# What is the probability distribution of the random variable X? Find mean, variance and standard deviation of X. - Mathematics

Sum

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.

#### 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 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) = 2/15, P(X =15) = 1/15 , P(X=16)= 2/15,

P(X=16)= 3/15,P(X=18)=1/15, P(X =19) = 2/15, P(X =20) = 3/15, P(X =21) = 1/15

Therefore, the probability distribution of random variable X is as follows.

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

Then, mean of X = E(X)

ΣXi  P(Xi)

=14xx2/15+15xx1/15+16xx2/15+17xx3/15+18xx1/15+19xx2/15+20xx3/15+21xx1/15

= 1/15(28 + 15 + 32 + 51+ 18 + 38 + 60 + 21)

= 263/15

= 17.53

E(X2) = ΣXi2  P(Xi)

= (14)^2xx2/15+(15)^2xx1/15+(16)^2xx2/15+(17)^2xx3/15+(18)^2xx1/15+(19)^2xx2/15+(20)^2xx3/15+(21)^2xx1/15

= 1/15(392 + 225 + 512 + 867 + 324 + 722 + 1200 + 441)

= 4683/15

= 312.2

Variance = V(X) = Σxi2 . P(xi) - [E(X)]2

= 312.2 - (17.53)2

= 312.2 - 307.3 = 4.9

Standard deviation = sqrt(V(X))= sqrt(4.9) = 2.21

Hence, mean = 17.53, variance = 4.9 and standard deviation = 2.21.

Is there an error in this question or solution?

#### APPEARS IN

NCERT Class 12 Maths
Chapter 13 Probability
Q 14 | Page 571