If the median of the data 6, 7, x-2, x, 18 an | vedclass

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(B) Given data: $6, 7, x-2, x, 18, 21$ (in ascending order).Number of observations $n = 6$ (even).Median = $\frac{(\frac{n}{2})^{	ext{th}} 	ext{ obs

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$31 \frac{1}{3}$

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(B) Given data: $6, 7, x-2, x, 18, 21$ (in ascending order).Number of observations $n = 6$ (even).Median = $\frac{(\frac{n}{2})^{	ext{th}} 	ext{ observation} + (\frac{n}{2}+1)^{	ext{th}} 	ext{ observation}}{2} = 16$.$\frac{(x-2) + x}{2} = 16 \Rightarrow 2x - 2 = 32 \Rightarrow 2x = 34 \Rightarrow x = 17$.The data set is $6, 7, 15, 17, 18, 21$.Mean $\bar{x} = \frac{6+7+15+17+18+21}{6} = \frac{84}{6} = 14$.Variance $\sigma^2 = \frac{1}{n} \sum (x_i - \bar{x})^2$.Calculations:$|\begin{array}{|c|c|c|} \hline x_i & x_i - \bar{x} & (x_i - \bar{x})^2 \ \hline 6 & -8 & 64 \ \hline 7 & -7 & 49 \ \hline 15 & 1 & 1 \ \hline 17 & 3 & 9 \ \hline 18 & 4 & 16 \ \hline 21 & 7 & 49 \ \hline 	ext{Sum} & & 188 \ \hline \end{array}|$Variance = $\frac{188}{6} = \frac{94}{3} = 31 \frac{1}{3}$.

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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