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A sample survey interviews an SRS of 267college women. Suppose (as is roughly true) that 70%of college women have been on a diet within the past12months. What is the probability that 75%or more of the women in the sample have been on a diet? Follow the four-step process.

Short Answer

Expert verified

The probability that 75%or more of the women in the sample have been on a dietP(p^75%)=0.0375.

Step by step solution

01

Given Information

Given

p=70%=0.70

p^=75%=0.75

n=267

02

Explanation

The mean of the sampling distribution of p^is equal to the population proportion p:

p^=p=0.70

The standard deviation of the sampling distribution of p^is:

p^=p(1-p)n=0.70(1-0.70)2670.0280449

The z-score is the value decreased by the mean, divided by the standard deviation:

z=x-=0.75-0.700.02804491.78

Determine the corresponding probability using table A:

P(p^75%)=P(z>1.78)=P(Z<-1.78)=0.0375

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Most popular questions from this chapter

More on insurance An insurance company knows that in the entire population of homeowners, the mean annual loss from 铿乺e is =250and the standard deviation of the loss is =300. The distribution of losses is strongly right-skewed: many policies have 0loss, but a few have large losses. If the company sells 10,000 policies, can it safely base its rates on the assumption that its average loss will be no greater than 275? Follow the four-step process

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(a) Less than10 minutes.
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(c) as we take larger and larger samples from this population, x will get closer and closer to .

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(e) the person measuring the children's weights does so without any systematic error.

Your mail-order company advertises that it ships 90%of its orders within three working days. You select an SRS of 100of the 5000orders received in the past week for an audit. The audit reveals that 86of these orders were shipped on time.

(a) If the company really ships 90%of its orders on time, what is the probability that the proportion in an SRS of 100orders is as small as the proportion in your sample or smaller? Follow the four-step process.

(b) A critic says, 鈥淎ha! You claim 90%, but in your sample, the on-time percentage is lower than that. So the 90%claim is wrong.鈥 Explain in simple language why your probability calculation in (a) shows that the result of the sample does not refute the 90%claim.

Thousands of travelers pass through the airport in Guadalajara, Mexico, each day. Before leaving the airport, each passenger must pass through the Customs inspection area. Customs agents want to be sure that passengers do not bring illegal items into the country. But they do not have time to search every traveler鈥檚 luggage. Instead, they require each person to press a button. Either a red or a green bulb lights up. If the red light shows, the passenger will be searched by Customs agents. A green light means 鈥済o ahead.鈥 Customs agents claim that the proportion of all travelers who will be stopped (red light) is0.30, because the light has probability 0.30of showing red on any push of the button. To test this claim, a concerned citizen watches a random sample of 100travelers push the button. Only 20get a red light.

(a) Assume that the Customs agents鈥 claim is true. Find the probability that the proportion of travelers who get a red light is as small as or smaller than the result in this sample. Show your work.

(b) Based on your results in (a), do you believe the Customs agents鈥 claim? Explain.

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