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(D) Let the observations be $x_i$. There are $30$ items in total.$10$ items have value $\frac{1}{2} - d$,$10$ items have value $\frac{1}{2}$,and $10$

Vedclass · Accepted Answer

$\sqrt{2}$

Vedclass · Answer

(D) Let the observations be $x_i$. There are $30$ items in total.$10$ items have value $\frac{1}{2} - d$,$10$ items have value $\frac{1}{2}$,and $10$ items have value $\frac{1}{2} + d$.The mean $\bar{x} = \frac{10(\frac{1}{2} - d) + 10(\frac{1}{2}) + 10(\frac{1}{2} + d)}{30} = \frac{15}{30} = \frac{1}{2}$.The variance $\sigma^2 = \frac{1}{n} \sum (x_i - \bar{x})^2$.$\sigma^2 = \frac{1}{30} [10(\frac{1}{2} - d - \frac{1}{2})^2 + 10(\frac{1}{2} - \frac{1}{2})^2 + 10(\frac{1}{2} + d - \frac{1}{2})^2]$.$\sigma^2 = \frac{1}{30} [10(-d)^2 + 10(0)^2 + 10(d)^2] = \frac{20d^2}{30} = \frac{2d^2}{3}$.Given $\sigma^2 = \frac{4}{3}$,we have $\frac{2d^2}{3} = \frac{4}{3}$.$2d^2 = 4 \Rightarrow d^2 = 2$.Therefore,$|d| = \sqrt{2}$.

Size $(x_i)$	$3.5$	$4.5$	$5.5$	$6.5$	$7.5$	$8.5$	$9.5$
Frequency $(f_i)$	$3$	$7$	$22$	$60$	$85$	$32$	$8$

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

The outcome of each of $30$ items was observed; $10$ items gave an outcome $\frac{1}{2} - d$ each,$10$ items gave an outcome $\frac{1}{2}$ each,and the remaining $10$ items gave an outcome $\frac{1}{2} + d$ each. If the variance of this outcome data is $\frac{4}{3}$,then $|d|$ equals:

Similar Questions

Find the variance of the following data:

Size $(x_i)$ $3.5$ $4.5$ $5.5$ $6.5$ $7.5$ $8.5$ $9.5$

Frequency $(f_i)$ $3$ $7$ $22$ $60$ $85$ $32$ $8$

The mean square deviation of a set of observations $x_1, x_2, \dots, x_n$ about a point $c$ is defined as $\frac{1}{n} \sum_{i=1}^n (x_i - c)^2$. If the mean square deviations about $-2$ and $2$ are $18$ and $10$ respectively,find the standard deviation of this set of observations.

If the standard deviation of data is $12$ and mean is $72$,then the coefficient of variation is: (in $\%$)

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

If the coefficient of variation and standard deviation are $ 60 $ and $ 21 $ respectively,the arithmetic mean of the distribution is:

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