The mean of the following data is 42. Find the missing frequencies x and y if the Sum of frequencies is 100.
Class interval | 0 - 10 | 10 - 20 | 20 - 30 | 30 - 40 | 40 - 50 | 50 - 60 | 60 - 70 | 70 - 80 |
Frequency | 7 | 10 | x | 13 | y | 10 | 14 | 9 |
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 - 10 | 5 | 7 | 35 |
10 - 20 | 15 | 10 | 150 |
20 - 30 | 25 | x | 25x |
30 - 40 | 35 | 13 | 455 |
40 - 50 | 45 | y | 45y |
50 - 60 | 55 | 10 | 550 |
60 - 70 | 65 | 14 | 910 |
70 - 80 | 75 | 9 | 675 |
TOTAL | 63 + x + y | 2775 + 25x + 45y |
We have got
Σfi = 63 + x + y and Σfixi = 2775 + 25x + 45y
∵ mean is given by
⇒ 
(∵ given: mean of pocket allowance is 42)
⇒ 2646 + 42x + 42y = 2775 + 25x + 45y
⇒ 42x – 25x + 42y – 45y = 2775 – 2646
⇒ 17x – 3y = 129 …(i)
As given in the question, frequency(Σfi) = 100
And as calculated by us, frequency (Σfi) = 63 + x + y
Equalizing them, we get
63 + x + y = 100
⇒ x + y = 37 …(ii)
We will now solve equations (i) and (ii), multiply eq.(ii) by 3 and then add it to eq.(i), we get
(17x – 3y) + [3(x + y)] = 129 + 111
⇒ 17x – 3y + 3x + 3y = 240
⇒ 20x = 240
⇒ x = 12
Substitute x = 12 in equation (ii),
12 + y = 37
⇒ y = 37 – 12
⇒ y = 25
Thus, x = 12 and y = 25.