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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

$\sqrt{2}$

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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 $10$ 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,$|d| = \sqrt{2}$.

$X$	$35$	$54$	$52$	$53$	$56$	$58$	$52$	$50$	$51$	$49$
$Y$	$108$	$107$	$105$	$105$	$106$	$107$	$104$	$103$	$104$	$101$

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:

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