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The outcome of each of 30 items was observed , 10 items gave an outcome `(1)/(2)-d` each, 10 items gave outcome `(1)/(2)` each and the remaining 10 items gave outcome `(1)/(2) +d` each. If the variance of this outcome data is `(4)/(3)`, then |d| equalsA. `(2)/(3)`B. `(sqrt(5))/(2)`C. `sqrt(2)`D. 2 |
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Answer» Correct Answer - C `sigma^(2)=(Sigmax^(2))/(n)-mu^(2)` `=(10((1)/(2)-d)^(2)+10xx(1)/(4)+10((1)/(2)+d)^(2))/(30)` `-((10((1)/(2)-d)+10xx(1)/(2)+10((1)/(2)+d))/(30))^(2)" "[because mu=(Sigma x_(i))/(n)]` `=(20((1)/(4)+d^(2))+5//2)/(30)-((1)/(4))` `=((15)/(2)+20d^(2))/(30)-(1)/(4)=(1)/(4)+(2d^(2))/(3)-(1)/(4)=(2)/(3)d^(2)` ` therefore (2)/(3) d^(2)=(4)/(3) rArr d^(2)=2 rArr |d|=sqrt(2)` |
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