Question

A box contains 12 similary shaped colored balls let the random variable X represents the number on the ball (numbered from 1-12) and a ball is drawn from a box What is the probobility distrubution of X.What mean and varian of X?

Answer 1

Since all the 12 similarly shaped colored balls are drawn at random, each of the numbers on the balls has equal chances of being selected, with each having a probability equal to $\frac{1}{12}$

Therefore, the probability distribution of X is:

X P(X)

1 $\frac{1}{12}$

2 $\frac{1}{12}$

3 $\frac{1}{12}$

3 $\frac{1}{12}$

4 $\frac{1}{12}$

5 $\frac{1}{12}$

6 $\frac{1}{12}$

7 $\frac{1}{12}$

8 $\frac{1}{12}$

9 $\frac{1}{12}$

10 $\frac{1}{12}$

11 $\frac{1}{12}$

12 $\frac{1}{12}$

The mean of X

μ_X  = Σ X_iP_i 

= 1($\frac{1}{12}$) + 2($\frac{1}{12}$) + 3($\frac{1}{12}$) +4($\frac{1}{12}$) + 5($\frac{1}{12}$) + 6($\frac{1}{12}$) + 7($\frac{1}{12}$) +8($\frac{1}{12}$)+ 9($\frac{1}{12}$)

+10($\frac{1}{12}$) + 11($\frac{1}{12}$) +12($\frac{1}{12}$)

= $\frac{78}{12}$

= 6.5

Answer: Mean of X = 6.5

The variance X

σ_X² = Σ(x – μ)²⋅ P(x)

=$\frac{1}{12}$(1-6.5)² + $\frac{1}{12}$(2-6.5)² + $\frac{1}{12}$(3-6.5)²+ $\frac{1}{12}$(4-6.5)² + $\frac{1}{12}$(5-6.5)² + $\frac{1}{12}$(6-6.5)² +

$\frac{1}{12}$(7-6.5)²+ $\frac{1}{12}$(8-6.5)²+ $\frac{1}{12}$(9-6.5)² + $\frac{1}{12}$(10-6.5)² + $\frac{1}{12}$(11-6.5)²+ $\frac{1}{12}$(12-6.5)²

=  143/12

= 11.9167

Answer: Variance of X = 11.9167