Instant dinner comes in packages with weights that are normally distributed with a standard deviation of 0.3 oz. If 2.3 percent of the dinners weigh more than 13.5 oz, what is the mean weight?

Answer 1

12.9 oz, which is two standard deviations below 13.5 oz
See http://psych.colorado.edu/~mcclella/java/normal/accurateNormal.html for a tool to do problems like this

Answer 2

2.3% = .023

so that is the probability that it will be over 13.5 oz, or .977 is the prob that it will be below 13.5

normal distribution charts show the probability below a given z-score.

z-score = (real score - mean)/standard deviation

I used this chart
http://www.math.unb.ca/~knight/utility/NormTble.htm

I found .977 at a z-score of 2.00

so 2.00 = (13.5-m)/.3
m = 12.9

Using this data in
http://davidmlane.com/hyperstat/z_table.html
and clicking on "above" at 13.5 gave me .02275, close enough for .023