Question

In: Statistics and Probability

For Problems 8-10, throw a die twice and let X, Y be the results (each from...

For Problems 8-10, throw a die twice and let X, Y be the results (each from 1 to 6).

Problem 8. Find the expectation and variance of X.

Problem 9. Find the variance of X - 2Y.

Problem 10. Find the expected value of X^2Y.

Expert Solution

Problem 8.

X : Results of throwing a single die

P(X) : Probability of X

Expectation :

Variance : Var(X) = E(X²) - E(X)²

X	P(X)	P(X)	XP(X)	X²P(X)
1	1/6	0.1667	0.1667	0.1667
2	1/6	0.1667	0.3333	0.6667
3	1/6	0.1667	0.5000	1.5000
4	1/6	0.1667	0.6667	2.6667
5	1/6	0.1667	0.8333	4.1667
6	1/6	0.1667	1.0000	6.0000
	Total	1.0000	3.5000	15.1667

Expectation

Expectation = 3.5

Variance : Var(X) = E(X²) - E(X)² = 15.1667 - 3.5² = 2.9167

Problem 9

Variance of X-2Y

The following table provides, All possibles of X,Y and their probability.

X	Y	P(X,Y)
1	1	1/36
1	2	1/36
1	3	1/36
1	4	1/36
1	5	1/36
1	6	1/36
2	1	1/36
2	2	1/36
2	3	1/36
2	4	1/36
2	5	1/36
2	6	1/36
3	1	1/36
3	2	1/36
3	3	1/36
3	4	1/36
3	5	1/36
3	6	1/36
4	1	1/36
4	2	1/36
4	3	1/36
4	4	1/36
4	5	1/36
4	6	1/36
5	1	1/36
5	2	1/36
5	3	1/36
5	4	1/36
5	5	1/36
5	6	1/36
6	1	1/36
6	2	1/36
6	3	1/36
6	4	1/36
6	5	1/36
6	6	1/36

The following table extends the above by calculating X-2Y

X	Y	X-2Y
1	1	-1
1	2	-3
1	3	-5
1	4	-7
1	5	-9
1	6	-11
2	1	0
2	2	-2
2	3	-4
2	4	-6
2	5	-8
2	6	-10
3	1	1
3	2	-1
3	3	-3
3	4	-5
3	5	-7
3	6	-9
4	1	2
4	2	0
4	3	-2
4	4	-4
4	5	-6
4	6	-8
5	1	3
5	2	1
5	3	-1
5	4	-3
5	5	-5
5	6	-7
6	1	4
6	2	2
6	3	0
6	4	-2
6	5	-4
6	6	-6

The above table is sorted on X-2Y, so that same value of X-2Y are together

X	Y	X-2Y
1	6	-11
2	6	-10
1	5	-9
3	6	-9
2	5	-8
4	6	-8
1	4	-7
3	5	-7
5	6	-7
2	4	-6
4	5	-6
6	6	-6
1	3	-5
3	4	-5
5	5	-5
2	3	-4
4	4	-4
6	5	-4
1	2	-3
3	3	-3
5	4	-3
2	2	-2
4	3	-2
6	4	-2
1	1	-1
3	2	-1
5	3	-1
2	1	0
4	2	0
6	3	0
3	1	1
5	2	1
4	1	2
6	2	2
5	1	3
6	1	4

X	Y	X-2Y	X-2Y	P(X-2Y)
1	6	-11	-11	1/36
2	6	-10	-10	1/36
1	5	-9	-9	2/36
3	6	-9
2	5	-8	-8	2/36
4	6	-8
1	4	-7	-7	3/36
3	5	-7
5	6	-7
2	4	-6	-6	3/36
4	5	-6
6	6	-6
1	3	-5	-5	3/36
3	4	-5
5	5	-5
2	3	-4	-4	3/36
4	4	-4
6	5	-4
1	2	-3	-3	3/36
3	3	-3
5	4	-3
2	2	-2	-2	3/36
4	3	-2
6	4	-2
1	1	-1	-1	3/36
3	2	-1
5	3	-1
2	1	0	0	3/36
4	2	0
6	3	0
3	1	1	1	2/36
5	2	1
4	1	2	2	2/36
6	2	2
5	1	3	3	1/36
6	1	4	4	1/36

Z:(X-2Y)	P(Z)=P(X-2Y)	ZP(Z)	Z²P(Z)
-11	0.0278	-0.3056	3.3611
-10	0.0278	-0.2778	2.7778
-9	0.0556	-0.5000	4.5000
-8	0.0556	-0.4444	3.5556
-7	0.0833	-0.5833	4.0833
-6	0.0833	-0.5000	3.0000
-5	0.0833	-0.4167	2.0833
-4	0.0833	-0.3333	1.3333
-3	0.0833	-0.2500	0.7500
-2	0.0833	-0.1667	0.3333
-1	0.0833	-0.0833	0.0833
0	0.0833	0.0000	0.0000
1	0.0556	0.0556	0.0556
2	0.0556	0.1111	0.2222
3	0.0278	0.0833	0.2500
4	0.0278	0.1111	0.4444
	1.0000	-3.5000	26.8333

Let Z= X-2Y

Variance of X-2T i.e Var(Z)

Var(Z) = E(Z²) - E(Z)²

E(Z) = -3.5

E(Z²) = 26.8333

Var(Z) = E(Z²) - E(Z)² = 26.8333 - (-3.5)² = 14.5833

Variance of X-2Y = 14.5833

Problem 10:

Following the above procedure,

W=X²Y

W	P(W)	WP(W)
1	0.0278	0.0278
2	0.0278	0.0556
3	0.0278	0.0833
4	0.0556	0.2222
5	0.0278	0.1389
6	0.0278	0.1667
8	0.0278	0.2222
9	0.0278	0.2500
12	0.0278	0.3333
16	0.0556	0.8889
18	0.0278	0.5000
20	0.0278	0.5556
24	0.0278	0.6667
25	0.0278	0.6944
27	0.0278	0.7500
32	0.0278	0.8889
36	0.0556	2.0000
45	0.0278	1.2500
48	0.0278	1.3333
50	0.0278	1.3889
54	0.0278	1.5000
64	0.0278	1.7778
72	0.0278	2.0000
75	0.0278	2.0833
80	0.0278	2.2222
96	0.0278	2.6667
100	0.0278	2.7778
108	0.0278	3.0000
125	0.0278	3.4722
144	0.0278	4.0000
150	0.0278	4.1667
180	0.0278	5.0000
216	0.0278	6.0000
	1.0000	53.0833

orchestra answered 1 year ago

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