Mathematics › Statistics › Measures of Dispersion
The variance of the following continuous frequency distribution is \(\begin{array}{lllll} \hline \text {…
The variance of the following continuous frequency distribution is
\(\begin{array}{lllll} \hline \text { Class Interval } & 0-10 & 10-20 & 20-30 & 30-40 \\ \hline \text { Frequency } & 2 & 3 & 4 & 1 \\ \hline \end{array}\)
201 62 19 84 Solution
Given
\(\begin{array}{llllll}
\hline \begin{array}{l}
\text { Class } \\
\text { Interval }
\end{array} & \begin{array}{c}
\text { Frequency } \\
\left(\boldsymbol{f}_{\boldsymbol{i}}\right)
\end{array} & \mathbf{x}_{\boldsymbol{i}} & \mathbf{x}_{\boldsymbol{i}} \boldsymbol{f}_{\boldsymbol{i}} & \left(\overline{\boldsymbol{x}}-\mathbf{x}_{\boldsymbol{i}}\right)^2 & \boldsymbol{f}_{\boldsymbol{i}}\left(\overline{\boldsymbol{x}}-\mathbf{x}_{\boldsymbol{i}}\right)^2 \\
\hline 0-10 & 2 & 5 & 10 & 196 & 392 \\
\hline 10-20 & 3 & 15 & 45 & 16 & 48 \\
\hline 20-30 & 4 & 25 & 100 & 36 & 144 \\
\hline 30-40 & 1 & 35 & 35 & 256 & 256 \\
\hline & N=\Sigma f_i=10 & & \Sigma \times f_i=190 & \Sigma f_i\left(\overline{\mathbf{x}}-x_i\right)^2=840 \\
\hline
\end{array}\)
\(\begin{array}{ll}
\because & \bar{x}=\frac{\Sigma x_i f_i}{N}=\frac{190}{10}=19 \\
\therefore \text { Variance }(\sigma)^2 & =\frac{1}{N} \Sigma f_i\left(\bar{x}-x_i\right)^2=\frac{1}{10}(840)=84
\end{array}\)
Hence, option (4) is correct.
Asked in: AP EAMCET 2019 (20 Apr Shift 1)
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