Assume that the president is correct and p = .30. What is the sampling distribut
ID: 3434690 • Letter: A
Question
Assume that the president is correct and p = .30. What is the sampling distribution of for this study?
- Select your answer -A normal distribution because np and n(1-p) are both greater than 5A normal distribution because np and n(1-p) are both less than 5A non normal distributionItem 1
What is the probability that the sample proportion will be between .20 and .40 (to 4 decimals)?
What is the probability that the sample proportion will be between .25 and .35 (to 4 decimals)?
The president of Doerman Distributors, Inc., believes that 30% of the firm's orders come from first-time customers. A random sample of 100 orders will be used to estimate the proportion of first-time customers.Assume that the president is correct and p = .30. What is the sampling distribution of for this study?
- Select your answer -A normal distribution because np and n(1-p) are both greater than 5A normal distribution because np and n(1-p) are both less than 5A non normal distributionItem 1
What is the probability that the sample proportion will be between .20 and .40 (to 4 decimals)?
What is the probability that the sample proportion will be between .25 and .35 (to 4 decimals)?
Explanation / Answer
the sampling distribution of for this study
Normal distribution with p=0.3 and standard deviation=sqrt(p*(1-p)/n)
=sqrt(0.3*0.7/100)
=0.04582576
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A normal distribution because np and n(1-p) are both greater than 5
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So the probability that the sample proportion will be between .20 and .40 is
P(0.2<phat<0.4)
= P((0.2-0.3)/sqrt(0.3*0.7/100) <(phat-p)/sqrt(p*(1-p)/n) <(0.4-0.3)/sqrt(0.3*0.7/100))
=P(-2.18<Z<2.18)
=0.9707 (from standard normal table)
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So the probability that the sample proportion will be between .25 and .35 is
P(0.25<phat<0.35)
=P((0.25-0.3)/sqrt(0.3*0.7/100) <Z<(0.35-0.3)/sqrt(0.3*0.7/100))
=P(-1.09<Z<1.09)
=0.7243 (from standard normal table)
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