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

Solution:

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(X)

$=\sum X_{i} P\left(X_{i}\right)$

$=14 \times \frac{2}{15}+15 \times \frac{1}{15}+16 \times \frac{2}{15}+17 \times \frac{3}{15}+18 \times \frac{1}{15}+19 \times \frac{2}{15}+20 \times \frac{3}{15}+21 \times \frac{1}{15}$

$=\frac{1}{15}(28+15+32+51+18+38+60+21)$

$=\frac{263}{15}$

$=17.53$

$\mathrm{E}\left(\mathrm{X}^{2}\right)=\sum \mathrm{X}_{i}^{2} \mathrm{P}\left(\mathrm{X}_{i}\right)$

$=(14)^{2} \cdot \frac{2}{15}+(15)^{2} \cdot \frac{1}{15}+(16)^{2} \cdot \frac{2}{15}+(17)^{2} \cdot \frac{3}{15}+$

$(18)^{2} \cdot \frac{1}{15}+(19)^{2} \cdot \frac{2}{15}+(20)^{2} \cdot \frac{3}{15}+(21)^{2} \cdot \frac{1}{15}$

$=\frac{1}{15} \cdot(392+225+512+867+324+722+1200+441)$

$=\frac{4683}{15}$

$=312.2$

$\therefore$ Variance $(X)=E\left(X^{2}\right)-[E(X)]^{2}$

$=312.2-\left(\frac{263}{15}\right)^{2}$

$=312.2-307.4177$

$=4.7823$

$\approx 4.78$

Standard derivation $=\sqrt{\operatorname{Variance}(\mathrm{X})}$

$=\sqrt{4.78}$

$=2.186 \approx 2.19$