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?

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

μX  = Σ XiPi 

= 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

σX2 = 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