\(\begin{array}{c|c} x & 1 & 2 & 3 & 4 & 5 \\ \hline f & 2 & 1 & 2 & 1 & 2\end{array}\)
Find the variance of the frequency distribution above
Find the variance of the frequency distribution above
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Correct Answer: Option B
Explanation:
\(\begin{array}{c|c} x & f & fx & \bar{x} - x & (\bar{x} - x)^2 & f(\bar{x} - x)^2 \\ \hline 1 & 2 & 2 & -2 & 4 & 8\\ 2 & 1 & 2 & -1 & 1 & 1\\ 3 & 2 & 6 & 0 & 0 & 0\\ 4 & 1 & 4 & 1 & 1 & 1\\ 2 & 2 & 10 & 2 & 4 & 8\\ \hline & 8 & 24 & & & 18 \end{array}\)
x = \(\frac{\sum fx}{\sum f}\)
= \(\frac{24}{8}\)
= 3
Variance (62) = \(\frac{\sum f(\bar{x} - x)^2}{\sum f}\)
= \(\frac{18}{8}\)
= \(\frac{9}{4}\)
\(\begin{array}{c|c} x & f & fx & \bar{x} - x & (\bar{x} - x)^2 & f(\bar{x} - x)^2 \\ \hline 1 & 2 & 2 & -2 & 4 & 8\\ 2 & 1 & 2 & -1 & 1 & 1\\ 3 & 2 & 6 & 0 & 0 & 0\\ 4 & 1 & 4 & 1 & 1 & 1\\ 2 & 2 & 10 & 2 & 4 & 8\\ \hline & 8 & 24 & & & 18 \end{array}\)
x = \(\frac{\sum fx}{\sum f}\)
= \(\frac{24}{8}\)
= 3
Variance (62) = \(\frac{\sum f(\bar{x} - x)^2}{\sum f}\)
= \(\frac{18}{8}\)
= \(\frac{9}{4}\)