Let x_1, x_2, dots, x_n be n observations,bar | vedclass

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(A) Given that the mean of $x_1, x_2, \dots, x_n$ is $\bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i$ and the variance is $\sigma^2 = \frac{1}{n} \sum_{i=1}

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Statement-$1$ is true,Statement-$2$ is false.

Vedclass · Answer

(A) Given that the mean of $x_1, x_2, \dots, x_n$ is $\bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i$ and the variance is $\sigma^2 = \frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{x})^2$.For the observations $2x_1, 2x_2, \dots, 2x_n$:The new mean $\bar{x}' = \frac{1}{n} \sum_{i=1}^{n} (2x_i) = 2 \left( \frac{1}{n} \sum_{i=1}^{n} x_i \right) = 2\bar{x}$.Thus,Statement-$2$ is false because the mean is $2\bar{x}$,not $4\bar{x}$.The new variance $\sigma'^2 = \frac{1}{n} \sum_{i=1}^{n} (2x_i - 2\bar{x})^2 = \frac{1}{n} \sum_{i=1}^{n} 4(x_i - \bar{x})^2 = 4 \left( \frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{x})^2 \right) = 4\sigma^2$.Thus,Statement-$1$ is true.Therefore,Statement-$1$ is true and Statement-$2$ is false.

$x_i$	$60$	$61$	$62$	$63$	$64$	$65$	$66$	$67$	$68$
$f_i$	$2$	$1$	$12$	$29$	$25$	$12$	$10$	$4$	$5$

Subject	Mathematics,Physics,Chemistry
Mean	$42, 32, 40.9$
Standard deviation	$12, 15, 20$

Let $x_1, x_2, \dots, x_n$ be $n$ observations,$\bar{x}$ be their mean,and $\sigma^2$ be their variance.
Statement-$1$: The variance of $2x_1, 2x_2, \dots, 2x_n$ is $4\sigma^2$.
Statement-$2$: The mean of $2x_1, 2x_2, \dots, 2x_n$ is $4\bar{x}$.

Similar Questions

The data is obtained in tabular form as follows:
$x_i$ $60$ $61$ $62$ $63$ $64$ $65$ $66$ $67$ $68$
$f_i$ $2$ $1$ $12$ $29$ $25$ $12$ $10$ $4$ $5$

Find the standard deviation of the given data. (in $.69$)

For $10$ observations $x_1, x_2, ..., x_{10}$, if $\sum_{i=1}^{10} (x_i + 2)^2 = 180$ and $\sum_{i=1}^{10} (x_i - 1)^2 = 90$, then their standard deviation is:

The sum of squares of deviations for $10$ observations taken from mean $50$ is $250$. The coefficient of variation is.....$\%$

What is the standard deviation of the numbers $31, 32, 33, \dots, 47$?

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Let $x_1, x_2, \dots, x_n$ be $n$ observations,$\bar{x}$ be their mean,and $\sigma^2$ be their variance.Statement-$1$: The variance of $2x_1, 2x_2, \dots, 2x_n$ is $4\sigma^2$.Statement-$2$: The mean of $2x_1, 2x_2, \dots, 2x_n$ is $4\bar{x}$.