Calculate the probability distribution of two team’s scores.

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 7.2, Problem 66E

(a)

To determine

Calculate the probability distribution of two team’s scores.

(a)

Expert Solution

Explanation of Solution

The probability distribution of scores is shown in below table.

Table 1

	Home team
Visiting team	0	1	2	3
0	0.14	0.11	0.09	0.10
1	0.12	0.10	0.05	0.02
2	0.09	0.07	0.04	0.01
3	0.03	0.02	0.01	0

Probability distribution of two team’s scoresis shown in below table.

P(T)	Probability distribution
P(T=0)	0.14
P(T=1)	0.23
P(T=2)	0.28
P(T=3)	0.25
P(T=4)	0.08
P(T=5)	0.02
P(T=6)	0

Economics Concept Introduction

Probability distribution: The probability distribution shows the probabilities of incidence of different likely outcomes in a test.

(b)

To determine

Mean,variance and standard deviation.

(b)

Expert Solution

Explanation of Solution

Mean of team’s scores is calculated as follows.

μ=E(X)=∑xP(x)=(0(0.14)+1(0.23)+2(0.28)+3(0.25)+4(0.08)+5(0.02)+6(0)) =1.96

Mean value is 0.96.

Variance of team’s scores is calculated as follows.

σ2=V(X)=∑(x−μ)2P(x)=((0−1.96)2(0.14) + (1−1.96)2(0.23) + (2−1.96)2(0.28)+(3−1.96)2(0.25)+(4−1.96)2(0.08)+(5−1.96)2(0.02)+(6−1.96)2(0)) = 1.538

Variance is 1.538.

Standard deviation (square root of variance) of team’s scores is calculated as follows.

σ=σ2=1.538=1.240

Standard deviation is 1.240.

(c)

To determine

Covariance and coefficient of correlation.

(c)

Expert Solution

Explanation of Solution

Covariance and coefficient of correlation of two team’s score is calculated as follows.

∑all x∑all yxyP(x,y)=((0)(0)(.14)+(0)(1)(.12)+(0)(2)(.09)+(0)(3)(.03)+(1)(0)(.11)+(1)(1)(.10)+(1)(2)(.07)+(1)(3)(.02)+(2)(0)(.09)+(2)(1)(.05)+(2)(2)(.04)+(2)(3)(.01)+(3)(0)(.10)+(3)(1)(.02)+(3)(2)(.01)++(3)(3)(0)) =1.76COV(X,Y)=∑all x∑all yxyP(x,y)−μxμy=0.74−(1.07)(0.89)=−0.212ρ=COV(X,Y)σxσy=−0.212(1.085)(0.937)=−0.217

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

You survey 5 members of a college class about how many hours they study each week: 12 hours, 18 hours, 7 hours, 3 hours, and 15 hours. The class has a total of 28 students. Based on the findings from the 5 students you interviewed, what is the standard deviation for the entire class? What is the range of hours you can expect students to study based on one standard deviation? (Hint: Think Normal Distribution Curve).

In one of its Spring catalogs, L.L. Bean advertised footwear on 29 of its 192 catalog pages. Suppose we randomly surgery 20 pages. We are interested in the number of pages that advertise footwear. Each page may be picked at most once. Calculate the standard deviation

An experiment consists of four outcomes with P(E1) = 0.2, P(E2) = 0.3, and P(E3) = 0.4. The probability of outcome E4 is Group of answer choices 0.900. 0.500. 0.024. 0.100.

Answer 2

Question

Chapter 7.2, Problem 66E

(a)

To determine

Calculate the probability distribution of two team’s scores.

(a)

Expert Solution

Explanation of Solution

The probability distribution of scores is shown in below table.

Table 1

	Home team
Visiting team	0	1	2	3
0	0.14	0.11	0.09	0.10
1	0.12	0.10	0.05	0.02
2	0.09	0.07	0.04	0.01
3	0.03	0.02	0.01	0

Probability distribution of two team’s scoresis shown in below table.

P(T)	Probability distribution
P(T=0)	0.14
P(T=1)	0.23
P(T=2)	0.28
P(T=3)	0.25
P(T=4)	0.08
P(T=5)	0.02
P(T=6)	0

Economics Concept Introduction

Probability distribution: The probability distribution shows the probabilities of incidence of different likely outcomes in a test.

(b)

To determine

Mean,variance and standard deviation.

(b)

Expert Solution

Explanation of Solution

Mean of team’s scores is calculated as follows.

μ=E(X)=∑xP(x)=(0(0.14)+1(0.23)+2(0.28)+3(0.25)+4(0.08)+5(0.02)+6(0)) =1.96

Mean value is 0.96.

Variance of team’s scores is calculated as follows.

