/*! 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} Q.7.4 Do you jog? The Gallup Poll once... [FREE SOLUTION] | 91影视

91影视

Do you jog? The Gallup Poll once asked a random sample of1540adults, "Do you happen to jog?" Suppose that in fact 15%of all adults jog.

(a) What is the mean of the sampling distribution of p^? Justify your answer.

(b) Find the standard deviation of the sampling distribution of p^. Check that the 10%condition is met.

(c) Is the sampling distribution of p^approximately Normal? Justify your answer.

(d) Find the probability that between 13%and 17%of a random sample of 1540adults are joggers. Show your work.

Short Answer

Expert verified

a). The mean is0.15.

b). The standard deviation is0.0091.

c). Normal approximation might be appropriate.

d). The probability is0.9722.

Step by step solution

01

Part (a) Step 1: Given Information

Population proportion (p)=0.15,

Sample size (n)=1540.

02

Part (a) Step 2: Explanation

The sample distribution's mean can be computed as:

p^=p

=0.15

03

Part (b) Step 1: Given Information

The standard deviation of p^'s sampling distribution must be determined. Make sure that the 10% criteria is met.

04

Part (b) Step 2: Explanation

The sample proportion's standard deviation is calculated as:

p^=p(1-p)n

=0.15(1-0.15)1540

=0.0091

05

Part (c) Step 1: Given Information

Given in the question that, The Gallup Poll once asked a random sample of 1540adults.

We have to find is the sampling distribution of p close to Normal? Justify your response.

06

Part (c) Step 2: Explanation

Here,

np=1540(0.15)

=231>10

n(1-p)=1540(1-0.15)

=1309>10

As a result, normal approximation might be appropriate.

07

Part (d) Step 1: Given Information

We need to calculate the likelihood that between 13 and 17 percent of a random sample of1540 people are joggers. Display your work.

08

Part (d) Step 2: Explanation

The probability that between 13%and 17%of the sample of 1540adults are joggers is calculated as:

P(0.13p^0.17)=P0.13-0.150.0091Z0.17-0.150.0091

=P(-2.20Z2.20) =0.9722

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

Suppose that you have torn a tendon and are facing surgery to repair it. The orthopedic surgeon explains the risks to you. Infection occurs in3%such operations, the repair fails in 14%, and both infection and failure occur together 1%at the time. What is the probability that the operation is successful for someone who has an operation that is free from infection?

(a) 0.0767

(b) 0.8342

(c) 0.8400

(d) 0.8660

(e)0.9900

Records from a random sample of dairy farms yielded the information below on the number of male and female calves born at various times of the day.

What is the probability that a randomly selected calf was
born in the night or was a female?
(a)369513

(b)485513

(c) 116513

(d)116252

(e) 116233

How many people are in a car? A study of rush-hour traf铿乧 in San Francisco counts the number of people in each car entering a freeway at a suburban interchange. Suppose that this count has a mean of 1.5. and a standard deviation of 0.75in the population of all cars that enter this interchange during rush hours.

(a) Could the exact distribution of the count be Normal? Why or why not?

(b) Traf铿乧 engineers estimate that the capacity of the interchange is 700cars per hour. Find the probability that 700 cars will carry more than 1075 people. Show your work. (Hint: Restate this event in terms of the mean number of people x per car.)

IQ tests The Wechsler Adult Intelligence Scale (WAIS) is a common "IQ test" for adults. The distribution of WAIS scores for persons over 16years of age is approximately Normal with mean 100 and standard deviation 15.

(a) What is the probability that a randomly chosen individual has a WAIS score of 105 or higher? Show your work.

(b) Find the mean and standard deviation of the sampling distribution of the average WAIS score x for an SRS of 60 people.

(c) What is the probability that the average WAIS score of an SRS of 60 people is 105 or higher? Show your work.

(d) Would your answers to any of parts (a), (b), or (c) be affected if the distribution of WAIS scores in the adult population were distinctly non-Normal? Explain.

AP2.22. A health worker is interested in determining if omega-3fish oil can help reduce cholesterol in adults. She obtains permission to examine the health records of200 people in a large medical clinic and classifies them
according to whether or not they take omega-3fish oil. She also obtains their latest cholesterol readings and finds that the mean cholesterol reading for those who are taking omega-3fish oil is 18points lower than the mean for the group not taking omega-3 fish oil.
(a) Is this an observational study or an experiment? Explain.
(b) Do these results provide convincing evidence that taking omega-3 fish oil lowers cholesterol?
(c) Explain the concept of confounding in the context of this study and give one example of a possible confounding variable.

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.