Question

The outcome of each of 30 items was observed; 10 items gave an outcome $\dfrac{1}{2}-d$ each, 10 items gave outcome $\dfrac{1}{2}$ each and the remaining 10 items gave outcome $\dfrac{1}{2}+d$ each. If the variance of this outcome data is $\dfrac{4}{3}$ then $\left| d \right|$ equals:-
A. $2$
B. $\dfrac{\sqrt{5}}{2}$
C. $\dfrac{2}{3}$
D. $\sqrt{2}$

Answer 1

Hint: We will be using the concepts of mean and deviation to solve the problem. We will be using the formula that variance$=\dfrac{\sum\limits_{r=1}{{{\left( {{x}_{1}}^{0}-\overline{x} \right)}^{2}}}}{n}$

Complete step-by-step answer:
Now, we have been given the outcome of 30 items.
It has been given to us that 10 items are $\dfrac{1}{2}-d$, also the other 10 items are $\dfrac{1}{2}$and remaining 10 items are $\dfrac{1}{2}+d$.
Now, we will find its variance according to the outcome and equate it to $\dfrac{4}{3}$.
Now, we know that variance is;
${{\sigma }^{2}}=\dfrac{\sum{{{x}^{2}}}}{N}-{{\left( \dfrac{\sum{x}}{N} \right)}^{2}}.................\left( 1 \right)$
Now, we have N = 30 and
$\begin{align}
  & \sum{x}=10\times \left( \dfrac{1}{2}-d \right)+10\times \dfrac{1}{2}+10\times \left( \dfrac{1}{2}+d \right) \\
 & =\dfrac{10}{2}-10d+\dfrac{10}{2}+\dfrac{10}{2}+10d \\
 & =\dfrac{3\times 10}{2} \\
 & \sum{x}=15 \\
 & \dfrac{\sum{x}}{N}=\dfrac{15}{30}=\dfrac{1}{2} \\
\end{align}$
So, we have $\overline{x}=\dfrac{1}{2}......................\left( 2 \right)$
Now, we have;
\[\begin{align}
  & \sum{{{x}^{2}}=10{{\left( \dfrac{1}{2}-d \right)}^{2}}+10\times {{\left( \dfrac{1}{2} \right)}^{2}}+10\times {{\left( \dfrac{1}{2}+d \right)}^{2}}} \\
 & =10\left( \dfrac{1}{4}+{{d}^{2}}-d \right)+10\times \dfrac{1}{4}+10\times \left( \dfrac{1}{4}+d+d \right) \\
 & =\dfrac{10}{4}+10{{d}^{2}}-10d+\dfrac{10}{4}+\dfrac{10}{4}+10{{d}^{2}}+10d \\
 & \sum{{{x}^{2}}=\dfrac{3\times 10}{4}+20{{d}^{2}}} \\
 & \sum{{{x}^{2}}}=\dfrac{15}{2}+20{{d}^{2}} \\
 & \dfrac{\sum{{{x}^{2}}}}{N}=\dfrac{15+40{{d}^{2}}}{2\times 30} \\
 & \dfrac{\sum{{{x}^{2}}}}{N}=\dfrac{15+40{{d}^{2}}}{60}................\left( 3 \right) \\
\end{align}\]
So, now from (2) and (3) we will have variance from (1) as;
$\begin{align}
  & =\dfrac{15+40{{d}^{2}}}{60}-{{\left( \dfrac{1}{2} \right)}^{2}} \\
 & =\dfrac{15+40{{d}^{2}}}{60}-\dfrac{1}{4} \\
 & =\dfrac{15+40{{d}^{2}}-15}{60} \\
 & =\dfrac{40{{d}^{2}}}{60} \\
 & =\dfrac{2{{d}^{2}}}{3} \\
\end{align}$
Now, we will equate it to $\dfrac{4}{3}$. So, we have,
\[\begin{align}
  & \dfrac{2}{3}{{d}^{2}}=\dfrac{4}{3} \\
 & {{d}^{2}}=2 \\
 & d=\pm \sqrt{2} \\
 & \left| d \right|=\left| \pm \sqrt{2} \right| \\
 & =\left| \sqrt{2} \right| \\
\end{align}\]
Hence, the correct option is (D).

Note: To solve these type of questions it is important to note that we have calculated variance as
${{\sigma }^{2}}=\dfrac{\sum{{{x}^{2}}}}{N}-{{\left( \dfrac{\sum{n}}{N} \right)}^{2}}$also one should remember identities like ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$.