Answer:
5.18oz.
Step-by-step explanation:
Problems of normally distributed samples can be 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]X = 6, \sigma = 0.4[/tex]
Overflow only 2% of the time?
Value of [tex]\mu[/tex] when Z has a pvalue of 1-0.02 = 0.98. So X when Z = 2.055.
[tex]Z = \frac{6 - \mu}{\sigma}[/tex]
[tex]2.055 = \frac{6 - \mu}{0.4}[/tex]
[tex]6 - \mu = 2.055*0.4[/tex]
[tex]\mu = 5.18[/tex]
So the correct answer is:
5.18oz.
