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

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

Solution:

Now, mean = = 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