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The outcome of each of 30 items was observed; 10 items
gave an outcome $\frac{1}{2}-\mathrm{d}$ each, 10 items gave outcome $\frac{1}{2}$
each and the remaining 10 items gave outcome $\frac{1}{2}+\mathrm{d}$ each.
If the variance of this outcome data is $\frac{4}{3}$ then $|\mathrm{d}|$ equals:
Options:
gave an outcome $\frac{1}{2}-\mathrm{d}$ each, 10 items gave outcome $\frac{1}{2}$
each and the remaining 10 items gave outcome $\frac{1}{2}+\mathrm{d}$ each.
If the variance of this outcome data is $\frac{4}{3}$ then $|\mathrm{d}|$ equals:
Solution:
2270 Upvotes
Verified Answer
The correct answer is:
$\sqrt{2}$
Outcomes are $\left(\frac{1}{2}-d\right),\left(\frac{1}{2}-d\right), 0 \ldots, 10$ times, $\frac{1}{2}, \frac{1}{2}$,
$\ldots, 10$ times, $\frac{1}{2}+d, \frac{1}{2}+d, \ldots, 10$ times
$$
\text { Mean }=\frac{1}{30}\left(\frac{1}{2} \times 30\right)=\frac{1}{2}
$$
Variance of the outcomes is,
$$
\begin{aligned}
\sigma^{2} &=\frac{1}{30} \Sigma x_{i}^{2}-(\bar{x})^{2} \\
&=\frac{1}{30}\left[\left(\frac{1}{2}-d\right)^{2} \times 10+\left(\frac{1}{2}\right)^{2} \times 10+\left(\frac{1}{2}+d\right)^{2} \times 10\right]-\frac{1}{4} \\
\Rightarrow & \frac{4}{3}=\frac{1}{30}\left[30 \times \frac{1}{4}+20 d^{2}\right]-\frac{1}{4} \\
\Rightarrow & \frac{4}{3}=\frac{1}{4}+\frac{2}{3} d^{2}-\frac{1}{4} \\
\Rightarrow & d^{2}=2 \Rightarrow|d|=\sqrt{2}
\end{aligned}
$$
$\ldots, 10$ times, $\frac{1}{2}+d, \frac{1}{2}+d, \ldots, 10$ times
$$
\text { Mean }=\frac{1}{30}\left(\frac{1}{2} \times 30\right)=\frac{1}{2}
$$
Variance of the outcomes is,
$$
\begin{aligned}
\sigma^{2} &=\frac{1}{30} \Sigma x_{i}^{2}-(\bar{x})^{2} \\
&=\frac{1}{30}\left[\left(\frac{1}{2}-d\right)^{2} \times 10+\left(\frac{1}{2}\right)^{2} \times 10+\left(\frac{1}{2}+d\right)^{2} \times 10\right]-\frac{1}{4} \\
\Rightarrow & \frac{4}{3}=\frac{1}{30}\left[30 \times \frac{1}{4}+20 d^{2}\right]-\frac{1}{4} \\
\Rightarrow & \frac{4}{3}=\frac{1}{4}+\frac{2}{3} d^{2}-\frac{1}{4} \\
\Rightarrow & d^{2}=2 \Rightarrow|d|=\sqrt{2}
\end{aligned}
$$
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