Class 10 RD Sharma Solutions – Chapter 7 Statistics – Exercise 7.1 | Set 1
Problem 1: Calculate the mean for the following distribution:
x: |
5 |
6 |
7 |
8 |
9 |
f: |
4 |
8 |
14 |
11 |
3 |
Solution:
x |
f |
fx |
5 |
4 |
20 |
6 |
8 |
48 |
7 |
14 |
98 |
8 |
11 |
88 |
9 |
3 |
27 |
|
N = 40 |
∑ fx = 281 |
We know that, Mean = ∑fx/ N = 281/40 = 7.025
Problem 2: Find the mean of the following data :
x: |
19 |
21 |
23 |
25 |
27 |
29 |
31 |
f: |
13 |
15 |
16 |
18 |
16 |
15 |
13 |
Solution:
x |
f |
fx |
19 |
13 |
247 |
21 |
15 |
315 |
23 |
16 |
368 |
25 |
18 |
450 |
27 |
16 |
432 |
29 |
15 |
435 |
31 |
13 |
403 |
|
N = 106 |
∑ fx = 2650 |
We know that, Mean = ∑fx/ N = 2650/106 = 25
Problem 3: If the mean of the following data is 20.6. Find the value of p.
x: |
10 |
15 |
p |
25 |
35 |
f: |
3 |
10 |
25 |
7 |
5 |
Solution:
x |
f |
fx |
10 |
3 |
30 |
15 |
10 |
150 |
p |
25 |
25p |
25 |
7 |
175 |
35 |
5 |
175 |
|
N = 50 |
∑ fx = 530 + 25p |
Given,
Mean = 20.6
We know that,
Mean = ∑fx/ N = (530 + 25p)/50
Now,
20.6 = (530 + 25p)/ 50
(20.6 × 50) – 530 = 25p
p = 500/25
So, p = 20
Problem 4: If the mean of the following data is 15, find p.
x: |
5 |
10 |
15 |
20 |
25 |
f: |
6 |
p |
6 |
10 |
5 |
Solution:
x |
f |
fx |
5 |
6 |
30 |
10 |
p |
10p |
15 |
6 |
90 |
20 |
10 |
200 |
25 |
5 |
125 |
|
N = p + 27 |
∑ fx = 445 + 10p |
Given,
Mean = 15
We know that,
Mean = ∑fx/ N = (445 + 10p)/(p + 27)
Now,
15 = (445 + 10p)/(p + 27)
15p + 405 = 445 + 10p
5p = 40
So, p = 8
Problem 5: Find the value of p for the following distribution whose mean is 16.6
x: |
8 |
12 |
15 |
p |
20 |
25 |
30 |
f: |
12 |
16 |
20 |
24 |
16 |
8 |
4 |
Solution:
x |
f |
fx |
8 |
12 |
96 |
12 |
16 |
192 |
15 |
20 |
300 |
p |
24 |
24p |
20 |
16 |
320 |
25 |
8 |
200 |
30 |
4 |
120 |
|
N = 100 |
∑ fx = 1228 + 24p |
Given,
Mean = 16.6
We know that,
Mean = ∑fx/ N = (1228 + 24p)/ 100
Now,
16.6 = (1228 + 24p)/ 100
1660 = 1228 + 24p
24p = 432
So, p = 18
Problem 6: Find the missing value of p for the following distribution whose mean is 12.58
x: |
5 |
8 |
10 |
12 |
p |
20 |
25 |
f: |
2 |
5 |
8 |
22 |
7 |
4 |
2 |
Solution:
x |
f |
fx |
5 |
2 |
10 |
8 |
5 |
40 |
10 |
8 |
80 |
12 |
22 |
264 |
p |
7 |
7p |
20 |
4 |
80 |
25 |
2 |
50 |
|
N = 50 |
∑ fx = 524 + 7p |
Given,
Mean = 12.58
We know that,
Mean = ∑fx/ N = (524 + 7p)/ 50
Now,
12.58 = (524 + 7p)/ 50
629 = 524 + 7p
7p = 105
So, p = 15
Problem 7: Find the missing frequency (p) for the following distribution whose mean is 7.68
x: |
3 |
5 |
7 |
9 |
11 |
13 |
f: |
6 |
8 |
15 |
p |
8 |
4 |
Solution:
x |
f |
fx |
3 |
6 |
18 |
5 |
8 |
40 |
7 |
15 |
105 |
9 |
p |
9p |
11 |
8 |
88 |
13 |
4 |
52 |
|
N = 41 + p |
∑ fx = 303 + 9p |
Given,
Mean = 7.68
We know that,
Mean = ∑fx/ N = (303 + 9p)/(41 + p)
Now,
7.68 = (303 + 9p)/(41 + p)
7.68(41 + p) = 303 + 9p
7.68p + 314.88 = 303 + 9p
1.32p = 11.88
So, p = 9
Problem 8: Find the value of p, if the mean of the following distribution is 20.
