NIMCET 2023 Question Paper with solutions

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Question : 45 of 120

Marks: +1, -0

Consider the following frequency distribution table: If the total frequency is 685 and the median is 42.6 then the possible values of

f_1

and

f_2

are:

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

82,23
80,25
83,22
79,26

Solution:

Class	Frequency	Cumulative Frequency
$10-20$	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	$580+f_1+f_2$

Total frequency

=685

580+f_1+f_2\;=685

f_1+f_2=105

Median

=42.6

(given) lies in

(40-50)

interval
Class

=40-50 .

Median

=l+\;{\;{ {\N}}/{2}-\;\text " C.f "\;}/{f} × h

\;l=40, h=10

\;\;\text " C.f "\;=214+f_1

42.6=40+\;{\;{685}/{2}- (214+f_1) }/{180} × 10

\;2.6 × 18=342.5-214-f_1

\;f_1=81.7 ≃ 82

\;f_2=105-82

\;=23 .

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