Question

Let x₁, x₂, ..., x_n be values taken by a variable X and y₁, y₂, ..., y_n be the values taken by a variable Y such that y_i = ax_i + b, i = 1, 2,..., n. Then,
(a) Var (Y) = a² Var (X)
(b) Var (X) = a² Var (Y)
(c) Var (X) = Var (X) + b
(d) none of these

Answer 1

$\left(a\right)Var\left(Y\right)=a^{2}Var\left(X\right)$

$$\begin{gathered} \mathrm{Var}\left(X\right)=\frac{{\displaystyle \sum _{i=1}^n}(x_i-{\displaystyle \stackrel{}{\overline{X}}}{)}^2}{n}\mathrm{where}\mathrm{Mean}\left(\overline{X}\right)=\frac{{\displaystyle \sum _{i=1}^n}{x}_i}{n} \\ \mathrm{Var}\left(Y\right)=\frac{{\sum }_{i=1}^n(y_i-{\displaystyle \overline{Y}{)}^2}}{n}\mathrm{and}\overline{Y}=\frac{{\displaystyle \sum _{i=1}^n}{y}_i}{n} \\ \mathrm{We}\mathrm{have}, \\ {y}_i=a{x}_i+b \\ \overline{Y}=\frac{{\sum }_{i=1}^n{y}_i}{n} \\ =\frac{{\sum }_{i=1}^{n}a{x}_i+b}{n} \\ =\frac{a{\sum }_{i=1}^n{x}_i}{n}+\frac{nb}{n} \\ =a\overline{X}+b \\ Var\left(Y\right)=\frac{{\displaystyle \sum _{i=1}^n}{\left(y_i-\overline{Y}\right)}^2}{n} \\ =\frac{\sum _{i=1}^n{\left\{a{x}_i+b-\left(a\overline{X}+b\right)\right\}}^2}{n} \\ =\frac{\sum _{i=1}^n(a{x}_i-a\overline{X}{)}^2}{n} \\ =a^2\frac{\sum _{i=1}^n(x_i-{\displaystyle \overline{X}{)}^2}}{n} \\ =a^{2}Var\left(X\right) \end{gathered}$$