Question 1119486: 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 3.4 weeks. Suppose you would like to select a random sample of 77 unemployed individuals for a follow-up study.
Find the probability that a single randomly selected value is greater than 15.6.
P(X > 15.6) = (Enter your answers as numbers accurate to 4 decimal places.)
Find the probability that a sample of size n=77 is randomly selected with a mean greater than 15.6.
P(M > 15.6) =
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! population mean = 16 weeks.
population stndard deviation = 3.4 weeks.
for a single randomly selected value from this population, the formula for z-score would be:
z-score = (x - m) / s
x is the raw score.
m is the mean of the population.
s is the standard deviation of the population.
use the z-score formula to get:
z = (x - m) / s becomes z = (15.6 - 16) / 3.4 = -.1176470588.
the area under the normal distribution curve to the left of that z-score is equal to .4531736114.
the area to the right of it would be equal to 1 - that = .5468263886.
that's the probability of getting a raw score greater than 15.6 when you randomly pick one sample from the population.
next you want to know the probability that the mean of a distribution of 77 randomly drawn samles is greater than 15.6.
the formula for the standard error is:
s = standard deviation / square root of sample size.
the s in this equation is the standard error, which is the standard deviation of the distribution of sample means.
same s, different meaning.
this happens a lot, so always be careful what the variable means in the context that it is being used.
we get:
s = 3.4 / sqrt(77) = .38746596
the z-score is now z = (x - m) / s = (15.6 - 16) / .38746596 = -1.032348751.
the area under the normal distribution curve to the left of that z-score is equal to .1509544032.
the area to the right of it would be equal to 1 - that = .8490455968.
that's the probability of getting the mean of a sample greater than 15.6 when you randomly pick a distribution of samples of size 77 from the population.
note that if the sample size is 1, s = standard deviation divided by sqrt(sample size) becomes s = 3.4 / sqrt(1) = 3.4.
as the sample size gets larger, the standard deviation of the mean of a sample of that size becomes smaller.
the biggest the sample size can get is the number of elements in the population itself.
in that case, the standard deviation of the distribution of sample means would essentially be equal to 0 because the sample is the population itself.
the formula doesn't allow it to become equal to 0, but it gets very very small and approaches 0 as the size of the sample approaches infinity.
in that case, the probability that the mean of the sample of the same size of the population being more than 15.6 is essentially 100%.
for an example, assume the population size if 5000.
if the sample size is 5000, then the standard error is 3.4 / sqrt(5000) which is equal to .0480832611.
the z-score is then (15.6 - 16) / .0480832611 = -8.318903308.
the probability of getting a z-score greater than that would be equal to 1, which is 100%.
this stands to reason since the mean of the population is 16 which will always be greater than 15.6.
the general rule for determining the standard error of a distribution of sample means is:
s = standard deviation of the population / sqrt(sample size).
the larger the sample size, the smaller the standard error.
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