Question

qυєsтıση :-
⠀
Given that $\sf{\overline{x}}$ is the mean and σ² is the variance of n observations $\bf{x_1 , x_2, ..., x_n}$.
Prove that the mean and variance of the observations $\bf{ax_1 , ax_2 , ax_3, ..., ax_n}$ are $\sf{a \overline{x}}$ and a²σ², respectively, (a ≠ 0).
⠀
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Answer 1

Given:

Mean $\bar{x}$ and variance $\sigma^2$ of following observations:

$x_1,x_2,x_3---------------------, x_n$

To Prove:

The mean and variance of the following observations :

$ax_1,ax_2,ax_3,------------, ax_n$

are $a\bar{x}\ and \, a^2\sigma^2$

Solution:

The solution is provided in the attached file

Concepts:

.$1. \,The\, mean \, of : x_1,x_2,x_3,---------------------, x_n\\\\\bar x = \frac {x_1+x_2+x_3,--------------+ x_n}{n}$

$2.\, The \, variance \, of : x_1,x_2,x_3,-------------- x_n\\\\ \sigma^2 = \frac {(x_1-\bar x)^2+(x_2-\bar x)^2+(x_3-\bar x)^2+------------------+( x_n-\bar x)^2}{n}$

Answer 2

$\large\underline\green{\bold{ \sf \: ANSWER}}$

$\begin{gathered}\begin{gathered}\bf \:Given - \begin{cases} &\sf{set \: of \: n \: observations \: \bf{x_1,x_2,x_3,- -, x_n}} \\ &\sf{having \: mean \: \overline{x} \: and \: variance \: { \sigma}^{2} } \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf \:To \: Prove - \begin{cases} &\sf{mean \: and \: variance \: of \: } \\ &\sf{\bf{ax_1 , ax_2 , ax_3, ..., ax_n} \: is}\\ &\sf{a\overline{x} \: and \: {a}^{2} {\sigma}^{2} } \end{cases}\end{gathered}\end{gathered}$

$\large\underline\purple{\bold{Solution :- }}$

We have given that

$\rm :\implies\:\overline{x} \: is \: mean \: of \: \bf{x_1,x_2,x_3, - - ,x_n} \: then$

$\rm :\implies\: \boxed{ \green{ \bf \: \overline{x} \: = \tt \: \dfrac{\sum_{i=1}^n \: x_i}{n} }}$

Also,

Given that

$\rm :\implies\: {\sigma}^{2} \: is \: variance \: of \: \bf{x_1,x_2,x_3, - - ,x_n} \: then$

$\rm :\implies\: \boxed{ \red{ \bf \: {\sigma}^{2} \: = \tt \:\dfrac{\sum_{i=1}^n \: {x_i}^{2} }{n} - {(\overline{x})}^{2} }}$

Now,

The new set of n observations are

$\bf{ax_1,ax_2,ax_3, - - - ,ax_n}$

$\bull \bf \: Let \: us \: assume \: that \: mean \: is \: \overline{ X}$

Then,

we know that

Mean is given by the formula

$\rm :\implies\:\overline{ X} \: = \: \dfrac{\sum_{i=1}^n \: ax_i}{n}$

$\rm :\implies\:\overline{ X} \: = \: \dfrac{a \: \sum_{i=1}^n \: x_i}{n}$

$\rm :\implies\:\overline{ X} \: = \: a \: \dfrac{\sum_{i=1}^n \: x_i}{n}$

$\rm :\implies\: \boxed{ \pink{ \bf \: \overline{ X} \: = \tt \: a \: \overline{x}}}$

${\boxed{\boxed{\bf{Hence, Proved}}}}$

Now,

We know that

variance is given by

$\rm :\implies\: {\sigma_{1}}^{2} = \dfrac{\sum_{i=1}^n \: {a}^{2} {x_i}^{2} }{n} - {(\overline{ X})}^{2}$

$\rm :\implies\: {\sigma_{1}}^{2} = \dfrac{ {a}^{2}\sum_{i=1}^n \: {x_i}^{2} }{n} - {(a\overline{ x})}^{2}$

$\rm :\implies\: {\sigma_{1}}^{2} = {a}^{2} \dfrac{ \sum_{i=1}^n \: {x_i}^{2} }{n} - {a}^{2} {(\overline{ x})}^{2}$

$\rm :\implies\: {\sigma_{1}}^{2} = {a}^{2} \bigg( \dfrac{ \sum_{i=1}^n \: {x_i}^{2} }{n} - {(\overline{ x})}^{2} \bigg)$

$\rm :\implies\: \boxed{ \pink{ \bf \: {\sigma_{1}}^{2} \: = \tt \: {a}^{2} {\sigma}^{2} }}$

${\boxed{\boxed{\bf{Hence, Proved}}}}$