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

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

$=\sqrt{4.78}$

$=2.186 \approx 2.19$