The outcome of each of 30 items was observed,10 items gave an outcome 12−d each, 10 items gave outcome 12+d each. Of remaining 10 items gave an outcome 12 each.If the variance of this outcome data is 43 then |d|=

∑x2=1012−d2+1a×10×10×12+d2=10×214+d2+104=304+20d2Variance = ∑xi2n−∑xi2n=43x¯=1012−d+10×12+10×12+d30=12 Now variance is 1012−d2+10×14+10×12+d230-14=43 ⇒d2=2⇒d=2

The outcome of each of 30 items was observed,10 items gave an outcome 12−d each, 10 items gave outcome 12+d each. Of remaining 10 items gave an outcome 12 each.If the variance of this outcome data is 43 then |d|=

∑x2=1012−d2+1a×10×10×12+d2=10×214+d2+104=304+20d2Variance = ∑xi2n−∑xi2n=43x¯=1012−d+10×12+10×12+d30=12 Now variance is 1012−d2+10×14+10×12+d230-14=43 ⇒d2=2⇒d=2

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

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$\frac{\sqrt{5}}{2}$

$\frac{2}{3}$

$\sqrt{2}$

answer is D.

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Detailed Solution

$\sum x^{2} = 10 {(\frac{1}{2} - d)}^{2} + \frac{1}{a} \times 10 \times 10 \times {(\frac{1}{2} + d)}^{2} = 10 \times 2 [\frac{1}{4} + d^{2}] + \frac{10}{4} = \frac{30}{4} + 20 d^{2}$
Variance = $\frac{\sum x_{i}^{2}}{n} - (\frac{\sum x_{i}^{2}}{n}) = \frac{4}{3}$
$\bar{x} = \frac{10 (\frac{1}{2} - d) + 10 \times \frac{1}{2} + 10 \times (\frac{1}{2} + d)}{30} = \frac{1}{2} N o w v a r i a n c e i s \frac{10 {(\frac{1}{2} - d)}^{2} + 10 \times \frac{1}{4} + 10 \times {(\frac{1}{2} + d)}^{2}}{30} - \frac{1}{4} = \frac{4}{3} \Rightarrow d^{2} = 2 \Rightarrow |d| = \sqrt{2}$