SOLUTION: A leading magazine (like Barron's) reported at one time that the average number of weeks an individual is unemployed is 16 weeks. Assume that for the population of all unemployed i
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Question 1205242: A leading magazine (like Barron's) reported at one time that the average number of weeks an individual is unemployed is 16 weeks. Assume that for the population of all unemployed individuals the population mean length of unemployment is 16 weeks and that the population standard deviation is 2.9 weeks. Suppose you would like to select a random sample of 95 unemployed individuals for a follow-up study.
Find the probability that a single randomly selected value is greater than 16. Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website! population mean is 16 weeks
population standard deviation is 2.9 weeks.
sample size is 95 unemployed individuals.
the probability that a single randomly selected value from the sample is greater than 16 should be .5, assuming a normal distribution.
that's because in a normal distribution, the mean and the median and the mode are in the middle of the distributionand the probability of getting a value greater than that is .5 and the probability of getting a value less than that is also .5.
here's what it looks like on a normal distribution graph.
using raw scores, you enter the mean and standard deviation of the population and look for a raw score greater than 16.
using the z-score, you enter the mean and and standard deviation of the population and look for a z-score greater than 0.
that's because the z-score formula is z = (x-m)/s where x is the raw score and m is the population mean and s is the standard deviation of the population.
you get z = (16 - 16) / 2.9 = 0.
note that the random sample could be higher or lower than 16, but, with enough sample draws, the average should be about half above and half below the population mean of 16.