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

Let, X be the age of the selected student.

Consider, the age of the student selected can be 14, 17, 15, 21, 19, 16, 18, 20.

So, the age of the selected student X can be 14, 15, 16, 17, 18, 19, 20.

Consider X be the difference between the number draw a table for the different sample space:

X	Number of students	P( X )
14	2	2 15
15	1	1 15
16	2	2 15
17	3	3 15
18	1	1 15
19	2	2 15
20	3	3 15
21	1	1 15

So the probability distribution is:

X	14	15	16	17	18	19	20	21
P( X )	2 15	1 15	2 15	3 15	1 15	2 15	3 15	1 15

The expression of the variance is,

Var( X )=E( X 2 )− [ E( X ) ] 2 (1)

Here, the mean is given by,

E( X )= ∑ i=1 n X i P i =( 14× 2 15 )+( 15× 1 15 )+( 16× 2 15 )+( 17× 3 15 )+( 18× 1 15 )+( 19× 2 15 ) +( 20× 3 15 )+( 21× 1 15 ) = 263 15 =17.53

Calculate the value of E( X 2 ),

E( X 2 )= ∑ i=1 n X i 2 P i =( 14 2 × 2 15 )+( 15 2 × 1 15 )+( 16 2 × 2 15 )+( 17 2 × 3 15 )+( 18 2 × 1 15 )+( 19 2 × 2 15 ) +( 20 2 × 3 15 )+( 21 2 × 1 15 ) = 4683 15

Substitute the values in equation (1),

Var( X )=E( X 2 )− [ E( X ) ] 2 = 4683 15 − ( 17.53 ) 2 =4.78

And standard deviation σ x is,

σ x = Var( x ) = 4.78 =2.18

Thus, the variance is 4.78 and standard deviation is 2.18.