NIMCET 2021 Question Paper with solutions

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Question : 22 of 120

Marks: +1, -0

Consider the following frequency distribution tables.

Class Interval	10-20	20-30	30-40	40-50	50-60	60-70	70-80
Frequency	180	$f_1$	34	180	136	$f_2$	50

If the total frequency is 686 and the median is

42.6

, then the value of

f_{1}

and

{f}_{2}

are

81,25
82,24
83,23
84,22

Solution:

Class	Frequency	C.F.
10-20	180	180
20-30	$f_1$	$180+F_{1}$
30-40	34	$214+F_{1}$
40-50	180	$394+F_{1}$
50-60	136	$530+F_{1}$
60-70	$f_2$	$530+F_{1}+F_{2}$
70-80	50	$580+F_{1}+F_{2}$
	686

Now formula Median

=L_{1}+{{N}/{2}-C}/{F} \×i

42.6=40+{{686}/{2}-(214+F_{1})}/{180} \×10

2.6={343-214-F_{1}}/{18}

46.8=129-F_{1}

F_{1}=82.2

{F}_{1}=82

(approx)
Also

{F}_{1}+{F}_{2}=106

{F}_{2}=106-82

{F}_{2}=24

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