σ2=V(X)=∑(x−μ)2P(x)=((0−1.96)2(0.14) + (1−1.96)2(0.23) + (2−1.96)2(0.28)+(3−1.96)2(0.25)+(4−1.96)2(0.08)+(5−1.96)2(0.02)+(6−1.96)2(0)) = 1.538

Variance is 1.538.

Standard deviation (square root of variance) of team’s scores is calculated as follows.

σ=σ2=1.538=1.240

Standard deviation is 1.240.

(c)

To determine

Covariance and coefficient of correlation.

(c)

Expert Solution

Explanation of Solution

Covariance and coefficient of correlation of two team’s score is calculated as follows.

∑all x∑all yxyP(x,y)=((0)(0)(.14)+(0)(1)(.12)+(0)(2)(.09)+(0)(3)(.03)+(1)(0)(.11)+(1)(1)(.10)+(1)(2)(.07)+(1)(3)(.02)+(2)(0)(.09)+(2)(1)(.05)+(2)(2)(.04)+(2)(3)(.01)+(3)(0)(.10)+(3)(1)(.02)+(3)(2)(.01)++(3)(3)(0)) =1.76COV(X,Y)=∑all x∑all yxyP(x,y)−μxμy=0.74−(1.07)(0.89)=−0.212ρ=COV(X,Y)σxσy=−0.212(1.085)(0.937)=−0.217

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 7.2, Problem 66E

(a)

To determine

Calculate the probability distribution of two team’s scores.

(a)

Expert Solution

Explanation of Solution

The probability distribution of scores is shown in below table.

Table 1

	Home team
Visiting team	0	1	2	3
0	0.14	0.11	0.09	0.10
1	0.12	0.10	0.05	0.02
2	0.09	0.07	0.04	0.01
3	0.03	0.02	0.01	0

Probability distribution of two team’s scoresis shown in below table.

P(T)	Probability distribution
P(T=0)	0.14
P(T=1)	0.23
P(T=2)	0.28
P(T=3)	0.25
P(T=4)	0.08
P(T=5)	0.02
P(T=6)	0

Economics Concept Introduction

Probability distribution: The probability distribution shows the probabilities of incidence of different likely outcomes in a test.

(b)

To determine

Mean,variance and standard deviation.

(b)

Expert Solution

Explanation of Solution

Mean of team’s scores is calculated as follows.

μ=E(X)=∑xP(x)=(0(0.14)+1(0.23)+2(0.28)+3(0.25)+4(0.08)+5(0.02)+6(0)) =1.96

Mean value is 0.96.

Variance of team’s scores is calculated as follows.

σ2=V(X)=∑(x−μ)2P(x)=((0−1.96)2(0.14) + (1−1.96)2(0.23) + (2−1.96)2(0.28)+(3−1.96)2(0.25)+(4−1.96)2(0.08)+(5−1.96)2(0.02)+(6−1.96)2(0)) = 1.538

Variance is 1.538.

Standard deviation (square root of variance) of team’s scores is calculated as follows.

σ=σ2=1.538=1.240

Standard deviation is 1.240.

(c)

To determine

Covariance and coefficient of correlation.

(c)

Expert Solution

Explanation of Solution

Covariance and coefficient of correlation of two team’s score is calculated as follows.

∑all x∑all yxyP(x,y)=((0)(0)(.14)+(0)(1)(.12)+(0)(2)(.09)+(0)(3)(.03)+(1)(0)(.11)+(1)(1)(.10)+(1)(2)(.07)+(1)(3)(.02)+(2)(0)(.09)+(2)(1)(.05)+(2)(2)(.04)+(2)(3)(.01)+(3)(0)(.10)+(3)(1)(.02)+(3)(2)(.01)++(3)(3)(0)) =1.76COV(X,Y)=∑all x∑all yxyP(x,y)−μxμy=0.74−(1.07)(0.89)=−0.212ρ=COV(X,Y)σxσy=−0.212(1.085)(0.937)=−0.217

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 7.2, Problem 66E

(a)

To determine

Calculate the probability distribution of two team’s scores.

(a)

Expert Solution

Explanation of Solution

The probability distribution of scores is shown in below table.

Table 1

	Home team
Visiting team	0	1	2	3
0	0.14	0.11	0.09	0.10
1	0.12	0.10	0.05	0.02
2	0.09	0.07	0.04	0.01
3	0.03	0.02	0.01	0

Probability distribution of two team’s scoresis shown in below table.

P(T)	Probability distribution
P(T=0)	0.14
P(T=1)	0.23
P(T=2)	0.28
P(T=3)	0.25
P(T=4)	0.08
P(T=5)	0.02
P(T=6)	0

Economics Concept Introduction

Probability distribution: The probability distribution shows the probabilities of incidence of different likely outcomes in a test.

(b)

To determine

Mean,variance and standard deviation.

