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
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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 = ∑fi xi/ ∑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 = ∑fi xi/ ∑fi = 5520/80 = 69
(ii) Short – cut method
|
Weekly wages (Rs) |
No. of workers (fi) |
Mid-value xi |
A = 72.5 di = x – 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 – 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 – 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 – 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 – 148f = 9324 – 8645
f = 679/97
Thus, f = 7
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