Question

A class has 15 students whose ages are 14, 17, 15, 14, 21, 17, 19, 20, 16, 18, 20, 17, 16, 19, and 20yr. One student is selected in such a manner that each has the same chance of being of 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 (SD) of X.

Answer 1

Here, total students=15

The ages of students in ascending order are 14, 14, 15, 16, 16, 17, 17, 17, 18, 19, 19, 20, 20, 20, 21

Now, $P(X=14)=\frac{2}{15},P(X=15)=\frac{1}{15},P(X=16)=\frac{2}{15},P(X=17)=\frac{3}{15}$

$P(X=18)=\frac{1}{15},P(X=19)=\frac{2}{15},P(X=20)=\frac{3}{15},P(X=21)=\frac{1}{15}$

Therefore, the probability distribution of random variable X is as follows:

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

The third row gives the probability distribution of X.

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

$=\frac{28+15+32+51+18+38+60+21}{15}=\frac{263}{15}=17.53$

Variance $X=\sum X^{2}P\left(X\right)-Mea{n}^2$

$\frac{\left[\begin{array}{c}(14{)}^2\times 2+(15{)}^2\times 1+(16{)}^2\times 2+(17{)}^2\times 3\\ +(18{)}^2\times 1+(19{)}^2\times 2+(20{)}^2\times 3+(21{)}^2\times 1\end{array}\right]}{15}-{\left(\frac{263}{15}\right)}^2=\frac{392+225+512+867+324+722+1200+441}{15}-{\left(\frac{263}{15}\right)}^2=\frac{4683}{15}-{\left(\frac{263}{15}\right)}^2=312.2-307.4=4.8$

SD of $X=\sqrt{Variance}=\sqrt{4.8}=2.19$