Question

2. Two fair six-sided dice are tossed independently. Let M = the maximum of the two tosses (so M (1,5) = 5, M (3,3) = 3, etc). (a) I4 ptsl Find the probability mass function of M. (b) 14 pts] Find the cumulative distribution function of M and graph it. (c) 12 pts] Find the expected value of M (d) 12 pts] Find the variance of M. (e) 12 pts] Find the standard deviation of M.

Answer 1

Solution:

   we are given that : two fair six sided dice are tossed independently.

M = Maximum of the two tosses.

Thus we get:

First Die	Second Die	M = Maximum of two dice
1	1	1
1	2	2
1	3	3
1	4	4
1	5	5
1	6	6
2	1	2
2	2	2
2	3	3
2	4	4
2	5	5
2	6	6
3	1	3
3	2	3
3	3	3
3	4	4
3	5	5
3	6	6
4	1	4
4	2	4
4	3	4
4	4	4
4	5	5
4	6	6
5	1	5
5	2	5
5	3	5
5	4	5
5	5	5
5	6	6
6	1	6
6	2	6
6	3	6
6	4	6
6	5	6
6	6	6

Part a) Probability mass function of M.

We need to find frequencies of each possible values of M and then divide each frequency by N = 36

M	f = frequency	P(M)
1	1	0.0278
2	3	0.0833
3	5	0.1389
4	7	0.1944
5	9	0.2500
6	11	0.3056
	N = 36

Part b) Cumulative distribution function of M and Graph it.

To get cumulative distribution function , we need to find F(M) column

That is find cumulative sum of probabilities of each value of M.

Thus we get :

M	P(M)	Calculations of F(M)	F(M)
1	0.0278	0.0278	0.0278
2	0.0833	0.0278+0.0833=0.1111	0.1111
3	0.1389	0.1111+0.1389=0.2500	0.2500
4	0.1944	0.2500+0.1944=0.4444	0.4444
5	0.2500	0.4444+0.2500=0.6944	0.6944
6	0.3056	0.6944+0.3056=1.0000	1.0000

CDF of F(M) 1.2 1.0000 0.8 0.6944 2 0.6 0.4444 0.4 0.2500 0.0278 4

Part c) Expected value of M

$E(M)= \sum M \times P(M)$

M	P(M)	M * P(M)
1	0.0278	0.0278
2	0.0833	0.1667
3	0.1389	0.4167
4	0.1944	0.7778
5	0.2500	1.2500
6	0.3056	1.8333
		$\sum M \times P(M)=4.4722$

Thus

$E(M)= \sum M \times P(M)$

$E(M)=4.4722$

Part d) Variance of M

$Var(M)=E(M^{2})-(E(M))^{2}$

Where

$E(M^{2})=\sum M^{2} \times P(M)$

M	P(M)	M * P(M)	M^2 * P(M)
1	0.0278	0.0278	0.0278
2	0.0833	0.1667	0.3333
3	0.1389	0.4167	1.2500
4	0.1944	0.7778	3.1111
5	0.2500	1.2500	6.2500
6	0.3056	1.8333	11.0000
		$\sum M \times P(M)=4.4722$	$\sum M^{2} \times P(M)=21.9722$

Thus $E(M^{2})=\sum M^{2} \times P(M)$

$E(M^{2})=21.9722$

$Var(M)=E(M^{2})-(E(M))^{2}$

$Var(M)=1.9715$

Part e) Standard deviation of M

$SD(M)=\sqrt{Var(M)}$

$SD(M)=\sqrt{1.9715 }$

$SD(M)=1.4041$