How many women must be randomly selected to estimate the mean weight of women in one age group. we want 90% confidence that the sample mean is within 2.7 lb of the population mean, and the population standard deviation is known to be 22 lb?

That is a very good question

Look Grank

you know that:

confidence interval = mean +or- Margin of Error

and

Margin of Error = (z)*(standard deviation) / (sqrt of n)
where n is the number of sample records

So we need to calculate z-value firstly

It says: "we want 90% confidence"

So that:

confidence90% corresponds to z-value of 1.645

Plug that into our formula of Margin of Error:

Margin of Error = (1.645)*(22) / (sqrt of n)

"The sample mean is within 2.7 lb of the population mean" means that Margin of Error is 2.7

Margin of Error:

2.7 = (1.645)*(22) / (sqrt of n)

Now solve for n:

n=179.66~180

SO that 180 women must be randomly selected to estimate the mean weight of women in one age group.