Help with a statistics question....

[Otrant]

Hi,

Worth a shot asking this here, as there may be some people out there that have taken statistics courses. I need to answer the following question:

An exam consists of 3 questions. In the past the instructor has learned that
the time students take to complete the first question can be approximated by a
normal random variable with a mean of 20 minutes and a variance of 40. The time
needed for the second question is approximated by a normal random variable with
a mean of 80 minutes and a variance of 70. Finally, the time needed for the third
question is approximated by a normal random variable with a mean of 40 minutes
and a variance of 115.
a. What is the expected time to complete the exam?
b. What is the variance in time needed given the questions are all
independent of each other?
c. How much time should be given by the instructor in order that 90% of the
students will have enough time to complete the exam?

Now, given the rules for mean and the rules for variance the answer to a) seems to be the summation of the means, while the answer to b) is the summation of the variances. (Someone feel free to correct me if I'm wrong).

However, I am at a loss as to how to answer part c). Any help is appreciated, and thanks in advance.

 

[Hector13]

You are on the right track... just remember that the sum of normal random variables is usually a normal variable also (and you already have stated its mean and variance in parts a & b).

 

[Otrant]

Ah...

If this is a normal distribution, then one could use the table of standard normal probabilities, correct?

In which case, I would simply need to locate (roughly) 90%, find the corresponding z-value, and then multiply that z-value by the standard deviation (which is SD sqrt(225) = 15 in this case) to find my answer?

If you could follow my logic and confirm this to be correct, it would put me at ease. I think I had trouble with part c) because I did not immediately realize this distribution was "normal", doh!

Thanks again!