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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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Verified Answer
The correct answer is:
225
Let $\bar{x}$ be the mean of 4 number
$$
\begin{aligned}
& \bar{x}=\frac{w+x+y+z}{4} \\
& \text { New Mean }=\frac{5 w+5 x+5 y+5 z}{4} \\
& =5\left[\frac{w+x+y+z}{4}\right]=5 \bar{x} \\
& \text { Variance }=\frac{\sum x^2}{n}-\left(\frac{\bar{x}}{n}\right)^2 \\
& a=\frac{w^2+x^2+y^2+z^2}{4}-\left(\frac{\bar{x}}{4}\right)^2 \\
& \text { New variance }=\frac{25 w^2+25 x^2+25 y^2+25 z^2}{4}-\frac{25 \bar{x}^2}{16} \\
& =25\left[\frac{w^2+x^2+y^2+z^2}{4}-\left(\frac{\bar{x}}{4}\right)^2\right] \\
& =25 \times 9=225 \\
&
\end{aligned}
$$
$$
\begin{aligned}
& \bar{x}=\frac{w+x+y+z}{4} \\
& \text { New Mean }=\frac{5 w+5 x+5 y+5 z}{4} \\
& =5\left[\frac{w+x+y+z}{4}\right]=5 \bar{x} \\
& \text { Variance }=\frac{\sum x^2}{n}-\left(\frac{\bar{x}}{n}\right)^2 \\
& a=\frac{w^2+x^2+y^2+z^2}{4}-\left(\frac{\bar{x}}{4}\right)^2 \\
& \text { New variance }=\frac{25 w^2+25 x^2+25 y^2+25 z^2}{4}-\frac{25 \bar{x}^2}{16} \\
& =25\left[\frac{w^2+x^2+y^2+z^2}{4}-\left(\frac{\bar{x}}{4}\right)^2\right] \\
& =25 \times 9=225 \\
&
\end{aligned}
$$
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