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.

Class Interval	Frequency
$0 - 6$	$10$
$6 - 12$	$8$
$12 - 18$	$6$
$18 - 24$	$4$
$24 - 30$	$2$

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

For a set of observations,if the coefficient of variation is $25$ and the mean is $44$,then the variance is:

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 $a x_{1}, a x_{2}, \ldots, a x_{n}$ are $a \bar{x}$ and $a^{2} \sigma^{2}$ respectively,where $a \neq 0$.

The variance of the following frequency distribution is
Class Interval Frequency
$0 - 6$ $10$
$6 - 12$ $8$
$12 - 18$ $6$
$18 - 24$ $4$
$24 - 30$ $2$

If the variance of $x_1, x_2, \ldots, x_n$ is $\sigma_x^2$,then the variance of $\lambda x_1, \lambda x_2, \ldots, \lambda x_n$ (where $\lambda \neq 0$) is:

The standard deviation of the first $10$ multiples of $4$ 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 $4\bar{x}$.