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

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

Question

(C) The variance of a set of observations is always non-negative,which implies that the mean of the squares is greater than or equal to the square of

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(C) The variance of a set of observations is always non-negative,which implies that the mean of the squares is greater than or equal to the square of the mean: $\frac{\sum x_i^2}{n} \geq \left(\frac{\sum x_i}{n}\right)^2$.Substituting the given values $\sum x_i^2 = 300$ and $\sum x_i = 60$:$\frac{300}{n} \geq \left(\frac{60}{n}\right)^2$$\frac{300}{n} \geq \frac{3600}{n^2}$Since $n > 0$,we can multiply both sides by $n^2$:$300n \geq 3600$$n \geq 12$.Among the given options,only $15$ satisfies the condition $n \geq 12$.

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

Let $x_1, x_2, \dots, x_n$ be $n$ observations such that $\sum x_i^2 = 300$ and $\sum x_i = 60$. Which of the following is a possible value for $n$?

Similar Questions

$A$ data consists of $n$ observations $x_1, x_2, ......, x_n$. If $\sum_{i=1}^n (x_i + 1)^2 = 9n$ and $\sum_{i=1}^n (x_i - 1)^2 = 5n$,then the standard deviation of this data is

Let one angle of a triangle be $60^{\circ}$. If the variance of the three angles of the triangle is $4614^{\circ}$,then the other two angles are

Find the mean and variance for the following data:

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

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

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

The mean and standard deviation of $100$ items are $50$ and $4$, respectively. Then the sum of the squares of all the items is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module