**Question:**

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

**Solution:**

The given $n$ observations are $x_{1}, x_{2} \ldots x_{n}$.

Mean $=\bar{x}$

Variance $=\sigma^{2}$

$\therefore \sigma^{2}=\frac{1}{n} \sum_{t=1}^{n} y_{i}\left(x_{i}-\bar{x}\right)^{2}$ $\ldots(1)$

If each observation is multiplied by a and the new observations are $y_{i,}$ then

$y_{1}=a x_{i}$ i.e., $x_{i}=\frac{1}{a} y_{1}$

$\therefore \bar{y}=\frac{1}{n} \sum_{i=1}^{g} y_{i}=\frac{1}{n} \sum_{i=1}^{e} a x_{i}=\frac{a}{n} \sum_{i=1}^{n} x_{i}=a \bar{x}$ $\left(\bar{x}=\frac{1}{n} \sum_{i=1}^{n} x_{i}\right)$

Therefore, mean of the observations, $a x_{1}, a x_{2} \ldots a x_{n}$, is $a \bar{x}$.

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

$\sigma^{2}=\frac{1}{n} \sum_{i=1}^{n}\left(\frac{1}{a} y_{i}-\frac{1}{a} y\right)^{2}$

$\Rightarrow a^{2} \sigma^{2}=\frac{1}{n} \sum_{i=1}^{n}\left(y_{i}-\bar{y}\right)^{2}$

Thus, the variance of the observations, $a x_{1}, a x_{2} \ldots a x_{n}$, is $a^{2} \sigma^{2}$.

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