(b)

Expert Solution

Explanation of Solution

Mean of team’s scores is calculated as follows.

μ=E(X)=∑xP(x)=(0(0.14)+1(0.23)+2(0.28)+3(0.25)+4(0.08)+5(0.02)+6(0)) =1.96

Mean value is 0.96.

Variance of team’s scores is calculated as follows.

σ2=V(X)=∑(x−μ)2P(x)=((0−1.96)2(0.14) + (1−1.96)2(0.23) + (2−1.96)2(0.28)+(3−1.96)2(0.25)+(4−1.96)2(0.08)+(5−1.96)2(0.02)+(6−1.96)2(0)) = 1.538

Variance is 1.538.

Standard deviation (square root of variance) of team’s scores is calculated as follows.

σ=σ2=1.538=1.240

Standard deviation is 1.240.

(c)

To determine

Covariance and coefficient of correlation.

(c)

Expert Solution

Explanation of Solution

Covariance and coefficient of correlation of two team’s score is calculated as follows.

∑all x∑all yxyP(x,y)=((0)(0)(.14)+(0)(1)(.12)+(0)(2)(.09)+(0)(3)(.03)+(1)(0)(.11)+(1)(1)(.10)+(1)(2)(.07)+(1)(3)(.02)+(2)(0)(.09)+(2)(1)(.05)+(2)(2)(.04)+(2)(3)(.01)+(3)(0)(.10)+(3)(1)(.02)+(3)(2)(.01)++(3)(3)(0)) =1.76COV(X,Y)=∑all x∑all yxyP(x,y)−μxμy=0.74−(1.07)(0.89)=−0.212ρ=COV(X,Y)σxσy=−0.212(1.085)(0.937)=−0.217

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 7.2, Problem 66E

(a)

To determine

Calculate the probability distribution of two team’s scores.

(a)

Expert Solution

Explanation of Solution

The probability distribution of scores is shown in below table.

Table 1

	Home team
Visiting team	0	1	2	3
0	0.14	0.11	0.09	0.10
1	0.12	0.10	0.05	0.02
2	0.09	0.07	0.04	0.01
3	0.03	0.02	0.01	0

Probability distribution of two team’s scoresis shown in below table.

P(T)	Probability distribution
P(T=0)	0.14
P(T=1)	0.23
P(T=2)	0.28
P(T=3)	0.25
P(T=4)	0.08
P(T=5)	0.02
P(T=6)	0

Economics Concept Introduction

Probability distribution: The probability distribution shows the probabilities of incidence of different likely outcomes in a test.

(b)

To determine

Mean,variance and standard deviation.

(b)

Expert Solution

Explanation of Solution

Mean of team’s scores is calculated as follows.

μ=E(X)=∑xP(x)=(0(0.14)+1(0.23)+2(0.28)+3(0.25)+4(0.08)+5(0.02)+6(0)) =1.96

Mean value is 0.96.

Variance of team’s scores is calculated as follows.

σ2=V(X)=∑(x−μ)2P(x)=((0−1.96)2(0.14) + (1−1.96)2(0.23) + (2−1.96)2(0.28)+(3−1.96)2(0.25)+(4−1.96)2(0.08)+(5−1.96)2(0.02)+(6−1.96)2(0)) = 1.538

Variance is 1.538.

Standard deviation (square root of variance) of team’s scores is calculated as follows.

σ=σ2=1.538=1.240

Standard deviation is 1.240.

(c)

To determine

Covariance and coefficient of correlation.

(c)

Expert Solution

Explanation of Solution

Covariance and coefficient of correlation of two team’s score is calculated as follows.

∑all x∑all yxyP(x,y)=((0)(0)(.14)+(0)(1)(.12)+(0)(2)(.09)+(0)(3)(.03)+(1)(0)(.11)+(1)(1)(.10)+(1)(2)(.07)+(1)(3)(.02)+(2)(0)(.09)+(2)(1)(.05)+(2)(2)(.04)+(2)(3)(.01)+(3)(0)(.10)+(3)(1)(.02)+(3)(2)(.01)++(3)(3)(0)) =1.76COV(X,Y)=∑all x∑all yxyP(x,y)−μxμy=0.74−(1.07)(0.89)=−0.212ρ=COV(X,Y)σxσy=−0.212(1.085)(0.937)=−0.217

Answer 6

Chapter 7.2, Problem 66E

(a)

To determine

Calculate the probability distribution of two team’s scores.

(a)

Expert Solution

Explanation of Solution

The probability distribution of scores is shown in below table.

Table 1

	Home team
Visiting team	0	1	2	3
0	0.14	0.11	0.09	0.10
1	0.12	0.10	0.05	0.02
2	0.09	0.07	0.04	0.01
3	0.03	0.02	0.01	0

Probability distribution of two team’s scoresis shown in below table.

