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

There are 15 students in the class. Each student has the same chance to be chosen.
Therefore, the probability of each student to be selected is $\frac{1}{15}$.
The given information can be compiled in the frequency table as follows.
$P(X=14)=\frac{2}{15},P(X=15)=\frac{1}{15},P(X=16)=\frac{2}{15},P(X=16)=\frac{3}{15},P(X=18)=\frac{1}{15},P(X=19)=\frac{2}{15},P(X=20)=\frac{3}{15},P(X=21)=\frac{1}{15}$
Therefore, the probability distribution of random variable $X$ is as follows.
Then, mean of $X=E\left(X\right)$
$=\sum X_{i}P\left(X_i\right)$
$=14\times \frac{2}{15}+15\times \frac{1}{15}+16\times \frac{2}{15}+17\times \frac{3}{15}+18\times \frac{1}{15}+19\times \frac{2}{15}+20\times \frac{3}{15}+21\times \frac{1}{15}$
$=\frac{1}{15}(28+15+32+51+18+38+60+21)$
$=\frac{263}{15}$
$=17.53$
$E\left(X^2\right)=\sum X_i^{2}P\left(X_i\right)$
$={\left(14\right)}^2\cdot \frac{2}{15}+{\left(15\right)}^2\cdot \frac{1}{15}+{\left(16\right)}^2\cdot \frac{2}{15}+{\left(17\right)}^2\cdot \frac{3}{15}+{\left(18\right)}^2\cdot \frac{1}{15}+{\left(19\right)}^2\cdot \frac{2}{15}+{\left(20\right)}^2\cdot \frac{3}{15}+{\left(21\right)}^2\cdot \frac{1}{15}$
$=\frac{1}{15}\cdot (392+225+512+867+324+722+1200+441)$
$=\frac{4683}{15}$
$=312.2$
$\therefore$ Variance $\left(X\right)=E\left(X^2\right){\left[E\left(X\right)\right]}^2$
$=312.2-{\left(\frac{263}{15}\right)}^2$
$=312.2-307.4177$
$=4.7823$
$\approx 4.78$
Standard deviation $=\sqrt{Variance\left(X\right)}$
$=\sqrt{4.78}$
$=2.186$
$\approx 2.19$