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.

x	14	15	16	17	18	19	20	21
f	2	1	2	3	1	2	3	1

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.

x	14	15	16	17	18	19	20	21
f	$\frac{2}{15}$	$\frac{1}{15}$	$\frac{2}{15}$	$\frac{3}{15}$	$\frac{1}{15}$	$\frac{2}{15}$	$\frac{3}{15}$	$\frac{1}{15}$

Then, mean of X = E(X)
= $\sum X_tP(X_t)$
=14 × $\frac{2}{15}$+15 × $\frac{1}{15}$+16 × $\frac{2}{15}$ +17 × $\frac{3}{15}$+18 × $\frac{1}{15}$ + 19 × $\frac{2}{15}$+20 × $\frac{3}{15}$+21 × $\frac{1}{15}$

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

= $\frac{263}{15}$
= 17.53
E(X2 ) = $\sum X^2_iP(X_i)$