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

• Last Updated : 25 Jan, 2021

### Problem 1: Calculate the mean for the following distribution:

Solution:

We know that, Mean = ∑fx/ N = 281/40 = 7.025

### Problem 2: Find the mean of the following data :

Solution:

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.

Solution:

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.

Solution:

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

Solution:

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

Solution:

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

Solution:

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.

Solution:

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.

Solution:

We know that, Mean = ∑fx/ N = 698/40 = 17.45

So, the mean age of the students is 17.45 years.

### 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

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

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