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

How to calculate total cost?

Problem 1-Experiments/Randomized Control TrialSuppose you are interested in studying the effect of being a part of the labor union on anindividual’s hourly wage.You collect data on 1000 people and run the following regression. where is a dummy variable which is equal t to 1 for people who are a part of labor union and0 for others. a. Suppose you estimate . Interpret b. Do you think is biased or unbiased? Explain. Now suppose you conduct a randomized control trial to answer the same question. Yourandomly assign some individuals to be a part of the labor union and others to not be a partof the labor union. The first step you take is to ensure that the randomization was donecorrectly. Then you estimate the following equation: =1 if the student is assigned to be a part of Labor Union=0 if the student is assigned to not be a part of Labor Union c. Why is it important to ensure that the randomization is done correctly? d. Name any two variables that you can use to test if Treatment and…

What is kiosk?

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