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Question

Given the following frequency distribution with some missing frequencies
Class10−2020−3030−4040−5050−6060−7070−80Frequency1803418013650
If the total frequency is 685 and approximate value of median is 42.6, then the approximate values for missing frequencies are

A
87,23
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B
80,25
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C
82,24
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D
83,22
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Solution

The correct option is C 82,24
Let missing frequincies be x,y
class1020203030304050506060707080Frequency180x34180136y50C.F180180+x214+x394+x530+x530+x+y580+x+y
Given: 580+x+y=685 x+y=105 (1)
Now,
N2=6852=342.5
Median=l+⎜ ⎜ ⎜ ⎜(N2)Ff⎟ ⎟ ⎟ ⎟×h
Here, Median class is 4050
l= lower limit of median class=40
N= total frequency=685
F= C.F of class just preceeding median class=214+x
f= frequency of median class=180
h= width of median class=10
putting these values in above equation
42.6=40+(342.5214x)180×10
x=81.7
y=10581.7=23.3
As frequencies are always whole numbers ,
missing frequencies are 82,24

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