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Q. 13    Let X denote the sum of the numbers obtained when two fair dice are rolled. Find the variance and standard deviation ofX.

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X denote the sum of the numbers obtained when two fair dice are rolled.

Total observations = 36

X can be 2,3,4,5,6,7,8,9,10,11,12

P(X=2)=P(1,1)=\frac{1}{36}

P(X=3)=P(1,2)+P(2,1)=\frac{2}{36}=\frac{1}{18}

P(X=4)=P(1,3)+P(3,1)+P(2,2)=\frac{3}{36}=\frac{1}{12}

P(X=5)=P(1,4)+P(4,1)+P(2,3)+P(3,2)=\frac{4}{36}=\frac{1}{9}

P(X=6)=P(1,5)+P(5,1)+P(2,4)+P(4,2)+P(3,3)=\frac{5}{36}

P(X=7)=P(1,6)+P(6,1)+P(2,5)+P(5,2)+P(3,4)+P(4,3)=\frac{6}{36}=\frac{1}{6}

P(X=8)=P(2,6)+P(6,2)+P(3,5)+P(5,3)+P(4,4)=\frac{5}{36}P(X=9)=P(3,6)+P(6,3)+P(4,5)+P(5,4)=\frac{4}{36}=\frac{1}{9}

P(X=10)=P(4,6)+P(6,4)+P(5,5)=\frac{3}{36}=\frac{1}{12}

P(X=11)=P(5,6)+P(6,5)=\frac{2}{36}=\frac{1}{18}

P(X=12)=P(6,6)=\frac{1}{36}

Probability distribution is as follows : 

X 2 3 4 5 6 7 8 9 10 11 12
P(X) \frac{1}{36} \frac{1}{18} \frac{1}{12} \frac{1}{9} \frac{5}{36} \frac{1}{6} \frac{5}{36} \frac{1}{9} \frac{1}{12} \frac{1}{18} \frac{1}{36}

   E(X)=2\times \frac{1}{36}+3\times \frac{1}{18}+4\times \frac{1}{12}+5\times \frac{1}{9}+6\times \frac{5}{36}+7\times \frac{1}{6}+8\times \frac{5}{36}+9\times \frac{1}{9}+10\times \frac{1}{12}+11\times \frac{1}{18}+12\times \frac{1}{36}

E(X)=\frac{1}{18}+\frac{1}{6}+\frac{1}{3}+\frac{5}{9}+\frac{5}{6}+\frac{7}{6}+\frac{10}{9}+1+\frac{5}{6}+\frac{11}{18}+\frac{1}{3}

E(X)=7

E(X^2)=4\times \frac{1}{36}+9\times \frac{1}{18}+16\times \frac{1}{12}+25\times \frac{1}{9}+36\times \frac{5}{36}+49\times \frac{1}{6}+64\times \frac{5}{36}+81\times \frac{1}{9}+100\times \frac{1}{12}+121\times \frac{1}{18}+144\times \frac{1}{36}

E(X^2)=\frac{987}{18}=\frac{329}{6}=54.833

Variance = E(X^2)-(E(X))^2

                      =54.833-7^2

                      =54.833-49

                       =5.833

Standard deviation ==\sqrt{5.833}=2.415

Posted by

seema garhwal

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