This is probably easy for you but very hard for me!

#1
Can someone please help me figure out how to do this problem?

A professor knows that from past experience that the time for students to complete a quiz has an N(19, 3) distribution.
a. If he allows 20 minutes for the quiz, what percent of the students will not complete the quiz?

b. Suppose that he wants to allow sufficient time so that 95% of the students will complete the quiz in the allotted time.
How much time should he allow for the quiz?
 
#2
A) Do you ever use R? This would easily be computed using pnorm(20,19,sd=3), assuming the 3 is a SD. If it is a variance, sd=sqrt(3)

You can also calculate a z-value by taking (20 - 19)/3. This gives a z-value of 0.33, from here, you can look at z-tables (http://www.stat.ufl.edu/~athienit/Tables/Ztable.pdf)

B) Similar calculation can easily be done in R with the following command: qnorm(.95, 19, 3) -- again assuming 3 is the SD.
 
#3
Thank you for responding so quickly! I am not familiar with qnorm and cannot find it in Excel. Can I find this solution using a Z table? Z=1.65 but I am unsure how to set up the problem. If Z=1.65 for 95% and its a normal distribution with a mean of 19 and SD of 3, how do I calculate the following:
Suppose that he wants to allow sufficient time so that 95% of the students will complete the quiz in the allotted time.
How much time should he allow for the quiz? N(19, 3)
 
#5
Think about what a Z is. It is the standardized value, which means you subtract the mean and divide by the standard deviation. So, to get your answer, work it in reverse.
Multiply the 1.65 by the sd (3), and then add the location parameter (19) to give you the answer.