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Class 10 RD Sharma Solutions – Chapter 7 Statistics – Exercise 7.1 | Set 1

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Problem 1: Calculate the mean for the following distribution:

x: 5 6 7 8 9
f: 8 14 11 3

Solution:

f fx
5 4 20
6 8 48
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: 

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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