The mean and the standard deviation of 10 obs | vedclass

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

Question

(A) Given: Original mean $\bar{x} = 20$,original standard deviation $\sigma = 2$. If each observation $x_i$ is transformed to $y_i = p x_i - q$,then t

Vedclass · Accepted Answer

$-20$

Vedclass · Answer

(A) Given: Original mean $\bar{x} = 20$,original standard deviation $\sigma = 2$. If each observation $x_i$ is transformed to $y_i = p x_i - q$,then the new mean $\bar{y} = p \bar{x} - q$ and the new standard deviation $\sigma_y = |p| \sigma$. Given that the new mean is half of the original mean: $\bar{y} = \frac{20}{2} = 10$. So,$p(20) - q = 10 \implies 20p - q = 10$ $(i)$. Given that the new standard deviation is half of the original standard deviation: $\sigma_y = \frac{2}{2} = 1$. So,$|p| 	imes 2 = 1 \implies |p| = \frac{1}{2} \implies p = \pm \frac{1}{2}$. Case $1$: If $p = \frac{1}{2}$,then $20(\frac{1}{2}) - q = 10 \implies 10 - q = 10 \implies q = 0$. This contradicts the condition $q 
eq 0$. Case $2$: If $p = -\frac{1}{2}$,then $20(-\frac{1}{2}) - q = 10 \implies -10 - q = 10 \implies q = -20$. Thus,$q = -20$.

$x$	$2$	$6$	$8$	$9$
$f$	$4$	$4$	$\alpha$	$\beta$

Similar Questions

If $65$ is the range of the ungrouped data $50, 70, 60, B, 20, 40$,then the absolute difference of the possible values of $B$ is

For a frequency distribution,the $7^{th}$ decile is computed by the formula:

Let the mean and variance of the frequency distribution be $6$ and $6.8$ respectively.

$x$ $2$ $6$ $8$ $9$

$f$ $4$ $4$ $\alpha$ $\beta$

If $x_{3}$ is changed from $8$ to $7$,then the mean for the new data will be:

Let $r$ be the range and $S^2 = \frac{1}{n - 1} \sum_{i = 1}^n (x_i - \bar{x})^2$ be the variance of a set of observations $x_1, x_2, \dots, x_n$. Then:

Find the range of the given numbers: $10, 8, 12, 11, 14, 9, 6$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module