$\begin{array}{c|c} x & 1 & 2 & 3 & 4 & 5 \\ \hline f & 2 & 1 & 2 & 1 & 2\end{array}$
Find the variance of the frequency distribution above

$\frac{3}{2}$

$\frac{9}{4}$

$\frac{5}{2}$

Correct answer is B

$\begin{array}{c|c} x & f & fx & \bar{x} - x & (\bar{x} - x)^2 & f(\bar{x} - x)^2 \\ \hline 1 & 2 & 2 & -2 & 4 & 8\\ 2 & 1 & 2 & -1 & 1 & 1\\ 3 & 2 & 6 & 0 & 0 & 0\\ 4 & 1 & 4 & 1 & 1 & 1\\ 2 & 2 & 10 & 2 & 4 & 8\\ \hline & 8 & 24 & & & 18 \end{array}$

x = $\frac{\sum fx}{\sum f}$

= $\frac{24}{8}$

= 3

Variance (6²) = $\frac{\sum f(\bar{x} - x)^2}{\sum f}$

= $\frac{18}{8}$

= $\frac{9}{4}$

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