Question

The probability distribution of a random variable ${\text{X}}$ is given below

Find the value of $k$ and the mean and variance of ${\text{X}}$.

$X = {x_i}$	$P\left( {X = {x_i}} \right)$
1	$k$
2	$2k$
3	$3k$
4	$4k$
5	$5k$

Answer 1

First, we will find the value of $k$ using the formula $\sum {{P_i}} = 1$. Then we will now find the value of ${{\text{X}}^2}{\text{P}}$ and ${\text{XP}}$ from the above table for mean, using the formula, $\sum {{\text{XP}}}$ and variance, using the formula, ${\text{Variance}} = \sum {{{\text{X}}^2}{\text{P}}} - {\left( {\sum {{\text{XP}}} } \right)^2}$

Complete step by step answer:

We will now form a table to find the value of the product $\sum {{P_i}}$ for the mean.

$X = {x_i}$	$P\left( {X = {x_i}} \right)$
1	$k$
2	$2k$
3	$3k$
4	$4k$
5	$5k$
$\sum P = 15k$

We know that the sum of $P\left( {X = {x_i}} \right)$is always equals to 1, that is, $\sum {{P_i}} = 1$.

Substituting the values of $\sum {{P_i}}$ to find the value of $k$ from the above formula, we get

$\Rightarrow 15k = 1$

Dividing the above equation by 15 on both sides, we get

$$\Rightarrow \dfrac{{15k}}{{15}} = \dfrac{1}{{15}} \\\ \Rightarrow k = \dfrac{1}{{15}} \\\$$

Hence, the value of $k$ is $\dfrac{1}{{15}}$.

We will now find the value of ${{\text{X}}^2}{\text{P}}$ ${\text{XP}}$from the above table for median and mode.

${\text{X}}$	${\text{P}}$	${\text{XP}}$	${{\text{X}}^2}{\text{P}}$
1	$k$	$k$	$k$
2	$2k$	$4k$	$8k$
3	$3k$	$9k$	$27k$
4	$4k$	$16k$	$64k$
5	$5k$	$25k$	$125k$
$\sum {\text{P}} = 15k$	$\sum {{\text{XP}}} = 55k$	$\sum {{{\text{X}}^2}{\text{P}}} = 225k$

Substituting the value of $k$ in $\sum {{\text{XP}}}$ and $\sum {{{\text{X}}^2}{\text{P}}}$, we get

$$\Rightarrow \sum {{\text{XP}}} = 55 \times \dfrac{1}{{15}} \\\ \Rightarrow \sum {{\text{XP}}} = 3.66 \\\$$ $$\Rightarrow \sum {{{\text{X}}^2}{\text{P}}} = 225 \times \dfrac{1}{{15}} \\\ \Rightarrow \sum {{{\text{X}}^2}{\text{P}}} = 15 \\\$$

Therefore, the mean of the given data is $3.66$.
We know that the formula to calculate variance is ${\text{Variance}} = \sum {{{\text{X}}^2}{\text{P}}} - {\left( {\sum {{\text{XP}}} } \right)^2}$.

Substituting these values in the above formula of variance, we get

$$\Rightarrow {\text{Variance}} = 15 - {\left( {3.66} \right)^2} \\\ \Rightarrow {\text{Variance}} = 15 - 13.4 \\\ \Rightarrow {\text{Variance}} = 1.6 \\\$$

Hence, the variance of the given data is $1.6$.

Note: In solving these types of questions, students should know the formulae of mean, median, and mode. The question is really simple, students should note down the values from the problem really carefully, else the answer can be wrong. One should take care while finding the mean, variance, and avoid calculation mistakes.