Question

A soft-drink machine can be regulated so that it discharges an average of μ ounces per cup. If the ounces of fill are normally distributed with standard deviation 0.7 ounce, give the setting for μ so that 32-ounce cups will overflow only 4% of the time. (Round your answer to three decimal places.)

Answer 1

`\mu = 30.775`

Explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

`\sigma = 0.7`

If the ounces of fill are normally distributed with standard deviation 0.7 ounce, give the setting for μ so that 32-ounce cups will overflow only 4% of the time.

This is
`\mu` for which Z when
`X = 32` has a pvalue of 1-0.04 = 0.96. So
`\mu` when Z = 1.75.

`Z = (X - \mu)/(\sigma)`

`1.75 = (32 - \mu)/(0.7)`

`32 - \mu = 1.75*0.7`

`\mu = 30.775`