Question

If both the standard deviation and mean of the data set ${{x}_{1}},{{x}_{2}},{{x}_{3}},...,{{x}_{50}}$ are $16$. Then the mean if the data set ${{\left( {{x}_{1}}-4 \right)}^{2}},{{\left( {{x}_{2}}-4 \right)}^{2}},{{\left( {{x}_{3}}-4 \right)}^{2}},...,{{\left( {{x}_{50}}-4 \right)}^{2}}$ is?
A. $200$
B. $100$
C. $400$
D. $1600$

Answer 1

From the given data that the mean and standard deviation of ${{x}_{1}},{{x}_{2}},{{x}_{3}},...,{{x}_{50}}$ is $16$, we will use basic formulas of mean and standard deviation and we will equate the given values to find the values of $\dfrac{\sum{{{x}_{i}}}}{50},\dfrac{\sum{{{x}_{i}}^{2}}}{50}$. Now using the both obtained values we will find the mean of the data set ${{\left( {{x}_{1}}-4 \right)}^{2}},{{\left( {{x}_{2}}-4 \right)}^{2}},{{\left( {{x}_{3}}-4 \right)}^{2}},...,{{\left( {{x}_{50}}-4 \right)}^{2}}$ from the basic formula of mean i.e. ratio of the sum of the variables to number of variables.

Complete step by step answer:
Given that,
The mean and standard deviation of ${{x}_{1}},{{x}_{2}},{{x}_{3}},...,{{x}_{50}}$ is $16$. Mathematically we will write it as
$\begin{aligned} & \text{Mean}=\dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}+...+{{x}_{50}}}{50} \\\ & \Rightarrow M=16....\left( \text{i} \right) \\\ \end{aligned}$
Standard deviation is
$\begin{aligned} & \text{S}\text{.D}=\sqrt{\dfrac{{{\left( {{x}_{1}}-M \right)}^{2}}+{{\left( {{x}_{2}}-M \right)}^{2}}+{{\left( {{x}_{3}}-M \right)}^{2}}+...+{{\left( {{x}_{50}}-M \right)}^{2}}}{50}} \\\ & \Rightarrow 16=\sqrt{\dfrac{{{\sum{{{x}_{i}}^{2}-50M}}^{2}}}{50}} \\\ &\Rightarrow 16=\sqrt{\dfrac{\sum{{{x}_{i}}^{2}}}{50}-{{M}^{2}}} \\\ \end{aligned}$
Squaring on both sides, we will have
$\begin{aligned} & {{16}^{2}}=\dfrac{\sum{{{x}_{i}}^{2}}}{50}-{{M}^{2}} \\\ & \Rightarrow 256=\dfrac{\sum{{{x}_{i}}^{2}}}{50}-{{M}^{2}} \\\ \end{aligned}$
From equation $\left( \text{i} \right)$ we have the value $M=16$, then
$\begin{aligned} & 256=\dfrac{\sum{{{x}_{i}}^{2}}}{50}-{{16}^{2}} \\\ &\Rightarrow 512=\dfrac{\sum{{{x}_{i}}^{2}}}{50}....\left( \text{ii} \right) \\\ \end{aligned}$
Now the mean of the data set ${{\left( {{x}_{1}}-4 \right)}^{2}},{{\left( {{x}_{2}}-4 \right)}^{2}},{{\left( {{x}_{3}}-4 \right)}^{2}},...,{{\left( {{x}_{50}}-4 \right)}^{2}}$ is
$\begin{aligned} & Mean=\dfrac{{{\left( {{x}_{1}}-4 \right)}^{2}}+{{\left( {{x}_{2}}-4 \right)}^{2}}+{{\left( {{x}_{3}}-4 \right)}^{2}}+...+{{\left( {{x}_{50}}-4 \right)}^{2}}}{50} \\\ & =\dfrac{\sum{{{\left( {{x}_{i}}-4 \right)}^{2}}}}{50} \\\ & =\dfrac{\sum{\left( {{x}_{i}}^{2}-2.4.{{x}_{i}}+16 \right)}}{50} \\\ & =\dfrac{\sum{{{x}_{i}}^{2}}}{50}-8\dfrac{\sum{{{x}_{i}}}}{50}+\dfrac{50\times 16}{50} \\\ \end{aligned}$
From equations $\left( \text{i} \right)$ and $\left( \text{ii} \right)$, we have
$\begin{aligned} & Mean=512-8\left( 16 \right)+16 \\\ & =400 \\\ \end{aligned}$
Hence the mean of the dataset ${{\left( {{x}_{1}}-4 \right)}^{2}},{{\left( {{x}_{2}}-4 \right)}^{2}},{{\left( {{x}_{3}}-4 \right)}^{2}},...,{{\left( {{x}_{50}}-4 \right)}^{2}}$ is $400$.

So, the correct answer is “Option C”.

Note: We can also solve the above problem in other method i.e. if the mean of the variables ${{x}_{1}},{{x}_{2}},{{x}_{3}},...,{{x}_{50}}$ is ${{M}_{1}}$, then the mean of the variables ${{\left( {{x}_{1}}-4 \right)}^{2}},{{\left( {{x}_{2}}-4 \right)}^{2}},{{\left( {{x}_{3}}-4 \right)}^{2}},...,{{\left( {{x}_{50}}-4 \right)}^{2}}$ is given by ${{\left( {{M}_{1}}-4 \right)}^{2}}$ and the variance of the both the data set is same. Then
Variance of ${{\left( {{x}_{1}}-4 \right)}^{2}},{{\left( {{x}_{2}}-4 \right)}^{2}},{{\left( {{x}_{3}}-4 \right)}^{2}},...,{{\left( {{x}_{50}}-4 \right)}^{2}}$ is
$\begin{aligned} & \sigma _{1}^{2}=\dfrac{\sum{{{\left( {{x}_{i}}-4 \right)}^{2}}-{{\left( 16-4 \right)}^{2}}}}{50} \\\ & 256=\dfrac{\sum{{{\left( {{x}_{i}}-4 \right)}^{2}}}}{50}-\dfrac{50\times {{12}^{2}}}{50} \\\ & \dfrac{\sum{{{\left( {{x}_{i}}-4 \right)}^{2}}}}{50}=400 \\\ \end{aligned}$
Hence the mean of the data set ${{\left( {{x}_{1}}-4 \right)}^{2}},{{\left( {{x}_{2}}-4 \right)}^{2}},{{\left( {{x}_{3}}-4 \right)}^{2}},...,{{\left( {{x}_{50}}-4 \right)}^{2}}$ is $400$.
From both the methods we get the same answer.