Normal Distribution

Question

The labels on boxes of Cheesy Poofs say that the box contains 16 ounces of Cheesy Poofs, but that doesn’t guarantee that each box contains exactly 16.00 ounces. The machine that fills the boxes doesn’t always put out the exact same amount. The amount in each box is Normally distributed, and has a standard deviation is .40 ounces. To help assure that most boxes contain as much or more than the label says, the machine is set so that the mean will be 16.28 ounces. 

i) The filling machine will always be this inconsistent, in that the amount in each box is Normally distributed, and has a standard deviation is .40 ounces. However, the average amount in each box can be adjusted. Where should the average be set to assure that 90% of the boxes sold contain at least 16.00 ounces?

j) An improved filling machine will always be less inconsistent, in that the amount in each box is still Normally distributed, and now has a standard deviation is .10 ounces. Where should the average be set to assure that 90% of the boxes sold contain at least 16.00 ounces when this new filling machine is used?

Answer 1

We are going to determine the average value using the standard normal distribution (mean = 0, sd = 1)

The formula for the standardized z-score is:

z = (x - mean)/sd

We need to look up the z-score that is associated with 90th percentile, which is 1.28.

so,using the values given in the problem in the formula for the z-score

1.28 = (x - 16.28)/0.4

x = 16.792

the second part of the problem will use the same equation with 0.1 substituted for 0.4 as the sd:

1.28 = (x - 16.28)/0.1

x = 16.408