Question 1181631: A leading magazine (like Barron's) reported at one time that the average number of weeks an individual is unemployed is 18 weeks. Assume that for the population of all unemployed individuals the population mean length of unemployment is 18 weeks and that the population standard deviation is 3.2 weeks. Suppose you would like to select a random sample of 73 unemployed individuals for a follow-up study.
Find the probability that a single randomly selected value is greater than 17.6.
P(X > 17.6) =
(Enter your answers as numbers accurate to 4 decimal places.)
Find the probability that a sample of size n=73 is randomly selected with a mean greater than 17.6.
P(M > 17.6) =
(Enter your answers as numbers accurate to 4 decimal places.)
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! z=(x-mean)/sd
for one individual, z > (17.6-18)/3.2=-0.4/3.2 or -0.125.
the probability z > -0.125 is 0.5497
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for the mean's being > 17.6 the sd becomes 3.2/sqrt(73)=0.3745
z >(17.6-18)*sqrt(73)/3.2=-1.068
the probability z > -1.068 is 0.8572
The distribution of the samples of 73 is much narrower, so that the same value to the left o the mean is much closer to the left side of the distribution, and since the probability we want is to the right of that, that probability is more.
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