Let X denote the sum of the numbers obtained when two fair dice are rolled. Find the variance and standard deviation of X.

Answer

N/a

Explanation

Let `X` repesents the sum of the numbers obtained when two unbiased dice are thrown. Since the sum of two numbers obtained cannot be equal to 1 when two unbiased dice are thrown. Therefore, the values of `X` can be 2,3,4,5,6,7,8,9,10,11,12.
`P(X=2)=P[{1,1}]=1/36`
`P(X=3)=P[{(1,2),(2,1)}]=2/36`
`P(X=4)=P[{(1,3),(2,2),(3,1)}]=3/36`
`P(X=5)=P[{(1,4),(2,3),(3,2),(4,1)}]=4/36`
`P(X=6)=P[{(1,5),(2,4),(3,3),(4,2),(5,1)}]`
`=5/36`
`P(X=7)=P[{(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)}]`
`=6/36`
`P(X=8)=P[{(2,6),(3,5),(4,4),(5,3),(6,2)}]`
`=5/36`
`P(X=9)=P[{(3,6),(4,5),(5,4),(6,3)}]=4/36`
`P(X=10)=P[{(4,6),(5,5),(6,4)}]=3/36`
`P(X=11)=P[{(5,6),(6,5)}]=2/36`
`P(X=12)=P[(6,6)]=1/36`

Mean of random variable `X`
`=sumXP(x)`
`=([2xx1+3xx2+4xx3+5xx+6xx5+7xx6+8xx5+9xx4+10xx3+11xx2+12xx1])/36`
`=([2+6+12+20+30+42+40+36+30+22+12])/36=252/36=7`
Variance of random variable `X=sumX^(2)P(X)-("mean")^(2)`
`=([2^(2)xx1+3^(2)xx2+4^(2)xx3+5^(2)xx4+6^(2)xx5+7^(2)xx6+8^(2)xx5+9^(2)xx4+10^(2)xx3+11^(2)xx2+12^(2)xx1])/36-7^(2)`
`=([4+18+48+100+180+294+320+324+300+242+144])/36`
`=1974/36-49=(1974-1764)/36=210/36=35/6`
Therefoe, standard deviation,
`SD=sqrt("variance")=sqrt(35/6)=sqrt(5.833)=2.415`