\(\begin{array}{c|c} \text{Class Interval} & Frequency & \text{Class boundaries} & Class Mid-point \\ \hline 1.5 - 1.9 & 2 & 1.45 - 1.95 & 1.7\\ 2.0 - 2.4 & 21 & 1.95 - 2.45 & 2.2\\ 2.5 - 2.9 & 4 & 2.45 - 2.95 & 2.7 \\ 3.0 - 2.9 & 15 & 2.95 - 3.45 & 3.2\\ 3.5 - 3.9 & 10 & 3.45 - 3.95 & 3.7\\ 4.0 - 4.4 & 5 & 3.95 - 4.45 & 4.2\\ 4.5 - 4.9 & 3 & 4.45 - 4.95 & 4.7\end{array}\)
Find the mode of the distribution above to find the mode of the distribution.
A. 3.2
B. 3.3
C. 3.7
D. 4.2
Correct Answer: B
Explanation
Mode = a + (b - a)(fm - Fb)
2Fm - Fa - Fb
= 3.0 + \(\frac{(3.4 - 3)(15 - 4)}{2(15) - 4 - 10}\)
= 3 + \(\frac{(6.4)(11)}{30 - 14}\)
= 3 + \(\frac{4.4}{16}\)
= 3 + 0.275
= 3.275
= 3.3cm
