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

Answer

M” method Given the following model, solve by the method of “M”. (see image)

As indicated in the attached image, U.S. earnings for high- and low-skill workers as measured by educational attainment began diverging in the 1980s. The remaining questions in this problem set use the model for the labor market developed in class to walk through potential explanations for this trend. 1. Assume that there are just two types of workers, low- and high-skill. As a result, there are two labor markets: supply and demand for low-skill workers and supply and demand for high-skill workers. Using two carefully drawn labor-market figures, show that an increase in the demand for high skill workers can explain an increase in the relative wage of high-skill workers. 2. Using the same assumptions as in the previous question, use two carefully drawn labor-market figures to show that an increase in the supply of low-skill workers can explain an increase in the relative wage of high-skill workers.

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