# Class 9 RD Sharma Solutions – Chapter 24 Measures of Central Tendency – Exercise 24.2

• Last Updated : 18 Mar, 2021

### Question 1. Calculate the mean for the following distribution:

Solution:

Now, mean = Attention reader! All those who say programming isn't for kids, just haven't met the right mentors yet. Join the  Demo Class for First Step to Coding Coursespecifically designed for students of class 8 to 12.

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= 281/40

= 7.025

### Question 2. Calculate the mean for the following distribution:

Solution:

Now, mean = = 2650/106

= 25

### Question 3. The mean of the following data is 20.6. Find the value of p.

Solution:

Now, mean = = (25p + 530)/50

Given,

Mean = 20.6

Solving, we get,

20.6 = (25p + 530)/50

25p + 530 = 1030

25p = 1030 − 530 = 500

that is, p = 20

### Question 4. If the mean of the following data is 15, find p.

Solution:

Mean = = (10p + 445)/(p + 27)

Give,

Mean = 15

Solving, (10p + 445)/(p + 27) = 15

10p + 445 = 15(p + 27)

10p – 15p = 405 – 445 = -40

-5p = -40

that is, p = 8

### Question 5. Find the value of p for the following distribution whose mean is 16.6.

Solution:

Now, mean = = (24p + 1228)/100

Given,

Mean = 16.6

Solving, (24p + 1228)/100 = 16.6

24p + 1228 = 1660

24p = 1660 – 1228 = 432

p = 432/24

= 18

### Question 6. Find the missing value of p for the following distribution whose mean is 12.58.

Solution:

Mean = = (7p + 524)/50

Given,

Mean = 12.58

Solving, (7p + 524)/50 = 12.58

7p + 524 = 12.58 x 50

7p + 524 = 629

7p = 629 – 524 = 105

p = 105/7

= 15

### Question 7. Find the missing frequency (p) for the following distribution whose mean is 7.68.

Solution:

Mean = = (9p + 303)/(p+41)

Given,

Mean = 7.68

Solving we get, (9p + 303)/(p+41) = 7.68

9p + 303 = 7.68 (p + 41)

9p + 303 = 7.68p + 314.88

9p − 7.68p = 314.88 − 303

1.32p = 11.88

that is, p = (11.881)/(1.32) = 9

Question 8. Find the missing value of p for the following distribution whose mean is 12.58.

Solution:

Given,

Mean = 12.58

=> 7p + 524/50 = 12.58

=> 7p + 524 = 629

=> 7p = 629 – 524

Solving for p, we get,

=> 7p = 105

that is, p = 105/7 = 15

### Question 9. Find the missing frequency (p) for the following distribution whose mean is 7.68.

Solution:

Mean = Given,

Mean = 7.68

Now,

9p + 303/ (p +41) = 7.68

Solving, we get,

9p + 303 = 7.68 + 314.88

9p-7.68p = 314.88 – 303

1.32p = 11.88

p = 9

### Question 10. Find the value of p, if the mean of the following distribution is 20.

Solution:

Given,

Mean = 20

Mean =  If p+1 = 0 or p-1 =0

p=-1

### 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 us assume the number of candidates in school III to be p.

Therefore,

Total number of candidates in all the four schools = 60 + 48 + p + 40 = 148 + p

Average score of four schools = 66

∴Computing total score of the candidates = (148 + p) x 66

Now,

The mean score of 60 in school I is equivalent to 75 .

Total in school I = 60 x 75 = 4500

The mean score of 48 in school II is equivalent to 80 .

Total in school II = 48 x 80 = 3840

In school III, mean of p = 55

Total in school III= 55 x p = 55p

and in school IV, mean of 40 = 50

Total in school IV = 40 x 50 = 2000

Since, total of the candidates is 148+p.

Also,

Total score = 4500 + 3840 + 55p + 2000 = 10340 + 55p

∴10340 + 55p = (148 + p) x 66 = 9768 + 66p

=> 10340 – 9768 = 66p – 55p

=> 572 = 11p

∴ p = 572/11

Therefore,

The number of candidates in school III = 52

### Question 12. Find the missing frequencies in the following frequency distribution if it is known that the mean of the distribution is 50.

Solution:

Given,

Mean = 50 And, given value of N = 120 Subtracting (ii) from (i) , Substituting f2  in (i) My Personal Notes arrow_drop_up