If the mean of the following frequency distribution is 54, find the value of p.
Class | 0 - 20 | 20 - 40 | 40 - 60 | 60 - 80 | 80 - 100 |
Frequency | 7 | p | 10 | 9 | 13 |
For equal class intervals, we will solve by finding mid points of these classes using direct method.
CLASS | MID - POINT(xi) | FREQUENCY(fi) | fixi |
0 - 20 | 10 | 7 | 70 |
20 - 40 | 30 | p | 30p |
40 - 60 | 50 | 10 | 500 |
60 - 80 | 70 | 9 | 630 |
80 - 100 | 90 | 13 | 1170 |
TOTAL | 39 + p | 2370 + 30p |
We have got
Σfi = 39 + p and Σfixi = 2370 + 30p
∵ mean is given by
⇒ 
(∵ given: mean of pocket allowance is 54)
⇒ 2106 + 54p = 2370 + 30p
⇒ 54p – 30p = 2370 – 2106
⇒ 24p = 264
⇒ p = 11
Thus, p is 11.
10