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Question

If the arithmetic mean of the following frequency distribution is 57.6 and the sum of the observations is 50, then the missing frequencies x and y are

Classes 0 - 20 20 - 40 40 - 60 60 - 80 80 - 100 100 - 120
Frequency 7 x 12 7 y 5

A
x=10, y=8
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B
x=8, y=10
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C
x=9, y=10
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D
x=10, y=9
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Solution

The correct option is C x=9, y=10
Classes Frequency fi Mid-points xi ui fiui
0 - 20 7 10 -2 -14
20 - 40 x 30 -1 -x
40 - 60 12 50 0 0
60 - 80 7 70 1 7
80 - 100 y 90 2 2y
100 - 120 5 110 3 15
Total 31 + x + y 8 - x + 2y

Let the assumed mean, A = 50
No. of all observations = 50
⇒ 31 + x + y = 50
∴ x + y = 19 …(1)

MEAN =57.6A+6i=1fiui6i=1fi×h=57.6

50+(8x+2y)×2050=57.6

(8x+2y)×25=7.6

8x+2y=19

x+2y=11 ...(2)

Adding (1) and (2), we get:

x+y=19
x+2y=11
3y=30y=10

Substituting y = 10 in (1), we get:

x+y=19

x+10=19

x=9

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