Question #330176

Two dice are rolled. Let X be a random variable denoting the sum of the numbers


on the two dice.


i) Give the probability distribution of X


ii) Compute the expected value of X and its variance

1
Expert's answer
2022-04-27T10:58:16-0400

i)


123456123456723456783456789456789105678910116789101112\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c:c} & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hdashline 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hdashline 3 &4 & 5 & 6 & 7 & 8 & 9 \\ \hdashline 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ \hdashline 5 & 6 & 7 & 8 & 9 & 10 & 11 \\ \hdashline 6 & 7 & 8 & 9 & 10 & 11 & 12 \\ \hdashline \end{array}


xp(x)21/3631/1841/1251/965/3671/685/3691/36101/12111/18121/36\def\arraystretch{1.5} \begin{array}{c:c} x & p(x) \\ \hline 2 & 1/36 \\ \hdashline 3 & 1/18 \\ \hdashline 4 & 1/12 \\ \hdashline 5 & 1/9 \\ \hdashline 6 & 5/36 \\ \hdashline 7 & 1/6 \\ \hdashline 8 & 5/36 \\ \hdashline 9 & 1/36 \\ \hdashline 10 & 1/12 \\ \hdashline 11 & 1/18 \\ \hdashline 12 & 1/36 \\ \hdashline \end{array}

ii)


E(X)=2(136)+3(236)+4(336)+5(436)E(X)=2(\dfrac{1}{36})+3(\dfrac{2}{36})+4(\dfrac{3}{36})+5(\dfrac{4}{36})

+6(536)+7(636)+8(536)+9(436)+6(\dfrac{5}{36})+7(\dfrac{6}{36})+8(\dfrac{5}{36})+9(\dfrac{4}{36})

+10(336)+11(236)+12(136)=7+10(\dfrac{3}{36})+11(\dfrac{2}{36})+12(\dfrac{1}{36})=7

Var(X)=σ2=E(X2)(E(X))2Var(X)=\sigma^2=E(X^2)-(E(X))^2


=22(136)+32(236)+42(336)+52(436)=2^2(\dfrac{1}{36})+3^2(\dfrac{2}{36})+4^2(\dfrac{3}{36})+5^2(\dfrac{4}{36})

+62(536)+72(636)+82(536)+92(436)+6^2(\dfrac{5}{36})+7^2(\dfrac{6}{36})+8^2(\dfrac{5}{36})+9^2(\dfrac{4}{36})

+102(336)+112(236)+122(136)72+10^2(\dfrac{3}{36})+11^2(\dfrac{2}{36})+12^2(\dfrac{1}{36})-7^2

=356=\dfrac{35}{6}

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