Answer :
Answer:
The mean should be 64.63 ounces.
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 = 0.38[/tex]
The quality control inspector wants to adjust the machine such that at least 95% of the jugs have more than 64 ounces of detergent. What should the mean amount of detergent poured by this machine into these jugs be?
This is [tex]\mu[/tex], for which X = 64 will have a pvalue of 1-0.95 = 0.05. So when X = 64, Z = -1.645.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-1.645 = \frac{64 - \mu}{0.38}[/tex]
[tex]64 - \mu = -1.645*0.38[/tex]
[tex]\mu = 64 + 1.645*0.38[/tex]
[tex]\mu = 64.63[/tex]
The mean should be 64.63 ounces.