The coefficient of variation of the following distribution is \(\begin{array}{l|c|c|c|c|c} \hline \text…
The coefficient of variation of the following distribution is
\(\begin{array}{l|c|c|c|c|c}
\hline \text {Class interval } & 0-5 & 5-10 & 10-15 & 15-20 & 20-25 \\
\hline \text {Frequency } & 4 & 1 & 10 & 3 & 2 \\
\hline
\end{array}\)
- \(\frac{25 \sqrt{139}}{12}\)
- \(\frac{25 \sqrt{139}}{6}\)
- \(\frac{139}{6}\)
- \(\frac{25 \times 139}{12}\)
Solution
According to given data,
\(\begin{array}{ccccc}
\hline \begin{array}{c}
\text {Class } \\
\text {Interval }
\end{array} & \begin{array}{c}
\text {Frequency } \\
\boldsymbol{f}_{\boldsymbol{i}}
\end{array} & \boldsymbol{x}_{\boldsymbol{i}} & \boldsymbol{x}_{\boldsymbol{i}} \boldsymbol{f}_{\boldsymbol{i}} & \left|\overline{\boldsymbol{x}}-\boldsymbol{x}_{\boldsymbol{i}}\right| \\
\hline 0-5 & 4 & 2.5 & 10 & 9.5 \\
\hline 5-10 & 1 & 7.5 & 7.5 & 4.5 \\
\hline 10-15 & 10 & 12.5 & 125 & 0.5 \\
\hline 15-20 & 3 & 17.5 & 52.5 & 5.5 \\
\hline 20-25 & 2 & 22.5 & 45 & 10.5 \\
\hline & N=\Sigma f_i & & \Sigma x_i f_i & \\
& =20 & & =240 & \\
\hline
\end{array}\)
\(\begin{aligned}
\therefore \quad \bar{x} & =\frac{\Sigma x_i f_i}{N}=\frac{240}{20}=12 \\
\therefore \quad \sigma^2 & =\frac{\Sigma f_i\left(\bar{x}-x_i\right)^2}{N} \\
& =\frac{4(9.5)^2+1(4.5)^2+10(0.5)^2+3(5.5)^2+2(10.5)^2}{20} \\
& =\frac{139}{4} \\
\therefore \quad \mathrm{CV} & =\frac{\sigma}{\bar{x}} \times 100=\frac{\sqrt{139}}{2 \times 12} \times 100=\frac{25 \sqrt{139}}{6}
\end{aligned}\)
Hence, option (b) is correct.
Asked in: AP EAMCET 2019 (23 Apr Shift 1)
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