The average age of residents in a large residential retirement community is 69 years with standard deviation 5.8 years. a simple random sample of 100 residents is to be selected, and the sample mean age of these residents is to be computed. the probability that the average age, of the 100 residents selected is less than 68.5 years is

To solve for the problem, we should get first the standard dev and it is computed by:

standard = 5.8 / sqrt (100) = -.58 

Then looking for the z value:
(68.5-69)/.58 = .86 

Looking at z table, the value with .86 is we get .195. This would be the answer.

283 square feet

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you cant answer this dummy

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