Let x1, x2,........., xn be n observations, and let [ x ] be their arithematic mean and [ …

Question

Let x1, x2,........., xn be n observations, and let $$\overline x $$ be their arithematic mean and $${\sigma ^2}$$ be their variance.

Statement 1 : Variance of 2x1, 2x2,......., 2xn is 4$${\sigma ^2}$$.

Statement 2 : : Arithmetic mean of 2x1, 2x2,......, 2xn is 4$$\overline x $$.

Accepted Answer

Given that,

for ${x_1},{x_2},....{x_n},$ $$A.M = \overline x $$

and variance $$ = {\sigma ^2}$$

Now A.M of

$$2{x_1},2x{}_2.....2{x_n} = {{2\left( {{x_1} + {x_2} + ....{x_n}} \right)} \over n} = 2\overline x $$

But given $$A.M = 4\overline x $$

$$\therefore\,\,\,$$ Statement $${\rm I}{\rm I}$$ is false.

Variance of $2{x_1},2{x_2}......2{x_n}$

$=$ Variance of $$\left\{ {2{x_i}} \right\}$$

$= {2^2}$ Variance of $$\left\{ {{x_i}} \right\} = 4{\sigma ^2}$$

So, statement $${\rm I}$$ is correct.

Solutions · 1