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If the variance of observations ${x_1},\,{x_2},\,......{x_n}$ is ${\sigma ^2}$, then the variance of $a{x_1},\,a{x_2}.......,\,a{x_n}$, $\alpha \ne 0$ is
${\sigma ^2}$
$a\,{\sigma ^2}$
${a^2}{\sigma ^2}$
$\frac{{{\sigma ^2}}}{{{a^2}}}$
Solution
Varivence of $x_1 \cdot x_2 \cdot \cdots \quad x_n=6^2$
Variane of $a x_1 a x_2, \ldots a x_n=$ ?
varience $=\sigma^2=\frac{1}{n} \sum \limits_{i=1}^r y_i\left(n_i-\bar{x}\right)^2$
If each obs is weltiplied $2 y$ a the $y_i=a x_i \quad i . e \quad x_i=\frac{1}{a} y_i$
$y_i=a x_i n$
$\therefore \bar{y}=\frac{1}{n} \sum\limits_{i=1}^n y_i=\frac{1}{n} \sum\limits_{i=1}^n a x_i=\frac{a}{n} \sum\limits_{i=1}^n x_i=a \bar{x} .$
${\left[\because \bar{x}=\frac{1}{n} \sum\limits_{i=1}^n x_i\right]}$
$(1) \Rightarrow \quad \sigma^2=\frac{1}{A} \sum_{i=1}^n\left(\frac{1}{a} y_i-\frac{1}{a} \bar{y}\right)^2$
$\Rightarrow a a^2 \sigma^2=\frac{1}{n} \sum_{i=1}^n\left(y_i-\bar{y}\right)^2$
These varieme of new obs' is $a^2 \sigma^2$
Similar Questions
Let $\mathrm{X}$ be a random variable with distribution.
| $\mathrm{x}$ | $-2$ | $-1$ | $3$ | $4$ | $6$ |
| $\mathrm{P}(\mathrm{X}=\mathrm{x})$ | $\frac{1}{5}$ | $\mathrm{a}$ | $\frac{1}{3}$ | $\frac{1}{5}$ | $\mathrm{~b}$ |
If the mean of $X$ is $2.3$ and variance of $X$ is $\sigma^{2}$, then $100 \sigma^{2}$ is equal to :
Find the standard deviation for the following data:
| ${x_i}$ | $3$ | $8$ | $13$ | $18$ | $25$ |
| ${f_i}$ | $7$ | $10$ | $15$ | $10$ | $6$ |
