Given that ( $¯ ¯ ¯ x$ ) is the mean and $σ^{2}$ is the variance of n observations $x_{1}, x_{2}, . . . . x_{n}$ . Prove that the mean and variance of the observations $a x_{1}, a x_{2}, a x_{3} . . . a x_{n}$ are a $¯ ¯ ¯ x$ and $a^{2} σ^{2}$ respectively $(a \neq 0)$

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Solution

Here $¯ x \frac{x_{1} + x_{2} + x_{3} + . . . . + x_{n}}{n} = \frac{\sum x}{n}$

Also $\frac{x_{1}^{2} + x_{2}^{2} + x_{3}^{2} + . . . . + x_{n}^{2}}{n} = \frac{\sum x^{2}}{n}$

New mean = $\frac{a x_{1} + a x_{2} + a x_{3} + . . . + a x_{n}}{n}$

Also $σ^{2} = \frac{n (x_{1}^{2} + x_{2}^{2} + x_{3}^{2} + . . . . x_{n}^{2}) - (x_{1} + x_{2} + x_{3} + . . . . + x_{n})}{n^{2}}$

New variance = $\frac{n (a^{2} x_{1}^{2} + a^{2} x_{2}^{2} + a^{2} x_{3}^{2} + . . . a^{2} x_{n}^{2}) - (a x_{1} + a x_{2} + a x_{3} + . . . + a x_{n}^{2})^{2}}{n^{2}}$

= $a^{2} [\frac{n (x_{1}^{2} + x_{2}^{2} + x_{3}^{2} + . . . . + x_{n}^{2}) - (x_{1} + x_{2} + x_{3} + . . . + x_{n})}{n^{2}}]$

= $a^{2} σ^{2} .$

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Similar questions

Q. Given that

¯ X

is the mean and

σ^{2}

is the variance of

n

observations

X_{1}, X_{2} . . . X_{n}

. Prove that the mean and variance of the observations

a X_{1}, a X_{2}, a X_{3} . . . . a X_{n}

are

¯ a x

and

a^{2} σ^{2}

respectively

(a \neq 0)

If the mean and variance of the observations $x_{1}, x_{2}, x_{3}, \dots, x_{n}$ are $¯ x$ and $σ^{2}$ respectively and a be a nonzero real number, then show that the mean and variance of $a x_{1}, a x_{2}, a x_{3}, \dots . a x_{n}$ are $¯ a x$ and $a^{2} σ^{2}$ respectively.