JAMB - Mathematics (1995 - No. 37)
\(\begin{array}{c|c} \text{Class Interval} & Frequency & \text{Class boundaries} & Class Mid-point \\ \hline 1.5 - 1.9 & 2 & 1.45 - 1.95 & 1.7\\ 2.0 - 2.4 & 1 & 1.95 - 2.45 & 2.2\\ 2.5 - 2.9 & 4 & 2.45 - 2.95 & 2.7 \\ 3.0 - 2.9 & 15 & 2.95 - 3.45 & 3.2\\ 3.5 - 3.9 & 10 & 3.45 - 3.95 & 3.7\\ 4.0 - 4.4 & 5 & 3.95 - 4.45 & 4.2\\ 4.5 - 4.9 & 3 & 4.45 - 4.95 & 4.7\end{array}\)
Find the mode of the distribution.
Find the mode of the distribution.
3.2
3.3
3.7
4.2
Explanation
Mode = a + \(\frac{ (b - a)(fm - Fb)}{2Fm - Fa - Fb}\)
= 3.0 + \(\frac{(3.45 - 2.95)(15 - 4)}{2(15) - 4 - 10}\)
= 3 + \(\frac{(0.5)(11)}{30 - 14}\)
= 3 + \(\frac{5.5}{16}\)
= 3 + 0.34375
= 3.34375
= 3.3cm
= 3.0 + \(\frac{(3.45 - 2.95)(15 - 4)}{2(15) - 4 - 10}\)
= 3 + \(\frac{(0.5)(11)}{30 - 14}\)
= 3 + \(\frac{5.5}{16}\)
= 3 + 0.34375
= 3.34375
= 3.3cm
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