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:

x:	5	6	7	8	9
f:	4	8	11	14	3

Solution:

x f fx
5 4 20
6 8 48
7 14 98
8 11 88
9 3 27
N = 40 $\sum f_x = 281$

Now, mean = $\overline{x} = \frac{\sum f_x}{N}$

= 281/40

= 7.025

Question 2. Calculate the mean for the following distribution:

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 $\sum f_x = 2650$

Now, mean = $\overline{x} = \frac{\sum f_x}{N}$

= 2650/106

= 25

Question 3. 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 $\sum f_x = 25p + 530$

Now, mean = $\overline{x} = \frac{\sum f_x}{N}$

= (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.

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 $\sum f_x=10p+445$

Mean = $\overline{x} = \frac{\sum f_x}{N}$

= (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.

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 $\sum f_x = 24p + 1228$

Now, mean = $\overline{x} = \frac{\sum f_x}{N}$

= (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.

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 $\sum f_x = 7p + 524$

Mean = $\overline{x} = \frac{\sum f_x}{N}$

= (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.

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 = p +41 $\sum f_x = 9p+303$

Mean = $\overline{x} = \frac{\sum f_x}{N}$

= (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.

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 $\sum f_x =7p + 524$

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.

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 = p + 41 $\sum f_x =9p + 303$

Mean = $\overline{x} = \frac{\sum f_x}{N}$

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.

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 = 15+5p $\sum f_x = 295+100p+5p^2$

Given,

Mean = 20

Mean = $\overline{x} = \frac{\sum f_x}{N}$

$\frac{295+100p+5p^2}{15+5p}=20 \\ 295 + 100p + 5p^2 = 300+100p \\ 295 + 100p + 5p^2 - 300 - 100p = 0 \\ 5p^2 - 5 =0 \\ p^2 -1 =0 \\(p+1)(p-1) = 0$

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

p=-1

Question 11. 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	Not available	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 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.

x:	10	30	50	70	90
f:	17	f₁	32	f₂	19	Total = 120

Solution:

x f fx
10 17 170
30 f₁ 30f₁
50 32 1600
70 f₂ 70f₂
90 19 1710
N = 120 $\sum f_x = 3480+30f_1 + 70f_2$

Given,

Mean = 50

$\frac{\sum f_x}{N} = 50 \\ \frac{30f_1 + 70f_2 + 3480}{120} = 50 \\ 30f_1 + 70f_2 + 3480 = 6000 \\ 30f_1 + 70f_2 = 6000- 3480 \\ 30f_1 + 70f_2 = 2520 \\ 3f_! + 7f_2 = 252 ....(i)$

And, given value of N = 120

$17 + f_1 + 32+ f_2 + 19 = 120 \\ 68 + f_1 + f_2 = 120 \\ f_1 + f_2 = 52 \\ 3f_1+3f_2 = 156 ......(ii)$

Subtracting (ii) from (i) ,

$3f_1+7f_2-3f_1-3f_2 = 252-156 \\ 4f_2 = 96 \\ f_2 = 24$

Substituting f₂in (i)

$3f_1 + 168 = 252 \\ 3f_1 = 252-168 = 84 \\ f_1 = 28$

Class 9 RD Sharma Solutions - Chapter 24 Measures of Central Tendency - Exercise 24.3

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Article Tags :

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	$\sum f_x = 2650$

x	f	fx
10	3	30
15	10	150
p	25	25p
25	7	175
35	5	175
	N = 50	$\sum f_x = 25p + 530$

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	$\sum f_x = 24p + 1228$

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	$\sum f_x = 7p + 524$

x	f	fx
3	6	18
5	8	40
7	15	105
9	p	9p
11	8	88
13	4	52
	N = p +41	$\sum f_x = 9p+303$

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	$\sum f_x =7p + 524$

x	f	fx
3	6	18
5	8	40
7	15	105
9	p	9p
11	8	88
13	4	52
	N = p + 41	$\sum f_x =9p + 303$

x	f	fx
15	2	30
17	3	51
19	4	76
20+p	5p	100p+5p²
23	6	138
	N = 15+5p	$\sum f_x = 295+100p+5p^2$

x	f	fx
10	17	170
30	f₁	30f₁
50	32	1600
70	f₂	70f₂
90	19	1710
	N = 120	$\sum f_x = 3480+30f_1 + 70f_2$

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

Question 1. Calculate the mean for the following distribution:

Question 2. Calculate the mean for the following distribution:

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

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

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

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

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

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

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

Question 11. Candidates of four schools appear in a mathematics test. The data were as follows:

If the average score of the candidates of all the four schools is 66, find the number of candidates that appeared from school III.

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

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