If sum_{i=1}^{18} (x_i - 8) = 9 and sum_{i=1} | vedclass

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

Question

(D) Let $d_i = x_i - 8$. The standard deviation of $x_i$ is the same as the standard deviation of $d_i$,denoted as $\sigma_d$. $\sigma_d = \sqrt{\frac

Vedclass · Accepted Answer

$3/2$

Vedclass · Answer

(D) Let $d_i = x_i - 8$. The standard deviation of $x_i$ is the same as the standard deviation of $d_i$,denoted as $\sigma_d$. $\sigma_d = \sqrt{\frac{\sum d_i^2}{n} - \left( \frac{\sum d_i}{n} \right)^2}$ Given $n = 18$,$\sum d_i = 9$,and $\sum d_i^2 = 45$. $\sigma_d = \sqrt{\frac{45}{18} - \left( \frac{9}{18} \right)^2}$ $\sigma_d = \sqrt{\frac{5}{2} - \left( \frac{1}{2} \right)^2}$ $\sigma_d = \sqrt{\frac{5}{2} - \frac{1}{4}} = \sqrt{\frac{10 - 1}{4}} = \sqrt{\frac{9}{4}} = \frac{3}{2}$

$x_i$	$6$	$10$	$14$	$18$	$24$	$28$	$30$
$f_i$	$2$	$4$	$7$	$12$	$8$	$4$	$3$

If $\sum_{i=1}^{18} (x_i - 8) = 9$ and $\sum_{i=1}^{18} (x_i - 8)^2 = 45$,find the standard deviation of $x_1, x_2, \dots, x_{18}$.

Similar Questions

If $x_1, x_2, ..., x_n$ are $n$ observations such that $\sum_{i=1}^n x_i^2 = 400$ and $\sum_{i=1}^n x_i = 100$,then which of the following is a possible value of $n$?

The sum of $10$ values is $12$ and the sum of their squares is $16.9$,then their standard deviation $(S.D.)$ is

If variables $x$ and $u$ are related by $u = \frac{x - a}{h}$,then the correct relationship between $\sigma_x$ and $\sigma_u$ is:

Find the mean and variance for the following data:

$x_i$ $6$ $10$ $14$ $18$ $24$ $28$ $30$

$f_i$ $2$ $4$ $7$ $12$ $8$ $4$ $3$

All students in a classroom performed poorly in mathematics. The teacher decided to increase the marks of each student by $10$. Which of the following statistical measures will not change even after adding the extra marks?

Vedclass Test Series

Exam Paper Generator

Online Exam Module