Given that x̅ is the mean and σ2 is the variance of n observations x1, x2, …,xn .

From the question it is given that, n observations are x₁, x₂,…..x_n

Mean of the n observation = x̅

Variance of the n observation = σ²

As we know that,

Variance = \(σ^2= \dfrac{1}{n} \displaystyle\sum_{i=1}^{n} y_i (x_i-\overline {x} )^2 \)..............(i)

If each observation is multiplied by a and the new observations are y_i, then

y_i = ax_i

Thus, \( x_i =\dfrac{1}{a}y_i\) 

\( \overline{y}=\dfrac{1}{n} \displaystyle\sum_{i=1}^{a} y_i\)

\( \overline{y}=\dfrac{1}{n} \displaystyle\sum_{i=1}^{a} ax_i\)

\( \overline{y}=\dfrac{a}{n} \displaystyle\sum_{i=1}^{a} x_i\)

\( \overline{y}=a\overline x\)

Therefore, mean of the observations, ax₁, ax₂ … ax_n, is a  \( \overline{x}\) Substituting the values of x_i and \( \overline{x}\) in (1), we obtain

\( σ^2= \dfrac{1}{n} \displaystyle\sum_{i=1}^{n}(\dfrac{1}{a}y_1-\dfrac{1}{a}\overline{y})^2 \)

\( a^2σ^2= \dfrac{1}{n} \displaystyle\sum_{i=1}^{n}(y_1-\overline{y})^2\)

Thus, the variance of the observations, ax₁, ax₂ …ax_n, is a² σ².