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

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency Ex 24.2

Other Exercises

Question 1.
Calculate the mean for the following distribution
x 5 6 7 8 9
f 4 8 14 11 3
Solution:

Question 2.
Find the mean of the following data:
x 19 21 23 25 27 29 31
f 13 15 16 18 16 15 13
Solution:

Question 3.
Find the mean of the following distribution:
x 10 12 20 25 35
f 3 10 15 7 5
Solution:

Question 4.
Five coins were simultaneously tossed 1000 times and at each toss the number of heads were observed. The number of tosses during which 0, 1, 2, 3, 4 and 5 heads were obtained are shown in the table below. Find the mean number of heads per toss.

Solution:

Question 5.
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:
Mean = 20.6

Question 6.
If the mean of the following data is 15, find p?.
x 5 10 15 20 25
f 6 p 6 10 5
Solution:
Mean = 15

Question 7.
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:
Mean = 16.6

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:
Mean = 12.58

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:
Mean = 7.68

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:
Mean = 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.
Solution:
Let number of candidates in school III = p
Then total number of candidates in 4 schools = 60 + 48 + p + 40 = 148 + p
Average score of 4 schools = 66
∴Total score = (148 + p) x 66
Now mean score of 60 in school I = 75 .
∴Total = 60 x 75 = 4500
In school II, mean of 48 = 80
∴Total = 48 x 80 = 3840
In school III, mean of p = 55
∴Total = 55 x p = 55p
and in school IV, mean of 40 = 50
∴Total = 40 x 50 = 2000
Now total of candidates of 4 schools = 148 + p
and total score = 4500 + 3840 + 55p + 2000 = 10340 + 55p
∴10340 + 55p = (148 + p) x 66 = 9768 + 66p
=> 10340 – 9768 = 66p – 55p
=> 572 = 11p
∴ p = $$\frac { 572 }{ 11 }$$
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 f1 19
Total 120
Solution:
Mean = 50

Hope given RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency Ex 24.2 are helpful to complete your math homework.

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