The following information is provided: 
average weekly earnings of $750, a standard deviation of $40 and a
sample size of 64, what is the probability that the sample mean will
be less than $740?

Dell has determined that the average repair time for a PC is 130
minutes with a standard deviation of 26 minutes.  If a random sample
of 40 services is selected, what is the probability that the sample of
40 will have a mean service time less than 136 minutes?

At ABC University, 40% of students live in the dormitory.  If a random
sample of 80 students is selected, what is the probability that the
sample proportion is less than .30?

Macy?s in Persia Mall conducted a random sample of 81 credit sales. It
showed an average sale of $68.00.  From past data it is known that the
standard deviation of the population is $27.00.  Construct a 99%
confidence interval estimate of the population mean:


GIGO Computer Services, Inc. wants to determine a 92% confidence
interval for the average CPU time of their processing transactions. A
sample of 144 transactions produced a mean of 0.20 seconds and a
standard deviation of .05.  Construct a 92% confidence interval for
the average processing time.

I have more small problems, I will be posting them once I get answer
to the above questions.

Question 1:
This is your typical normal distribution problem.  sigma is 40, mu is
750.  To solve it, you need to convert from this distribution to the
standard normal.  The way to do this is with the formula:
z=(x-bar - mu)/(sigma/sqrt(n))

So, Prob(x-bar<740) = Prob(z< (740-750)/(40/sqrt(64))
=Prob(z< -10/5) = Prob (z<-2) = .0228

Question 2:
This follows the exact same format of Question 1.

Prob(x-bar<136) = Prob(z<(136-130)/(26/sqrt(4))
=Prob(z<1.46) = .9279

Question 3:
This sounds like it would be again the same format, but there's no
standard deviation given.  As it is, I don't believe it can be
answered.

Question 4:
Confidence intervals are much like the prevoius questions.  Comparing
the formulas should show you this.

CI: x-bar + or - z*(sigma/sqrt(n))
Z coems from the normal table.  Find a value for z that gives you a
probability equal to the desired confidence interval.

Here, this gives 68 +- 2.575*(27/sqrt(80) = 68 +- 7.77

Question 5:
Here, since the standard deviation came from the data, rather than
being a known number, you need to use a t-test.  Again, compare this
formula with the one in question 1 and you will see the similarities.

CI: x-bar +- t(s/sqrt(n)) where s is the estimated standard deviation
(Note: standard t-distribution tables do not give a value for 92%, I
estimated this using 90% and 95%)
Here this gives:

.2 +- 1.82(.05/sqrt(144)) = .2+-.0076

Posting 5 questions for such a small value is a bit much.  You may
want to increase your price if you plan to post another large set. 
Also, when questions all have the same format, it'd be useful if you
simply asked one, and used the answer given to figure out the rest.

Hi,

Thank you very much for your help. It is my first time using
Google-answers site so I am not too familiar with the process. I just
posted the first set of questions to see whether the site works or
not. I have 15 more small problems and if you interested in solving
them, please let me know and I don't know how the payment process
works with this site and which party is getting paid, so I would
appreciate your comments on that. I actually don't know much about
statistics. I have just been reading my daughter's book to help her.
She is weak in this subject so I am trying to help her and see if she
is answering her assigned problems properly. She gets one homework set
per week consisting 20 problems but I don't have the time myself to
sit down and figure them out on my own. Plese let me know if I can
post the other 15 problems.