Answer to Question #340138 in Statistics and Probability for Maneo

Question

In an experiment of rolling a balanced die twice, let X be the maximum of the two numbers obtained.

1) What are the possible values of X?

2) Write down the sample space

3) Wat is the probability function for X?

4) Determine and sketch the probability mass function of X.

5) Find the expected value of X.

6) Find the variance of X.

Answer 1

$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c:c} & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline 1 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hdashline 2 & 2 & 2 & 3 & 4 & 5 & 6 \\ \hdashline 3 & 3 & 3 & 3 & 4 & 5 & 6 \\ \hdashline 4 & 4 & 4 & 4 & 4 & 5 & 6 \\ \hdashline 5 & 5 & 5 & 5 & 5 & 5 & 6 \\ \hdashline 6 & 6 & 6 & 6 & 6 & 6 & 6 \\ \hdashline \end{array}$

1) The possible values of $X$ are $1, 2, 3, 4, 5,$ and $6.$

2)

$S=\{1,2,3,4,5,6\}$

3)

$\def\arraystretch{1.5} \begin{array}{c:c} x & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline p(x) & 1/36 & 3/36 & 5/36 & 7/36 & 9/36 & 11/36 \\ \end{array}$

4)

$f(x) = \begin{cases} 1/36 &\ x=1 \\ 3/36&\ x=2 \\ 5/36&\ x=3 \\ 7/36&\ x=4 \\ 9/36&\ x=5 \\ 11/36&\ x=6 \\ \end{cases}$

5)

$\mu=E(X)=\dfrac{1}{36}(1)+\dfrac{3}{36}(2)+\dfrac{5}{36}(3)$

$+\dfrac{7}{36}(4)+\dfrac{9}{36}(5)+\dfrac{11}{36}(6)=\dfrac{161}{36}$

6)

$E(X^2)=\dfrac{1}{36}(1)^2+\dfrac{3}{36}(2)^2+\dfrac{5}{36}(3)^2$

$+\dfrac{7}{36}(4)^2+\dfrac{9}{36}(5)^2+\dfrac{11}{36}(6)^2=\dfrac{791}{36}$

$Var(X)=\sigma^2=E(X^2)-(E(X))^2$

$=\dfrac{791}{36}-(\dfrac{161}{36})^2=\dfrac{2555}{1296}$