A coin tossed and die is rolled. The outcome of the coin is recoded "1" when it shows a head, and "0" when it shows a tail. The random variable gives the sum of the outcomes of coin and die. Compute the average value of the random variable?compute its variance and standard deviation

Solution:

$\def\arraystretch{1.5} \begin{array}{c:c:c} & 1 & 0 \\ \hline 1 & 1+1=2 & 1+0=1 \\ \hdashline 2 & 2+1=3 & 2+0=2 \\ \hdashline 3 & 3+1=4 & 3+0=3 \\ \hdashline 4 & 4+1=5 & 4+0=4 \\ \hdashline 5 & 5+1=6 & 5+0=5 \\ \hdashline 6 & 6+1=7 & 6+0=6 \\ \end{array}$$E(X)=mean=\dfrac{1}{12}(1)+\dfrac{2}{12}(2)+\dfrac{2}{12}(3)$$+\dfrac{2}{12}(4)+\dfrac{2}{12}(5)+\dfrac{2}{12}(6)+\dfrac{1}{12}(7)=4$

The average value of the random variable is 4.

$E(X^2)=\Sigma x^2.P(x) \\=1^2\times\dfrac1{12}+2^2\times\dfrac2{12}+3^2\times\dfrac2{12}+4^2\times\dfrac2{12}+5^2\times\dfrac2{12}+6^2\times\dfrac2{12}+7^2\times\dfrac1{12} \\=\dfrac{115}6$

Now, $Var(X)=E(X^2)-[E(X)]^2=\dfrac{115}6-4^2=\dfrac{19}6$

$S.D(X)=\sqrt{Var(X)}=\sqrt{\dfrac{19}6}=1.78$