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Let $x_1, x_2, x_3, x_4, .......... , x_n$ be $n$ observations and let $\bar x$ be their arithmetic mean and $\sigma ^2$ be their variance.
Statement $-1$ : Variance of observations $2x_1, 2x_2, 2x_3, ......, 2x_n$ is $4\sigma ^2$ .
Statement $-2$ : Arithmetic mean of $2x _1, 2x_2, 2x_3, ......, 2x_n$ is $4\bar x$ .
Statement $-1$ is true, statement $-2$ is true and statement $-2$ is $NOT$ the correct explanation for statement $-1$
Statement $-1$ is true, statement $-2$ is false
Statement $-1$ is false, stateemnt $-2$ is true
Statement $-1$ is true, statement $-2$ is true and statement $-2$ is correct explanation for statement $-1$
Solution
If each observations is multiplied by a constant $k$ then their mean is multiplied with $k$ and their variance is multiplied by $k^2$ .
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 :
The following values are calculated in respect of heights and weights of the students of a section of Class $\mathrm{XI}:$
| Height | Weight | |
| Mean | $162.6\,cm$ | $52.36\,kg$ |
| Variance | $127.69\,c{m^2}$ | $23.1361\,k{g^2}$ |
Can we say that the weights show greater variation than the heights?
