Answer to Question #208473 in Statistics and Probability for Ishanka Amarasingh

Question

7. Two fair cubical dice are thrown: one is red and one is blue. The random variable M represents the score on the red die minus the score on the blue die. (a) Find the distribution of M. (b) Write down E(M). (c) Find Var(M).

Answer 1

$$\def\arraystretch{1.5} \begin{array}{c:c:c} Sample & Red,Blue & M \\ \hline 1 & 1,1 & 0 \\ \hdashline 2 & 1,2 & -1 \\ \hdashline 3 & 1,3 & -2 \\ \hdashline 4 & 1,4 & -3\\ \hdashline 5 & 1,5 & -4 \\ \hdashline 6 & 1,6 & -5 \\ \hdashline 7 & 2,1 & 1 \\ \hdashline 8 & 2,2 & 0 \\ \hdashline 9 & 2,3 & -1 \\ \hdashline 10 & 2,4 & -2\\ \hdashline 11 & 2,5 & -3 \\ \hdashline 12 & 2,6 & -4 \\ \hdashline 13 & 3,1 & 2 \\ \hdashline 14 & 3,2 & 1 \\ \hdashline 15 & 3,3 & 0 \\ \hdashline 16 & 3,4 & -1\\ \hdashline 17 & 3,5 & -2 \\ \hdashline 18 & 3,6 & -3 \\ \hdashline 19 & 4,1 & 3 \\ \hdashline 20 & 4,2 & 2 \\ \hdashline 21 & 4,3 & 1 \\ \hdashline 22 & 4,4 & 0\\ \hdashline 23 & 4,5 & -1 \\ \hdashline 24 & 4,6 & -2 \\ \hdashline 25 & 5,1 & 4 \\ \hdashline 26 & 5,2 & 2 \\ \hdashline 27 & 5,3 & 2 \\ \hdashline 28 & 5,4 & 1\\ \hdashline 29 & 5,5 & 0 \\ \hdashline 30 & 5,6 & -1 \\ \hdashline 31 & 6,1 & 5 \\ \hdashline 32 & 6,2 & 4 \\ \hdashline 33 & 6,3 & 3 \\ \hdashline 34 & 6,4 & 2\\ \hdashline 35 & 6,5 & 1 \\ \hdashline 36 & 6,6 & 0 \\ \hline \end{array}$$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} M & f & f(M) & Mf(M)& M^2f(M) \\ \hline -5 & 1 & 1/36 & -5/36 & 25/36 \\ \hdashline -4 & 2 & 2/36 & -8/36 & 32/36 \\ \hdashline -3 & 3 & 3/36 & -9/36 & 27/36 \\ \hdashline -2 & 4 & 4/36 & -8/36 & 16/36 \\ \hdashline -1 & 5 & 5/36 & -5/36 & 5/36 \\ \hdashline 0 & 6 & 6/36 & 0 & 0 \\ \hdashline 1 & 5 & 5/36 & 5/36 & 5/36 \\ \hdashline 2 & 4 & 4/36 & 8/36 & 16/36 \\ \hdashline 3 & 3 & 3/36 & 9/36 & 27/36 \\ \hdashline 4 & 2 & 2/36 & 8/36 & 32/36 \\ \hdashline 5 & 1 & 1/36 & 5/36 & 25/36 \\ \hdashline Total & 36 & 1 & 0 & 35/6 \\ \hline \end{array}$$

1.

$$\begin{matrix} M & p(M) \\ \hline -5 & 1/36 \\ -4 & 2/36 \\ -3 & 3/36 \\ -2 & 4/36 \\ -1 & 5/36 \\ 0 & 6/36 \\ 1 & 5/36 \\ 2 & 4/36 \\ 3 & 3/36 \\ 4 & 2/36 \\ 5 & 1/36 \\ \end{matrix}$$

2.

$$E(M)=\sum Mp(M)=0$$

The mean of the sampling distribution of the sample means is equal to the

the mean of the population.

$$E(\bar{X})=0=\mu$$

3.

$$Var(M)=\sum M^2p(M)-(\sum Mp(M))^2$$

$$=\dfrac{35}{6}-(0)^2=\dfrac{35}{6}$$