Question

If the mean of the data : $7,8,9,7,8,7,\lambda ,8$ is 8, then the variance of this data is
A. $\dfrac{9}{8}$
B. $2$
C. $\dfrac{7}{8}$
D. $1$

Answer 1

Hint: In this question, first use the formula of mean to find the value of $\lambda $. Now, use the elements to find the value of variance, which is variance = $\dfrac{\sum {{x}_{i}}^{2}}{N}-{{\mu }^{2}}$.

Complete step-by-step solution:
Here, the given data is $7,8,9,7,8,7,\lambda ,8$. The mean of the given data is 8. We have to find the variance of the given data. We have, the mean $\left( \mu \right)=8$, and the number of elements $\left( N \right)=8$. Let us first use the basic principle of mean $\left( \mu \right)$ and find the value of $\lambda $. We know that mean is given by,
Mean $\left( \mu \right)=\dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}+{{x}_{5}}+{{x}_{6}}+{{x}_{7}}+{{x}_{8}}}{8}$
The above mean expression is for 8 elements, which is,
Mean = $\dfrac{\left( \text{sum of elements} \right)}{\left( \text{total number of elements} \right)}=\dfrac{\sum {{x}_{i}}}{N}$
So, we get the mean as,
$\begin{align}
  & \mu =\dfrac{7+8+9+7+8+7+\lambda +8}{8} \\
 & \Rightarrow 8=\dfrac{15+16+15+8+\lambda }{8} \\
 & \Rightarrow 8\times 8=15+16+15+8+\lambda \\
 & \Rightarrow 64=30+24+\lambda \\
 & \Rightarrow 64=54+\lambda \\
 & \Rightarrow \lambda =64-54 \\
 & \Rightarrow \lambda =10 \\
\end{align}$
So, here we have the values of $\lambda =10$, therefore, we have the elements as, 7, 8, 9, 7, 8, 7, 10, 8. We know that variance =$\dfrac{\sum {{x}_{i}}^{2}}{N}-{{\mu }^{2}}$ . So, let us first find $\sum {{x}_{i}}^{2}$ by squaring each element and adding them with each other. So, we will get,
${{x}_{1}}^{2}={{7}^{2}}=49,{{x}_{2}}^{2}=64,{{x}_{3}}^{2}=81,{{x}_{4}}^{2}=49,{{x}_{5}}^{2}=64,{{x}_{6}}^{2}=49,{{x}_{7}}^{2}=100,{{x}_{8}}^{2}=64$
We also need to find ${{\mu }^{2}}$, which will be, ${{\mu }^{2}}={{8}^{2}}=64$. Now let us add all these values to find $\sum {{x}_{i}}^{2}$. So, we get,
$\begin{align}
  & \sum {{x}_{i}}^{2}={{x}_{1}}^{2}+{{x}_{2}}^{2}+{{x}_{3}}^{2}+{{x}_{4}}^{2}+{{x}_{5}}^{2}+{{x}_{6}}^{2}+{{x}_{7}}^{2}+{{x}_{8}}^{2} \\
 & \Rightarrow \sum {{x}_{i}}^{2}=49+64+81+49+64+49+100+64 \\
 & \Rightarrow \sum {{x}_{i}}^{2}=520 \\
\end{align}$
Now, we have $\sum {{x}_{i}}^{2}=520$, $N=8$ and ${{\mu }^{2}}=64$, so let us substitute these values in the variance formula and calculate the required result. So, we have,
$\begin{align}
  & \dfrac{\sum {{x}_{i}}^{2}}{N}-{{\mu }^{2}} \\
 & =\dfrac{520}{8}-64 \\
 & =65-64=1 \\
\end{align}$
Therefore, we get the variance of the given data as 1.
Hence, the correct answer is option D.

Note: Here, we can also use the formula of variance as $\dfrac{\sum {{\left( {{x}_{i}}-\mu \right)}^{2}}}{N}$. Variance is denoted by $\left( {{\sigma }^{2}} \right)$, the square root of variance is known as standard deviation. For using this formula first we will have to find the value of $\left( {{x}_{i}}-\mu \right)$ for each element and then have to square of those term which would take much more time so we avoid using this formula.