short – cut method

(ii) Step – deviation method

Solution:

(i) Short – cut method

 Marks No. of boys (fi) Mid-value xi A = 65 di = x â€“ A fidi 30 – 40 10 35 -30 -300 40 – 50 12 45 -20 -240 50 – 60 14 55 -10 -140 60 – 70 12 A = 65 0 0 70 – 80 9 75 10 90 80 – 90 7 85 20 140 90 – 100 6 95 30 180 Total 70 -270

Here, A = 65

(ii) Step – deviation method

 Marks No. of boys (fi) Mid-value xi A = 65 ui = (xi â€“ A)/ h fiui 30 – 40 10 35 -3 -30 40 – 50 12 45 -2 -24 50 – 60 14 55 -1 -14 60 – 70 12 A = 65 0 0 70 – 80 9 75 1 9 80 – 90 7 85 2 14 90 – 100 6 95 3 18 Total 70 -27

Here, A = 65 and h = 10

4. Find mean by step – deviation method:

 C. I. 63-70 70-77 77-84 84-91 91-98 98-105 105-112 Freq 9 13 27 38 32 16 15

Solution:

 C. I. Frequency (fi) Mid-value xi A = 87.50 ui = (xi â€“ A)/ h fiui 63 – 70 9 66.50 -3 -27 70 – 77 13 73.50 -2 -26 77 – 84 27 80.50 -1 -27 84 – 91 38 A = 87.50 0 0 91 – 98 32 94.50 1 32 98 – 105 16 101.50 2 32 105 – 112 15 108.50 3 45 Total 150 29

Here, A = 87.50 and h = 7

5. The mean of the following frequency distribution is. Find the value of ‘f’.

 C. I. 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 freq 8 22 31 f 2

Solution:

Given,

 C. I. frequency Mid-value (xi) fixi 0-10 8 5 40 10-20 22 15 330 20-30 31 25 775 30-40 f 35 35f 40-50 2 45 90 Total 63 + f 1235 + 35f

9324 + 148f = 8645 + 245f

245f â€“ 148f = 9324 â€“ 8645

f = 679/97

Thus, f = 7