\(\begin{array}{c|c} \text{Class Interval} & Frequency & \text{Class boundaries} & Class Mid-point \\ \hline 1.5 -...

JAMB 1995

\(\begin{array}{c|c} \text{Class Interval} & Frequency & \text{Class boundaries} & Class Mid-point \\ \hline 1.5 - 1.9 & 2 & 1.45 - 1.95 & 1.7\\ 2.0 - 2.4 & 21 & 1.95 - 2.45 & 2.2\\ 2.5 - 2.9 & 4 & 2.45 - 2.95 & 2.7 \\ 3.0 - 2.9 & 15 & 2.95 - 3.45 & 3.2\\ 3.5 - 3.9 & 10 & 3.45 - 3.95 & 3.7\\ 4.0 - 4.4 & 5 & 3.95 - 4.45 & 4.2\\ 4.5 - 4.9 & 3 & 4.45 - 4.95 & 4.7\end{array}\)
The median of the distribution above is

A. 4.0
B. 3.4
C. 3.2
D. 3.0

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Correct Answer: Option B

Explanation

Median = \(\frac{a + b}{fm}\) (\(\frac{1}{2} \sum f - CF_b\))

= 2.95 + \(\frac{0.5}{15}\)(2.-7)

= 2.95 + \(\frac{0.5}{15}\) x 13

= 2.95 + 0. 43

= 3.38

= 3.4

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