Consider rolling two fair six sided dice with faces numbered 1, 2, 3, 4, 5, 6. (a) Write out the sample space for one roll of the two dice. (b) Let X be the result of the first die and Y be the result of the second die. Consider the random variable Z = |X − Y |, the absolute value of the difference of the first die minus the second die. What is the sample space for Z? (c) What is the expected value of Z? (d) Write out the cumulative distribution function in tabular form for the random variable Z

a. The sample space for [tex](X,Y)[/tex] has 36 possible outcomes,(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),(2, 1), (2, 2), (2, 3), ...,(3, 1), (3, 2), ...,and so on.b. In the array above, replace each coordinate pair with the absolute difference between the coordinates; you'd end up with an array like0, 1, 2, 3, 4, 5,1, 0, 1, 2, 3, 4,2, 1, 0, 1, 2, 3,3, 2, 1, 0, 1, 2,4, 3, 2, 1, 0, 1,5, 4, 3, 2, 1, 0so that the sample space for [tex]Z[/tex] is the set of integers, {0, 1, 2, 3, 4, 5}.c. The sample space above illustrates that[tex]P(Z=z)=\begin{cases}\frac16&\text{for }z=0\\\frac5{18}&\text{for }z=1\\\frac29&\text{for }z=2\\\frac16&\text{for }z=3\\\frac19&\text{for }z=4\\\frac1{18}&\text{for }z=5\\0&\text{otherwise}\end{cases}[/tex]so the expected value of [tex]Z[/tex] is[tex]E[Z]=\displaystyle\sum_{z=0}^5z\,P(Z=z)=\frac06+\frac5{18}+\frac49+\frac12+\frac49+\frac5{18}=\boxed{\frac{35}{18}}[/tex]d. Not sure what you mean by "tabular form", but you can obtain the CDF from the usual definition,[tex]F_Z(z)=P(Z\le z)=\begin{cases}0&\text{for }z<0\\\frac16&\text{for }z<1\\\frac49&\text{for }z<2\\\frac23&\text{for }z<3\\\frac56&\text{for }z<4\\\frac{17}{18}&\text{for }z<5\\1&\text{for }z\ge5\end{cases}[/tex]