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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:56789
f:4811143

Solution:

xffx
5420
6848
71498
81188
9327
 N = 40\sum f_x = 281

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

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= 281/40



= 7.025

Question 2. Calculate the mean for the following distribution:

x:19212325272931
f:13151618161513

Solution:

xffx
1913247
2115315
2316368
2518450
2716432
2915435
3113403
 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:1015p2535
f:3102575

Solution: 

xffx
10330
1510150
p2525p
257175
355175
 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:510152025
f:6105

Solution: 

xffx
5630
10p10p
15690
2010200
255125
 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:81215p202530
f:121620241684

Solution: 

xffx
81296
1216192
1520300
p2424p
2016320
258200
304120
 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:581012p2025
f:25822742

Solution:

xffx
5210
8540
10880
1222264
p77p
20480
25250
 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:35791113
f:6815p84

Solution: 

xffx
3618
5840
715105
9p9p
11888
13452
 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:581012p2025
f:25822742

Solution:

xffx
5210
8540
10880
1222264
p77p
20480
25250
 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:35791113
f:6815p84

Solution:

xffx
3618
5840
715105
9p9p
11888
13452
 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:15171920+p23
f:2345p6

Solution:

xffx
15230
17351
19476
20+p5p100p+5p2
236138
 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:

SchoolsNo. of CandidatesAverage Score
I6075
II4880
IIINot available55
IV4050

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:1030507090 
f:17f132f219Total = 120

Solution:

xffx
1017170
30f130f1
50321600
70f270f2
90191710
 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 f2  in (i)

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




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