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

Options

A. `(31)/6 and sqrt((31)/6)`
B. `(35)/6 and sqrt((35)/6)`
C. `(17)/6 and sqrt((17)/6)`
D. None of these

Answer

B

Explanation

Let X denotes the sum of the numbers obtained when two fair dice are rolled. So, X may have values,2,3,4,56,7,8,9,10,11 or 12
(as 1 can't be the sum of two numbers on fair dice)
`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 `X=SigmaXP(X)`
`([2xx1+3xx2+4xx3+5xx4+6xx5+7xx6+8xx5+9xx4+10xx3+11xx2+12xx1])/(36)`
`=(252)/(26)=7`
Variance `X=SigmaX^2P(X)-("Mean")^2`
`=([2^2xx1+3^2xx2+4^2xx3+5^2xx6^2xx5+7^2xx6+8^2xx5+9^2xx4+10^2xx3+11^2xx2+12^2xx1])/(36)-7^2`
`=(1974)/(36)-49=(1974-1764)/(36)=(210)/(36)=(35)/6`
Hence , `SD=sqrt("Variance")=sqrt((35)/6)`