Question

A class has $15$ students whose ages are $14,17,15,14,21,17,19,20,16,18,20,17,16,19$ and $20$ years. One student is selected in such a manner that each has the same chance of being chosen and the age $X$ of the selected student is recorded. What is the probability distribution of the random variable $X?$ Find mean, variance and standard deviation of $X.$

Answer 1

Given: Total no students $=15$

Whose ages are $14,17,15,14,21,17,19,20,$
$16,18,20,17,$
$16,19$ and $20$ years.
Let $X$: The age of student selected
The age of student selected can be
$14,17,15,21,19,16,18,20$
So,$X$ can have the value
$14,15,16,17,18,19,20,21$
From given data we can draw a table,

$\begin{array}{ccccccccc}X& 14& 15& 16& 17& 18& 19& 20& 21\\ F& 2& 1& 2& 3& 1& 2& 3& 1\end{array}$

Now, probabilities of ages of each student are
$P(X=14)=\frac{2}{15}$

$P(X=15)=\frac{1}{15}$

$P(X=16)=\frac{2}{15}$

$P(X=17)=\frac{3}{15}$

$P(X=18)=\frac{1}{15}$

$P(X=19)=\frac{2}{15}$

$P(X=20)=\frac{3}{15}$

$P(X=21)=\frac{1}{15}$

Hence, the required probability distribution is,

$\begin{array}{ccccccccc}X& 14& 15& 16& 17& 18& 19& 20& 21\\ P\left(X\right)& \frac{2}{15}& \frac{1}{15}& \frac{2}{15}& \frac{3}{15}& \frac{1}{15}& \frac{2}{15}& \frac{3}{15}& \frac{1}{15}\end{array}$

The Mean of $X$ is given by

$\mu =E\left(X\right)={\sum }_{i=1}^n{x}_i{p}_i$

$\Rightarrow E\left(X\right)=14\times \frac{2}{15}+15\times \frac{1}{15}+16\times \frac{2}{15}+$

$17\times \frac{3}{15}+18\times \frac{1}{15}+19\times \frac{2}{15}+$

$20\times \frac{3}{15}+21\times \frac{1}{15}$

$\Rightarrow E\left(X\right)=\frac{28+15+32+51+18+38+60+21}{15}$

$\Rightarrow E\left(X\right)=\frac{263}{15}$

$\therefore E\left(X\right)=17.53$

We know that variance of $X$ is

$Var\left(X\right)=E\left(X^2\right)-\left(E\right(X){)}^2$ ... $\left(1\right)$

Here, $E\left(X\right)=17.53$

Finding $E\left(X^2\right)$

$E\left(X^2\right)={\sum }_{i=1}^n{x}_i^2\cdot p\left(x_i\right)$

$\Rightarrow E\left(X^2\right)=(14{)}^2\times \frac{2}{15}+(15{)}^2\times \frac{1}{15}$
$+(16{)}^2\times \frac{2}{15}+(17{)}^2\times \frac{3}{15}+(18{)}^2\times \frac{1}{15}$
$+(19{)}^2\times \frac{2}{15}+(20{)}^2\times \frac{3}{15}+(21{)}^2\times \frac{1}{15}$

$\Rightarrow E\left(X^2\right)=\frac{392+225+512+867+324+722+1200+441}{15}$

$\Rightarrow E\left(X^2\right)=\frac{4683}{15}$

$\therefore E\left(X^2\right)=312.2$

Now putting the value of $E\left(X^2\right)\&E\left(X\right)$ in $\left(1\right)$

$\Rightarrow Var\left(X\right)=312.2-(17.53{)}^2$

$\therefore Var\left(X\right)=312.2-307.417\approx 4.78$

$\therefore$Standard deviation $=\sqrt{\left(Var\right(X\left)\right)}$

$=\sqrt{4.78}$

$\therefore$ Standard deviation $=2.19$