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  • Find the mean and variance of the frequency distribution given below:x 1lessthanequaltoxlessthan3 f 6

Find the mean and variance of the frequency distribution given below:

\begin{array}{|l|l|l|l|l|} \hline \mathrm{x} & 1 \leq \mathrm{x}<3 & 3 \leq \mathrm{x}<5 & 5 \leq \mathrm{x}<7 & 7 \leq \mathrm{x}<10 \\ \hline \mathrm{f} & 6 & 4 & 5 & 1 \\ \hline \end{array}

 

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Frequency distribution is given. We have to find the mean and variance

Converting the ranges of x to groups the given table can be re written as

\begin{array}{|l|l|l|l|l|} \hline \mathrm{X} \text { (Class) } & \text { Size }\left(x_{i}\right) & \text { Frequency }\left(f_{i}\right) & f_{i} x_{i} & f_{i} x_{i}^{2} \\ \hline 1-3 & 2 & 6 & 12 & 24 \\ \hline 3-5 & 4 & 4 & 16 & 64 \\ \hline 5-7 & 6 & 5 & 30 & 180 \\ \hline 7-10 & 8.5 & 1 & 8.5 & 72.25 \\ \hline \text { Total } & & \mathrm{N}=16 & =66.5 & 340.25 \\ \hline \end{array}

And, we know that variance is  \\\\ ~ \sigma ^{2}= \frac{ \Sigma f_{i}x_{i}^{2}}{n}- \left( \frac{ \Sigma f_{i}x_{i}}{n} \right) ^{2}~ \\\\


\\ \text{~ Substituting values from above table, we get } \sigma ^{2}=\frac{340.25}{16}- \left( \frac{66.5}{16} \right) ^{2}~~~ \\\\ \\ ~ \sigma ^{2}=21.265- \left( 4.16 \right) ^{2}~=21.265-17.305=3.96~ \\\\ \\ ~mean~\overline{X~}=\frac{ \Sigma f_{i}x_{i}}{ \Sigma f_{i}}=\frac{66.5}{16}=4.16~ \\\\ \\ \text{ Hence, the mean and variance of the given frequency distribution is 4.16 and 3.96 respectively } \\\\

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