If the variance of four numbers $w, x, y$ and $z$ is 9 , then the variance of $5 w, 5 x, 5 y$ and $5 z$ is

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Aicharya · Accepted Answer

Let $\bar{x}$ be the mean of 4 number $ \begin{aligned} & \bar{x}=\frac{w+x+y+z}{4} \ & 	ext { New Mean }=\frac{5 w+5 x+5 y+5 z}{4} \ & =5\left[\frac{w+x+y+z}{4}ight]=5 \bar{x} \ & 	ext { Variance }=\frac{\sum x^2}{n}-\left(\frac{\bar{x}}{n}ight)^2 \ & a=\frac{w^2+x^2+y^2+z^2}{4}-\left(\frac{\bar{x}}{4}ight)^2 \ & 	ext { New variance }=\frac{25 w^2+25 x^2+25 y^2+25 z^2}{4}-\frac{25 \bar{x}^2}{16} \ & =25\left[\frac{w^2+x^2+y^2+z^2}{4}-\left(\frac{\bar{x}}{4}ight)^2ight] \ & =25 	imes 9=225 \ & \end{aligned} $

If the variance of four numbers $w, x, y$ and $z$ is 9 , then the variance of $5 w, 5 x, 5 y$ and $5 z$ is

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