1.

Find the missing frequencies in the following frequency distribution if it is known that the mean of the distribution 50.x: 1030507090y: 17f132f219

Answer»
Xyyx
1017170
30f130f1
50321600
70f270f2
90191710
 N = 120 \(\sum\)yx = 30f1 + 70f2 + 3480

Given, 

mean = 50

\(\frac{\sum yx}{N}=50\)

\(\frac{30f1+70f2+3480}{120}=50\)

30f1 + 70f2 + 3480 = 50 x 120

30f1 + 70f2 + 3480 = 6000 (i)

Also, ∑y = 120

17 + f1 + 32 + f2 + 19 = 120

f1 + f2 = 52

f1 = 52 – f2

Substituting value of f1 in (i), we get

30 (52 – f2) + 70f2 + 3480 = 6000

40f2 = 960

f2 = 24

Hence, 

f1 = 52 – 24 = 28

Therefore, 

f1 = 28 and f2 = 24



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