Answer :
Answer:
[tex]\mu = 30.775[/tex]
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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:
[tex]\sigma = 0.7[/tex]
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 [tex]\mu[/tex] for which Z when [tex]X = 32[/tex] has a pvalue of 1-0.04 = 0.96. So [tex]\mu[/tex] when Z = 1.75.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.75 = \frac{32 - \mu}{0.7}[/tex]
[tex]32 - \mu = 1.75*0.7[/tex]
[tex]\mu = 30.775[/tex]