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If the median of the data $6,7, x-2, x, 18$ and 21 written in ascending order is 16 , then the variance of that data is
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Verified Answer
The correct answer is:
$31 \frac{1}{3}$
Since, there are six observations and are written in ascending order then
$\begin{aligned}
& \text { Median }=\left(\frac{n}{2}\right)^{\text {th }} \text { observation }+\left(\frac{n}{2}+1\right)^{\text {th }} \frac{\text { observation }}{2} \\
& \frac{x-2+x}{2}=16 \\
& 2 x-2=32 \\
& 2 x=34 \Rightarrow x=17
\end{aligned}$
So the observations are $6,7,15,17,18$ and 21 .
$\begin{aligned}
& \therefore \quad \text { Mean }=\frac{6+7+15+17+18+21}{6} \\
& \bar{x}=\frac{84}{6}=14
\end{aligned}$
$\therefore \quad$ Variance $=\frac{1}{n} \sum\left(x_i-x\right)^2=\frac{188}{6}=\frac{94}{3}=31 \frac{1}{3}$
$\begin{aligned}
& \text { Median }=\left(\frac{n}{2}\right)^{\text {th }} \text { observation }+\left(\frac{n}{2}+1\right)^{\text {th }} \frac{\text { observation }}{2} \\
& \frac{x-2+x}{2}=16 \\
& 2 x-2=32 \\
& 2 x=34 \Rightarrow x=17
\end{aligned}$
So the observations are $6,7,15,17,18$ and 21 .
$\begin{aligned}
& \therefore \quad \text { Mean }=\frac{6+7+15+17+18+21}{6} \\
& \bar{x}=\frac{84}{6}=14
\end{aligned}$
$\therefore \quad$ Variance $=\frac{1}{n} \sum\left(x_i-x\right)^2=\frac{188}{6}=\frac{94}{3}=31 \frac{1}{3}$
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