begin array c c text Class Interval amp 1 - 5 a... - JAMB Mathematics 1994 Question
(\begin{array}{c|c} \text{Class Interval} & 1 - 5 & 6 - 10 & 11 - 15 & 16 - 20 & 21 - 25 \ \hline Frequency & 6 & 15 & 20 & 7 & 2\end{array})
Estimate the median of the frequency distribution above
A
10\(\frac{1}{2}\)
B
11\(\frac{1}{2}\)
C
12
D
13
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Correct Option: C
Median = L + [(\frac{\frac{N}{2} - f}{fm})]h
N = Sum of frequencies
L = lower class boundary of median class
f = sum of all frequencies below L
fm = frequency of modal class and
h = class width of median class
Median = 11 + [(\frac{\frac{50}{2} - 21}{20})]5
= 11 + ((\frac{25 - 21}{20}))5
= 11 + ((\frac{(4)}{20}))
11 + 1 = 12
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