/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q1E The probability distribution sho... [FREE SOLUTION] | 91影视

91影视

The probability distribution shown here describes a population of measurements that can assume values of 0, 2, 4, and 6, each of which occurs with the same relative frequency:

  1. List all the different samples of n = 2 measurements that can be selected from this population. For example, (0, 6) is one possible pair of measurements; (2, 2) is another possible pair.
  2. Calculate the mean of each different sample listed in part a.
  3. If a sample of n = 2 measurements is randomly selected from the population, what is the probability that a specific sample will be selected.
  4. Assume that a random sample of n = 2 measurements is selected from the population. List the different values of x found in part b and find the probability of each. Then give the sampling distribution of the sample mean x in tabular form.
  5. Construct a probability histogram for the sampling distribution ofx.

Short Answer

Expert verified

a. The required answer is -

(0,2),(0,4),(0,6),(2,4),(2,6),(4,6),(0,0),(2,2),(4,4),(6,6),(2,0),(4,0),(6,0),(4,2),(6,2),(6,4).

b.

Samples

Mean

0,2

1

0,4

2

0,6

3

2,4

3

2,6

4

4,6

5

0.0

0

2,2

2

4,4

4

6,6

6

2,0

1

4,0

2

6,0

3

4,2

3

6,2

4

6,4

5

c.116

d.

Mean

Number of times

0

116

1

18

2

316

3

14

4

316

5

118

6

116

e.

Step by step solution

01

Formula forcalculating different samples

a.

The formula for calculating the total number of possible samples from the given values of x is shown below.


Numbersofsamples=(Totalnumberofvaluesofx)n=(4)2=16

02

Total number of samples

It has been found that only 16 different samples can be found, and accordingly, the possible samples are listed below.

(0,2),(0,4),(0,6),(2,4),(2,6),(4,6),(0,0),(2,2),(4,4),(6,6),(2,0),(4,0),(6,0),(4,2),(6,2),(6,4)

03

Total number of samples

b.

The table below shows the list of the means of different samples.

Samples

Mean

0,2

0+22=1

0,4

0+42=2

0,6

0+62=3

2,4

2+42=3

2,6

2+62=4

4,6

4+62=5

0.0

0+02=0

2,2

2+22=2

4,4

4+42=4

6,6

6+62=6

2,0

2+02=1

4,0

4+02=2

6,0

6+02=3

4,2

4+22=3

6,2

6+22=4

6,4

6+42=5

04

Formula for calculating different samples

c.

For each sample, the probability can be found by multiplying each of the probabilities of the respective values of x. Here, it can be seen that all the values have the same probabilities.

05

Total number of samples

The probability of each sample getting chosen is shown below.

Samples

Probability

0,2

1414=116

0,4

1414=116

0,6

1414=116

2,4

1414=116

2,6

1414=116

4,6

1414=116

0,0

1414=116

2,2

1414=116

4,4

1414=116

6,6

1414=116

2,0

1414=116

4,0

1414=116

6,0

1414=116

4,2

1414=116

6,2

1414=116

6,4

1414=116

Therefore, it can be observed that for all the samples, the possibility is 116.

06

Number of times the mean values are observed

d.

The number of times the mean values have been observed in Part b isshownbelow.

Mean

Number of times

0

1

1

2

2

3

3

4

4

3

5

2

6

1

07

Calculation of the probability

The probability of each mean ofgetting chosen is shown below

Mean

Number of times

Probability

0

1

1116=116

1

2

2116=116

2

3

3116=116

3

4

4116=116

4

3

3116=116

5

2

2116=116

6

1

1116=116

Therefore, it can be observed that for all the samples, the possibility for 0 and 6 is 116,for 1 and 3, it is 18, and for 2 and 4, it is.role="math" localid="1657972705521" 316

08

List of probabilities of x

The number of times the mean values have been observed in Part b isshownbelow.

Mean

Number of times

0

116

1

role="math" localid="1657972888816" 18

2

316

3

14

4

316

5

18

6

116

09

Elucidation on the graph

The diagram shows the probabilities of the respective values of the mean in the form of a bar graph.The mean value 3 in the diagram shows the maximum probability, which is 14.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Soft-drink bottles. A soft-drink bottler purchases glass bottles from a vendor. The bottles are required to have an internal pressure of at least 150 pounds per square inch (psi). A prospective bottle vendor claims that its production process yields bottles with a mean internal pressure of 157 psi and a standard deviation of 3 psi. The bottler strikes an agreement with the vendor that permits the bottler to sample from the vendor鈥檚 production process to verify the vendor鈥檚 claim. The bottler randomly selects 40 bottles from the last 10,000 produced, measures the internal pressure of each, and finds the mean pressure for the sample to be 1.3 psi below the process mean cited by the vendor.

a. Assuming the vendor鈥檚 claim to be true, what is the probability of obtaining a sample mean this far or farther below the process mean? What does your answer suggest about the validity of the vendor鈥檚 claim?

b. If the process standard deviation were 3 psi as claimed by the vendor, but the mean were 156 psi, would the observed sample result be more or less likely than in part a? What if the mean were 158 psi?

c. If the process mean were 157 psi as claimed, but the process standard deviation were 2 psi, would the sample result be more or less likely than in part a? What if instead the standard deviation were 6 psi?

Consider the following probability distribution:

  1. Findand2.
  2. Find the sampling distribution of the sample mean x for a random sample of n = 2 measurements from this distribution
  3. Show that xis an unbiased estimator of . [Hint: Show that.]x=xpx=.]
  4. Find the sampling distribution of the sample variances2for a random sample of n = 2 measurements from this distribution.

Length of job tenure. Researchers at the Terry College ofBusiness at the University of Georgia sampled 344 business students and asked them this question: 鈥淥ver the course of your lifetime, what is the maximum number of years you expect to work for any one employer?鈥 The sample resulted in x= 19.1 years. Assume that the sample of students was randomly selected from the 6,000 undergraduate students atthe Terry College and that = 6 years.

  1. Describe the sampling distribution of X.
  2. If the mean for the 6,000 undergraduate students is= 18.5 years, findPx>19.1.
  3. If the mean for the 6,000 undergraduate students is= 19.5 years, findPx>19.1.
  4. If,P(x>19.1)=0.5 what is?
  5. If,Px>19.1=0.2 isgreater than or less than 19.1years? Explain.

Consider the population described by the probability distribution shown below.

The random variable x is observed twice. If these observations are independent, verify that the different samples of size 2 and their probabilities are as shown below.

a. Find the sampling distribution of the sample meanx.

b. Construct a probability histogram for the sampling distribution ofx.

c. What is the probability thatxis 4.5 or larger?

d. Would you expect to observe a value ofxequal to 4.5 or larger? Explain.

Switching banks after a merger. Banks that merge with others to form 鈥渕ega-banks鈥 sometimes leave customers dissatisfied with the impersonal service. A poll by the Gallup Organization found 20% of retail customers switched banks after their banks merged with another. One year after the acquisition of First Fidelity by First Union, a random sample of 250 retail customers who had banked with First Fidelity were questioned. Letp^ be the proportion of those customers who switched their business from First Union to a different bank.

  1. Find the mean and the standard deviation of role="math" localid="1658320788143" p^.
  2. Calculate the interval Ep^2p^.
  3. If samples of size 250 were drawn repeatedly a large number of times and determined for each sample, what proportion of the values would fall within the interval you calculated in part c?
See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.