Question

Find the variance and standard deviation of the random variable X whose probability distribution is given below:

x	0	1	2	3
P(X=x)	$\dfrac{1}{8}$	$\dfrac{3}{8}$	$\dfrac{3}{8}$	$\dfrac{1}{8}$

Answer 1

First, before proceeding for this , we must know the following formulas for the variance which is denoted as ${{\sigma }^{2}}$and given by the formula as ${{\sigma }^{2}}=E\left( {{X}^{2}} \right)-{{\left( E\left( X \right) \right)}^{2}}$. Then, we need the second order mean $E\left( {{X}^{2}} \right)$ and first order mean $E\left( X \right)$as to get the variance. Then, to get the value of standard deviation, we must know the relationship that standard deviation is the square root of the variance.

Complete step-by-step answer:
In this question, we are supposed to find the variance and standard deviation of the random variable X whose probability distribution.

x	0	1	2	3
P(X=x)	$\dfrac{1}{8}$	$\dfrac{3}{8}$	$\dfrac{3}{8}$	$\dfrac{1}{8}$

So, before proceeding for this , we must know the following formulas for the variance which is denoted as ${{\sigma }^{2}}$and given by the formula:
${{\sigma }^{2}}=E\left( {{X}^{2}} \right)-{{\left( E\left( X \right) \right)}^{2}}$
Then, in the above formula, the term is $E\left( X \right)$ is the mean of the distribution given by:
$E\left( X \right)=\sum\limits_{i=0}^{3}{{{p}_{i}}{{x}_{i}}}$
So, by using the above formula to calculate mean as:
$\begin{aligned} & E\left( X \right)=0\times \dfrac{1}{8}+1\times \dfrac{3}{8}+2\times \dfrac{3}{8}+3\times \dfrac{1}{8} \\\ & \Rightarrow E\left( X \right)=0+\dfrac{3}{8}+\dfrac{6}{8}+\dfrac{3}{8} \\\ & \Rightarrow E\left( X \right)=\dfrac{12}{8} \\\ & \Rightarrow E\left( X \right)=\dfrac{3}{2} \\\ \end{aligned}$
Similarly, we need to the second order mean $E\left( {{X}^{2}} \right)$ to get the variance as:
$E\left( {{X}^{2}} \right)=\sum\limits_{i=0}^{3}{{{p}_{i}}{{x}_{i}}^{2}}$
Now, by using the above stated formula, we get the mean about second order as:

& E\left( {{X}^{2}} \right)={{0}^{2}}\times \dfrac{1}{8}+{{1}^{2}}\times \dfrac{3}{8}+{{2}^{2}}\times \dfrac{3}{8}+{{3}^{2}}\times \dfrac{1}{8} \\\ & \Rightarrow E\left( {{X}^{2}} \right)=0+\dfrac{3}{8}+\dfrac{12}{8}+\dfrac{9}{8} \\\ & \Rightarrow E\left( {{X}^{2}} \right)=\dfrac{24}{8} \\\ & \Rightarrow E\left( {{X}^{2}} \right)=3 \\\ \end{aligned}$$ Now, by substituting the value of the $E\left( {{X}^{2}} \right)$ as 3 and $E\left( X \right)$as $\dfrac{3}{2}$to get the variance as: $\begin{aligned} & {{\sigma }^{2}}=3-{{\left( \dfrac{3}{2} \right)}^{2}} \\\ & \Rightarrow {{\sigma }^{2}}=3-\dfrac{9}{4} \\\ & \Rightarrow {{\sigma }^{2}}=\dfrac{12-9}{4} \\\ & \Rightarrow {{\sigma }^{2}}=\dfrac{3}{4} \\\ \end{aligned}$ So, it gives the variance as $\dfrac{3}{4}$. Now, to get the value of standard deviation, we must know the relationship that standard deviation is the square root of the variance. So, standard deviation is given by: $\begin{aligned} & \sigma =\sqrt{{{\sigma }^{2}}} \\\ & \Rightarrow \sigma =\sqrt{\dfrac{3}{4}} \\\ & \Rightarrow \sigma =\dfrac{\sqrt{3}}{2} \\\ \end{aligned}$ So, the standard deviation is $\dfrac{\sqrt{3}}{2}$. **Hence, the variance of the distribution is $\dfrac{3}{4}$ and standard deviation is $\dfrac{\sqrt{3}}{2}$**. **Note:** Now, to solve these types of the questions we need to know some of the basic formulas of the distribution function so that we can proceed easily and find the required answer. Now, the basic formulas are: Mean is given by $E\left( X \right)=\sum\limits_{i=0}^{n}{{{p}_{i}}{{x}_{i}}}$ Variance is given by ${{\sigma }^{2}}=E\left( {{X}^{2}} \right)-{{\left( E\left( X \right) \right)}^{2}}$ Standard deviation is given by $\sigma =\sqrt{E\left( {{X}^{2}} \right)-{{\left( E\left( X \right) \right)}^{2}}}$.