P(T)	Probability distribution
P(T=0)	0.14
P(T=1)	0.23
P(T=2)	0.28
P(T=3)	0.25
P(T=4)	0.08
P(T=5)	0.02
P(T=6)	0

Economics Concept Introduction

Probability distribution: The probability distribution shows the probabilities of incidence of different likely outcomes in a test.

Answer 7

Chapter 7.2, Problem 66E

(a)

To determine

Calculate the probability distribution of two team’s scores.

Answer 8

(a)

Expert Solution

Explanation of Solution

The probability distribution of scores is shown in below table.

Table 1

	Home team
Visiting team	0	1	2	3
0	0.14	0.11	0.09	0.10
1	0.12	0.10	0.05	0.02
2	0.09	0.07	0.04	0.01
3	0.03	0.02	0.01	0

Probability distribution of two team’s scoresis shown in below table.

P(T)	Probability distribution
P(T=0)	0.14
P(T=1)	0.23
P(T=2)	0.28
P(T=3)	0.25
P(T=4)	0.08
P(T=5)	0.02
P(T=6)	0

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Economics Concept Introduction

Probability distribution: The probability distribution shows the probabilities of incidence of different likely outcomes in a test.

Answer 13

(b)

To determine

Mean,variance and standard deviation.

(b)

Expert Solution

Explanation of Solution

Mean of team’s scores is calculated as follows.

μ=E(X)=∑xP(x)=(0(0.14)+1(0.23)+2(0.28)+3(0.25)+4(0.08)+5(0.02)+6(0)) =1.96

Mean value is 0.96.

Variance of team’s scores is calculated as follows.

σ2=V(X)=∑(x−μ)2P(x)=((0−1.96)2(0.14) + (1−1.96)2(0.23) + (2−1.96)2(0.28)+(3−1.96)2(0.25)+(4−1.96)2(0.08)+(5−1.96)2(0.02)+(6−1.96)2(0)) = 1.538

Variance is 1.538.

Standard deviation (square root of variance) of team’s scores is calculated as follows.

σ=σ2=1.538=1.240

Standard deviation is 1.240.

Answer 14

(b)

To determine

Mean,variance and standard deviation.

Answer 15

(b)

Expert Solution

Explanation of Solution

Mean of team’s scores is calculated as follows.

μ=E(X)=∑xP(x)=(0(0.14)+1(0.23)+2(0.28)+3(0.25)+4(0.08)+5(0.02)+6(0)) =1.96

Mean value is 0.96.

Variance of team’s scores is calculated as follows.

σ2=V(X)=∑(x−μ)2P(x)=((0−1.96)2(0.14) + (1−1.96)2(0.23) + (2−1.96)2(0.28)+(3−1.96)2(0.25)+(4−1.96)2(0.08)+(5−1.96)2(0.02)+(6−1.96)2(0)) = 1.538

Variance is 1.538.

Standard deviation (square root of variance) of team’s scores is calculated as follows.

σ=σ2=1.538=1.240

Standard deviation is 1.240.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

(c)

To determine

Covariance and coefficient of correlation.

(c)

Expert Solution

Explanation of Solution

Covariance and coefficient of correlation of two team’s score is calculated as follows.

∑all x∑all yxyP(x,y)=((0)(0)(.14)+(0)(1)(.12)+(0)(2)(.09)+(0)(3)(.03)+(1)(0)(.11)+(1)(1)(.10)+(1)(2)(.07)+(1)(3)(.02)+(2)(0)(.09)+(2)(1)(.05)+(2)(2)(.04)+(2)(3)(.01)+(3)(0)(.10)+(3)(1)(.02)+(3)(2)(.01)++(3)(3)(0)) =1.76COV(X,Y)=∑all x∑all yxyP(x,y)−μxμy=0.74−(1.07)(0.89)=−0.212ρ=COV(X,Y)σxσy=−0.212(1.085)(0.937)=−0.217

Answer 20

(c)

To determine

Covariance and coefficient of correlation.

Answer 21

(c)

Expert Solution

Explanation of Solution

Covariance and coefficient of correlation of two team’s score is calculated as follows.

∑all x∑all yxyP(x,y)=((0)(0)(.14)+(0)(1)(.12)+(0)(2)(.09)+(0)(3)(.03)+(1)(0)(.11)+(1)(1)(.10)+(1)(2)(.07)+(1)(3)(.02)+(2)(0)(.09)+(2)(1)(.05)+(2)(2)(.04)+(2)(3)(.01)+(3)(0)(.10)+(3)(1)(.02)+(3)(2)(.01)++(3)(3)(0)) =1.76COV(X,Y)=∑all x∑all yxyP(x,y)−μxμy=0.74−(1.07)(0.89)=−0.212ρ=COV(X,Y)σxσy=−0.212(1.085)(0.937)=−0.217