x: |
15 |
17 |
19 |
20 + p |
23 |
f: |
2 |
3 |
4 |
5p |
6 |
Solution:
x |
f |
fx |
15 |
2 |
30 |
17 |
3 |
51 |
19 |
4 |
76 |
20 + p |
5p |
100p + 5p2 |
23 |
6 |
138 |
|
N = 5p + 15 |
∑ fx = 295 + 100p + 5p2 |
Given,
Mean = 20
We know that,
Mean = ∑fx/ N = (295 + 100p + 5p2)/(5p + 15)
Now,
20 = (295 + 100p + 5p2)/(5p + 15)
20(5p + 15) = 295 + 100p + 5p2
100p + 300 = 295 + 100p + 5p2
5p2 – 5 = 0
5(p2 – 1) = 0
p2 – 1 = 0
(p + 1)(p – 1) = 0
So, p = 1 or -1
Here, p = -1 (Reject as frequency of a number cannot be negative)
So, p = 1
Problem 9: The following table gives the number of boys of a particular age in a class of 40 students. Calculate the mean age of the students.
Age (in years): |
15 |
16 |
17 |
18 |
19 |
20 |
No. of students: |
3 |
8 |
10 |
10 |
5 |
4 |
Solution:
Age (in years)
(x)
|
No. of students
(f)
|
fx |
15 |
3 |
45 |
16 |
8 |
128 |
17 |
10 |
170 |
18 |
10 |
180 |
19 |
5 |
95 |
20 |
4 |
80 |
|
N = 40 |
∑ fx = 698 |
We know that, Mean = ∑fx/ N = 698/40 = 17.45
So, the mean age of the students is 17.45 years.
Problem 10: Candidates of four schools appear in a mathematics test. The data were as follows :
Schools |
No. of Candidates |
Average Score |
I |
60 |
75 |
II |
48 |
80 |
III |
NA |
55 |
IV |
40 |
50 |
If the average score of the candidates of all the four schools is 66, find the number of candidates that appeared from school III.
Solution:
Let the number of candidates that appeared from school III = p
Schools |
No. of Candidates
(f)
|
Average Score
(x)
|
fx |
I |
60 |
75 |
4500 |
II |
48 |
80 |
3840 |
III |
p |
55 |
55p |
IV |
40 |
50 |
2000 |
|
N = 148 + p |
|
∑ fx = 10340 + 55p |
Given,
Average score of the candidates of all the four schools = Mean = 66
We know that,
Mean = ∑fx/ N = (10340 + 55p)/(148 + p)
Now,
66 = (10340 + 55p)/(148 + p)
66(148 + p) = 10340 + 55p
66p + 9768 = 10340 + 55p
11p = 572
So, p = 52
So, the number of candidates that appeared from school III are 52
Last Updated :
25 Jan, 2021
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