The average weight of a cereal box is 20 ounces with a variance of 0.25 ounces. a sample of 25 boxes is randomly selected and weighed. what is the probability that the mean of the sample is less than 19.9 ounces?

The probability is 0.1587.

The z-score associated with this is given by the formula

z=(x-μ)/(σ/√n)

Since the variance, σ², is 0.25, the standard deviation, σ, is √0.25 = 0.05.  This gives us:
z = (19.9-20)/(0.05/√25) = -0.1/(0.05/5) = -1

Using a z-table (http://www.z-table.com) we see that the area to the left of this, less than, is 0.1587.