Question

Given that \(\bar{x}\) is the mean and \(σ2\) is the variance of n observations x₁,x₂,...,x_n. Prove that the mean and variance of the observations ax₁,ax₂,ax₃,...,ax_n are \(a\bar{x}\) and \(a^2σ2\), respectively,\((a ≠ 0).\)

Answer 1

The given n observations are x1, x2…x_n

\(Mean=\bar{x}\)

\(Varience= if^2\)

\(σ^2=\frac{1}{n}\sum_{i=1}^n(x_i-x-\bar{x})^2\,..........(1)\)

\(3×8\)

\(=24 ....[(Using(1)]\) 

\(Standard\,deviation\,σ=√\frac{1}{n}\sum_{ti1}^6(x_i-\bar{x})^2\)

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

\(y_i=ax,\,i.e.,\,x_i=\frac{1}{a}y_1\)

∴ \(\bar{y}\frac{1}{n}\sum_{i=1}^ny_i=\frac{1}{2}\sum_{i=1}^nax_i=\frac{a}{n}\sum_{i=1}^nx_i=a\bar{x}\,\,(\bar{x}=\frac{1}{n}\sum_{i=1}^nx_i)\)

Therefore, mean of the observations, ax₁, ax₂ ….ax_n, is \(a\bar{x}\),

Substituting the values of x and \(\bar{x}\) in (1), we obtain

\(σ^2=\frac{1}{2}\sum_{i=1}^n(\frac{1}{a}y_i-\frac{1}{a}y_i)^2\)

⇒ \(σ^2σ^2=\frac{1}{n}\sum_{i=1}^n(y_i-\bar{y})^2\)

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