```Question 382317
The average age of retirement residents is 69. The standard deviation is 5.8 years. A simple random sample of 100 residents is picked and the sample mean is computed.
1) What is the probability that the average age of the 100 residents is over 70?
t(70) = (70-69)/[5.8/sqrt(100)] = 1.7241
P(x-bar > 70) = P(t > 1.7241 when df=99) = 0.0439
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2) Now suppose the sample had a 90% confidence interval for the population mean "u". This means?
This means we are 90% confident that mean of the sample means
is within the intervals limits.
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Cheers,
Stan H.

For question #1, my answer was 0.4235. The second time I submitted the assignment, my answer was 0.0427. I used table a to come up with these answers.
For question #2 I said that it is an interval with a marginal error of greater than or less than 90%. I believe this is correct but is it also true that if this property was repeated, each time creating a new 90% interval, then in the long run 90% of those intervals would contain "u"?
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