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

Answer:

$\mu = 30.775$

Step-by-step 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:

$Z = \frac{X - \mu}{\sigma}$

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.

$Z = \frac{X - \mu}{\sigma}$

$1.75 = \frac{32 - \mu}{0.7}$

$32 - \mu = 1.75*0.7$

$\mu = 30.775$