Question

Find the variance and STD deviation of the r.v. $x$ whose probability distribution is given:

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

Answer 1

Hint: For solving this question in which we need to find the variance and STD deviation, we will draw a table and will find the some related values that are required to find the variance. Then, we will use the formula of variance $\sum{{{x}^{2}}\cdot P(x)}-{{\left( \sum{x\cdot P\left( x \right)} \right)}^{2}}$ and simplify it. Then, we will find the square root of variance to find the STD deviation of random variables.

Complete step by step solution:
Since, we have some given data. We will use given data and find the some value by using table as:

$x$	$P\left( X=x \right)$	$x\cdot P\left( x \right)$	${{x}^{2}}\cdot P\left( x \right)$
$0$	$\dfrac{1}{8}$	$0\times \dfrac{1}{8}=0$	${{0}^{2}}\times \dfrac{1}{8}=0$
$1$	$\dfrac{3}{8}$	$1\times \dfrac{3}{8}=\dfrac{3}{8}$	${{1}^{2}}\times \dfrac{3}{8}=1\times \dfrac{3}{8}=\dfrac{3}{8}$
$2$	$\dfrac{3}{8}$	$2\times \dfrac{3}{8}=\dfrac{6}{8}$	${{2}^{2}}\times \dfrac{3}{8}=4\times \dfrac{3}{8}=\dfrac{12}{8}$
$3$	$\dfrac{1}{8}$	$3\times \dfrac{1}{8}=\dfrac{3}{8}$	${{3}^{2}}\times \dfrac{1}{8}=9\times \dfrac{1}{8}=\dfrac{9}{8}$

Now, we will find the sum of $x\cdot P\left( x \right)$ as:
$\Rightarrow \sum{x\cdot P\left( x \right)}=0+\dfrac{3}{8}+\dfrac{6}{8}+\dfrac{3}{8}$
Since, the denominator of all the fractions is the same. So, we can add them directly as:
$\Rightarrow \sum{x\cdot P\left( x \right)}=\dfrac{3+6+3}{8}$
Now, we will do addition in numerator as:
$\Rightarrow \sum{x\cdot P\left( x \right)}=\dfrac{12}{8}$
Here, we will simplify the fraction into its simplest form as:
$\Rightarrow \sum{x\cdot P\left( x \right)}=\dfrac{3}{2}$
Now, we will calculate the sum of ${{x}^{2}}\cdot P\left( x \right)$ as:
$\Rightarrow \sum{{{x}^{2}}\cdot P\left( x \right)=0+\dfrac{3}{8}+\dfrac{12}{8}+\dfrac{9}{8}}$
Again, we will follow rule of addition in this step as the denominator is same:
$\Rightarrow \sum{{{x}^{2}}\cdot P\left( x \right)=\dfrac{3+12+9}{8}}$
Now, add the numerator as:
$\Rightarrow \sum{{{x}^{2}}\cdot P\left( x \right)=\dfrac{24}{8}}$
Here, we will simplify it as:
\[\Rightarrow \sum{{{x}^{2}}\cdot P\left( x \right)=3}\]
Since, we got the corresponding values of the variance formula. We will use it in the formula of variance as:
$\Rightarrow \operatorname{var}=\sum{{{x}^{2}}\cdot P(x)}-{{\left( \sum{x\cdot P\left( x \right)} \right)}^{2}}$
We will substitute the corresponding values as:
$\Rightarrow \operatorname{var}=3-{{\left( \dfrac{3}{2} \right)}^{2}}$
After squaring the fraction, we will have $\dfrac{9}{4}$ as:
$\Rightarrow \operatorname{var}=3-\dfrac{9}{4}$
Now, we will do the addition as:
$\Rightarrow \operatorname{var}=\dfrac{12-9}{4}$
After subtraction, we will get the variance.
$\Rightarrow \operatorname{var}=\dfrac{3}{4}$
Now, we can calculate STD deviation by taking square root of variance as:
$\sigma =\sqrt{\operatorname{var}}$
Here, we will apply the value of variance as:
$\sigma =\sqrt{\dfrac{3}{4}}$
We will simplify the above step as:
$\sigma =\dfrac{\sqrt{3}}{2}$
Hence, the variance of the random variables is $\dfrac{3}{4}$ and standard deviation is $\dfrac{\sqrt{3}}{2}$.

Note: The variance is used to measure the spread of date from mean value and it is calculated as the average squared deviation for each number from the mean of data. And the standard deviation is the square root of the variance. It is also used to measure the spread of data.