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 ) =  \(\sum X^2_iP(X_i)\)