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Maths-

General

Easy

Question

Compute the median from the following table

36.55
35.55
40.05
None of these

The correct answer is: 36.55

$n equals sum f equals 159$ . Here n = 159, which is odd
$therefore$ Median number $equals fraction numerator 1 over denominator 2 end fraction left parenthesis n plus 1 right parenthesis equals fraction numerator 1 over denominator 2 end fraction left parenthesis 159 plus 1 right parenthesis equals 80$ ,
which is in the class 30-40. (see the row of cumulative frequency 95, which contains 80).
Hence median class is 30-40.
\ We have,
l = Lower limit of median class = 30
f = Frequency of median class = 45
C = Total of all frequencies preceding median class
= 50
i = Width of class interval of median class = 10
\ Required median $equals l plus fraction numerator fraction numerator N over denominator 2 end fraction minus C over denominator f end fraction cross times i$
$equals 30 plus fraction numerator fraction numerator 159 over denominator 2 end fraction minus 50 over denominator 45 end fraction cross times 10 equals 30 plus fraction numerator 295 over denominator 45 end fraction equals 36.55$ .

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