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RD Chapter 24 Measures of Central Tendency Ex 24.2 Solutions

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

Answer - 1 : -


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

Answer - 2 : -

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

Answer - 3 : -

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.

Answer - 4 : -

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

Answer - 5 : - 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

Answer - 6 : - 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

Answer - 7 : - 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

Answer - 8 : - 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

Answer - 9 : - Mean = 7.68

Question - 10 : -
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.

Answer - 10 : -

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 = 572/11
Number of candidates in school III = 52

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