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NCERT Class XI Mathematics - Statistics - Solutions

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Question : 31 of 34
Marks: +1, -0
Given that x‾\overline{x} is the mean and σ2\sigma^2 is the variance of n observations x1,x2x_1, x_2, ......., xnx_n. Prove that the mean and variance of the observations ax1,ax2,ax3ax_1, ax_2, ax_3,......, axnax_n are ax and a2σ2a^2\sigma^2, respectively, (a ≠ 0).
Solution:  
Here x‾\overline{x} = x1+x2+x3+⋯+xnn\frac{x_1+x_2+x_3+\dots+x_n}{n} = Σxn\frac{\Sigma x}{n}
Also , x12+x22+x32+⋯+xn2n\frac{x_1^2+x_2^2+x_3^2+\dots+x_n^2}{n} = Σx2n\frac{\Sigma x^2}{n}
New mean = ax1+ax2+ax3+⋯+axnn\frac{ax_1+ax_2+ax_3+\dots+ax_n}{n}
= a(x1+x2+x3+⋯+xn)n\frac{a(x_1+x_2+x_3+\dots+x_n)}{n} = ax‾\overline{ax}
Also , σ2\sigma^2 =
n(x12+x22+⋯+xn2)−(x1+x2+⋯+xn)2n2\frac{n(x_1^2+x_2^2+\dots+x_n^2)-(x_1+x_2+\dots+x_n)^2}{n^2}
∴ New variance =
n(a2x12+a2x22+a2x32+⋯+a2xn2)−(ax1+ax2+ax3+⋯+axn)2n2\frac{n(a^2x_1^2+a^2x_{22}+a^2x_3^2+\dots+a^2x_n^2)-(ax_1+ax_2+ax_3+\dots+ax_n)^2}{n^2}
=
a2[n(x12+x22+x32+⋯+xn2)−(x1+x2+x3+⋯+xn)2n2]=a2σ2a^2\left[\frac{n(x_1^2+x_2^2+x_3^2+\dots+x_n^2)-(x_1+x_2+x_3+\dots+x_n)^2}{n^2}\right] = a^2\sigma^2
.
Hence proved
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