# Normal Distribution: Z- test

• Mar 25th 2013, 12:04 PM
daphnewoon
Normal Distribution: Z- test
A pharmaceutical company manufactures tubes of a certain skin cream. The tubes have a stated volume on the tubes of 125ml, where the content of the tubes has a Normal distribution with a mean mu and a known standard deviation of 0.6ml. If it is a legal requirement for 95% of all tubes sold to contain at least the stated volume, determine the minimum value that mu can take so that the legal requirement is met, giving your answer to the nearest 0.1ml.
• Mar 26th 2013, 05:19 AM
daphnewoon
Re: Normal Distribution: Z- test
Attachment 27697

Here is a copy of the Z-table.
Please try to help!! Urgent Assignment :(
• Mar 26th 2013, 07:24 AM
joeDIT
Re: Normal Distribution: Z- test
is this all the information you were given in the Q?
• Mar 26th 2013, 07:31 AM
daphnewoon
Re: Normal Distribution: Z- test
Yes, it is.
• Mar 26th 2013, 10:21 AM
HallsofIvy
Re: Normal Distribution: Z- test
Quote:

Originally Posted by daphnewoon
A pharmaceutical company manufactures tubes of a certain skin cream. The tubes have a stated volume on the tubes of 125ml, where the content of the tubes has a Normal distribution with a mean mu and a known standard deviation of 0.6ml. If it is a legal requirement for 95% of all tubes sold to contain at least the stated volume, determine the minimum value that mu can take so that the legal requirement is met, giving your answer to the nearest 0.1ml.

You want to determine $\displaystyle \mu$ so that $\displaystyle z= (125- \mu)/.6$ gives 0.95. What z does your table give?
(A very nice table of the normal distribution is at Standard Normal Distribution Table)

After you have found z from the table, solve $\displaystyle z= (125- \mu)/.6$ for $\displaystyle \mu$.
• Mar 26th 2013, 10:53 AM
daphnewoon
Re: Normal Distribution: Z- test
In the table, the z value for 0.95 is 1.645.

Let me just share with you my way of approach.

I actually did slightly differently.

I used 1.96 = (125- mu)/0.6 as I got mixed up with confidence interval.