Class 9 RD Sharma Solutions – Chapter 24 Measures of Central Tendency – Exercise 24.2
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 Now, mean =
= 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 Now, mean =
= 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 Now, mean =
= (25p + 530)/50
Given,
Mean = 20.6
Solving, we get,
20.6 = (25p + 530)/50
25p + 530 = 1030
25p = 1030 âˆ’ 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 Mean =
= (10p + 445)/(p + 27)
Give,
Mean = 15
Solving, (10p + 445)/(p + 27) = 15
10p + 445 = 15(p + 27)
10p â€“ 15p = 405 â€“ 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 Now, mean =
= (24p + 1228)/100
Given,
Mean = 16.6
Solving, (24p + 1228)/100 = 16.6
24p + 1228 = 1660
24p = 1660 â€“ 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 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 â€“ 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 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 âˆ’ 7.68p = 314.88 âˆ’ 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 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 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.
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+5p^{2} 23 6 138 N = 15+5p Given,
Mean = 20
Mean =
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
âˆ´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
âˆ´10340 + 55p = (148 + p) x 66 = 9768 + 66p
=> 10340 â€“ 9768 = 66p â€“ 55p
=> 572 = 11p
âˆ´ 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 | f_{1} | 32 | f_{2} | 19 | Total = 120 |
Solution:
x f fx 10 17 170 30 f_{1} 30f_{1} 50 32 1600 70 f_{2} 70f_{2} 90 19 1710 N = 120 Given,
Mean = 50
And, given value of N = 120
Subtracting (ii) from (i) ,
Substituting f_{2 }in (i)
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