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\(\begin{array}{c|c} \text{Class Interval} & Frequency & \te...

Class IntervalFrequencyClass boundariesClassMidpoint1.51.921.451.951.72.02.4211.952.452.22.52.942.452.952.73.02.9152.953.453.23.53.9103.453.953.74.04.453.954.454.24.54.934.454.954.7
The median of the distribution above is

A.

4.0

B.

3.4

C.

3.2

D.

3.0

Correct answer is B

Median = a+bfm (12fCFb)

= 2.95 + 0.515(2.-7)

= 2.95 + 0.515 x 13

= 2.95 + 0. 43

= 3.38

= 3.4