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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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}

Vedclass · Accepted Answer

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.

$x_i$	$4$	$3$	$1$
$f_i$	$1$	$3$	$5$

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

If the variance of the distribution is $45.8$,then find the variance of the distribution given below:

$x_i$ $4$ $8$ $11$ $17$ $20$ $24$ $32$

$f_i$ $3$ $5$ $9$ $5$ $4$ $3$ $1$

$y_i$ $10$ $18$ $24$ $36$ $42$ $50$ $66$

$f_i$ $3$ $5$ $9$ $5$ $4$ $3$ $1$

In a discrete data,$\frac{1}{4}$ of the observations are equal to $a$,another $\frac{1}{4}$ of the observations are equal to $-a$. Out of the remaining,half of them are equal to $b$ and the rest are equal to $-b$. If the variance of all the observations is $ab$,then:

Find the mean and variance for the first $n$ natural numbers.

The coefficient of variation for the frequency distribution is
$x_i$ $4$ $3$ $1$
$f_i$ $1$ $3$ $5$

The variance of the first $20$ natural numbers is

Vedclass Test Series

Exam Paper Generator

Online Exam Module

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 $4\bar{x}$.