\(\begin{array}{c|c} \text{Class Interval} & Frequency & \text{Class boundaries} & Class Mid-point \\ \hline 1.5 - 1.9 & 2 & 1.45 - 1.95 & 1.7\\ 2.0 - 2.4 & 21 & 1.95 - 2.45 & 2.2\\ 2.5 - 2.9 & 4 & 2.45 - 2.95 & 2.7 \\ 3.0 - 2.9 & 15 & 2.95 - 3.45 & 3.2\\ 3.5 - 3.9 & 10 & 3.45 - 3.95 & 3.7\\ 4.0 - 4.4 & 5 & 3.95 - 4.45 & 4.2\\ 4.5 - 4.9 & 3 & 4.45 - 4.95 & 4.7\end{array}\)
The median of the distribution above is
a
4.0
b
3.4
c
3.2
d
3.0
Explanation
Correct Option
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Davechill
4 weeks ago
Median = L + [((N/2) - c.f.) / f] * h
Where:
L = 2.95
c.f. = 27
f = 15
h = 0.5
Calculation:
Median = 2.95 + [(30 - 27) / 15] * 0.5
Median = 2.95 + [3 / 15] * 0.5
Median = 2.95 + [0.2] * 0.5
Median = 2.95 + 0.1
Median = 3.05
