Find the missing frequencies f1 and f2 in the table given below, it being given that the mean of the given frequency distribution is 50.
Class | 0 - 20 | 20 - 40 | 40 - 60 | 60 - 80 | 80 - 100 | Total |
Total Frequency | 17 | f1 | 32 | f2 | 19 | 120 |
For equal class intervals, we will solve by finding mid points of these classes using direct method.
CLASS | MID - POINT(xi) | TOTAL FREQUENCY(fi) | fixi |
0 - 20 | 10 | 17 | 170 |
20 - 40 | 30 | f1 | 30f1 |
40 - 60 | 50 | 32 | 1600 |
60 - 80 | 70 | f2 | 70f2 |
80 - 100 | 90 | 19 | 1710 |
TOTAL | 68 + f1 + f2 | 3480 + 30f1 + 70f2 |
We have got
Mean = 50 and N = 120 (as given in the question)
Σfi = 68 + f1 + f2 and Σfixi = 3480 + 30f1 + 70f2
∵ mean is given by
⇒ 
(∵ given: mean is 50)
⇒ 3400 + 50f1 + 50f2 = 3480 + 30f1 + 70f2
⇒ 50f1 – 30f1 + 50f2 – 70f2 = 3480 – 3400
⇒ 20f1 – 20f2 = 80
⇒ f1 – f2 = 4 …(i)
As given in the question, frequency(Σfi) = 120
And as calculated by us, frequency (Σfi) = 68 + f1 + f2
Equalizing them, we get
68 + f1 + f2 = 120
⇒ f1 + f2 = 120 – 68 = 52
⇒ f1 + f2 = 52 …(ii)
We will now solve equations (i) and (ii), adding them we get
(f1 + f2) + (f1 – f2) = 52 + 4
⇒ 2f1 = 56
⇒ f1 = 56/2
⇒ f1 = 28
Substitute f1 = 28 in equation (ii),
28 + f2 = 52
⇒ f2 = 52 – 28
⇒ f2 = 24
Thus, f1 = 28 and f2 = 24.