The monthly starting salaries of students who receive business degrees
have a standard deviation of $600. What size sample should be selected
so that there is 0.95 probability of estimating the mean monthly
income within $150 or less?

Hi again lanee19!!


Recall from the other exercise the formula for the extremes of the
confidence interval:

Extremes = Mean +/- (1.96 * STD)/sqrt(n)

Here you do not know Mean and n, 
but you know that the value of (1.96 * STD)/sqrt(n) is:

(1.96 * STD)/sqrt(n) = 150


Then the minimum sample size n to estimate Mean to within maximum
error of estimate $150 and with confidence level of 95%:

n >= [(1.96 * STD)/150]^2 = [(1.96 * 600)/150]^2 = [7.84]^2 = 61.4656

n = 62.

The size sample to be selected if you want that there is 0.95
probability of estimating the mean monthly income within $150 or less
is 62.


The same text of your other question can be used as reference (see page 2):
"Confidence Intervals":
http://math.cofc.edu/faculty/diamond/104confint.pdf


I hope that this helps you, and again feel free to use the
clarification feature to request for further assistance on this if you
need it.


Best regards.
livioflores-ga

everything was great except wouldn't he rounded answer be n=61? Other
than that everything was right........thank you!!!!!!!