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

#### Caroyer1

##### New Member

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?

#### simplemts

##### New Member
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.

#### Caroyer1

##### New Member
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)

#### ted00

##### New Member
yes, that's right, you have to standardize it then you can look it up in a z-table

#### simplemts

##### New Member
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.

#### ted00

##### New Member
I am not familiar with qnorm and cannot find it in Excel.
qnorm is the normal quantile function in R

in Excel it's NORMINV