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

Question

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

Question #208473

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).

Expert's answer

"\\def\\arraystretch{1.5}\n \\begin{array}{c:c:c}\n Sample & Red,Blue & M \\\\\n\\hline\n 1 & 1,1 & 0 \\\\\n \\hdashline\n 2 & 1,2 & -1 \\\\\n \\hdashline\n 3 & 1,3 & -2 \\\\\n \\hdashline\n 4 & 1,4 & -3\\\\\n \\hdashline\n 5 & 1,5 & -4 \\\\\n\\hdashline\n 6 & 1,6 & -5 \\\\\n \\hdashline\n7 & 2,1 & 1 \\\\\n \\hdashline\n8 & 2,2 & 0 \\\\\n \\hdashline\n9 & 2,3 & -1 \\\\\n \\hdashline\n10 & 2,4 & -2\\\\\n \\hdashline\n 11 & 2,5 & -3 \\\\\n\\hdashline\n 12 & 2,6 & -4 \\\\\n\\hdashline\n13 & 3,1 & 2 \\\\\n \\hdashline\n14 & 3,2 & 1 \\\\\n \\hdashline\n15 & 3,3 & 0 \\\\\n \\hdashline\n16 & 3,4 & -1\\\\\n \\hdashline\n17 & 3,5 & -2 \\\\\n\\hdashline\n18 & 3,6 & -3 \\\\\n \\hdashline\n19 & 4,1 & 3 \\\\\n \\hdashline\n20 & 4,2 & 2 \\\\\n \\hdashline\n 21 & 4,3 & 1 \\\\\n \\hdashline\n22 & 4,4 & 0\\\\\n \\hdashline\n23 & 4,5 & -1 \\\\\n\\hdashline\n24 & 4,6 & -2 \\\\\n \\hdashline\n25 & 5,1 & 4 \\\\\n \\hdashline\n26 & 5,2 & 2 \\\\\n \\hdashline\n 27 & 5,3 & 2 \\\\\n \\hdashline\n 28 & 5,4 & 1\\\\\n \\hdashline\n 29 & 5,5 & 0 \\\\\n\\hdashline\n 30 & 5,6 & -1 \\\\\n \\hdashline\n31 & 6,1 & 5 \\\\\n \\hdashline\n 32 & 6,2 & 4 \\\\\n \\hdashline\n 33 & 6,3 & 3 \\\\\n \\hdashline\n 34 & 6,4 & 2\\\\\n \\hdashline\n 35 & 6,5 & 1 \\\\\n\\hdashline\n 36 & 6,6 & 0 \\\\\n \n \\hline\n\\end{array}"

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

1.

"\\begin{matrix}\n M & p(M) \\\\\n\\hline \n -5 & 1\/36 \\\\\n-4 & 2\/36 \\\\\n -3 & 3\/36 \\\\\n-2 & 4\/36 \\\\\n -1 & 5\/36 \\\\\n0 & 6\/36 \\\\\n 1 & 5\/36 \\\\\n2 & 4\/36 \\\\\n 3 & 3\/36 \\\\\n4 & 2\/36 \\\\\n 5 & 1\/36 \\\\\n\\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}"