Determine Sample Size

Hi all,

The following has two parts and I *think* I have figured out the first part. The one I'm really troubled by is determining the sample size from Part B. Please advise on which formula or how I should go about finding the answer. Thank you very much.

The average price of a gallon of unleaded regular gasoline was reported to be $2.34 in northern Kentucky (The Cincinnati Enquirer, January 21, 2006). Use this price as the population mean, and assume the population standard deviation is $.20.

A. What is the probability that the mean price for a sample of 50 service stations is within $.03 of the population mean (to 4 decimals)?
I got:

Population mean = 2.34
Population standard deviation = .20
Sample size = 50
Standard deviation of the sample mean = (about) .0283

Using the Z-score, I know I'm looking for the area of

P(-1.06<z<1.06)

=0.8554-0.1446
= 0.7108

Did I do this right?

B. Calculate the sample size necessary to guarantee at least .95 probability that the sample mean is within $.03 of the population mean (0 decimals).

The only notes I can find on determining a sample size is when I'm given a margin of error, which deals with the confidence level; so I'm not sure whether I'm on the right track with this one. Hence, I'm utterly lost with how to go about this problem (no clue where to begin). Advise, please. Thank you!!!

minimum sample size is given by

\(\displaystyle n=\left(\frac{z\cdot \sigma}{E}\right)^{2}\)

Where \(\displaystyle z=1.96, \;\ {\sigma}=.20, \;\ E=.03\)

Always round up.

This formula is derived from the familiar formula:

\(\displaystyle E=z\cdot\frac{\sigma}{\sqrt{n}}\)

\(\displaystyle E\sqrt{n}=z\sigma\)

\(\displaystyle \sqrt{n}=\frac{z\sigma}{E}\)

\(\displaystyle n=\text{the formula above}\)

Hi there,

Thank you very much for your help. I did not realize the margin of error was given in the problem and that where I was constantly being stopped at.