Answer: At a value of 12.3 the mean should be set so that 99.9% of all cans exceed 12 fluid ounces.
Step-by-step explanation:
Let us assume that X is a normal rando variable with a mean value [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex].
Hence, random variable Z will be introduced as follows.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
(a) Set the value [tex]\sigma = 0.1[/tex] and the equation will be written down as follows.
[tex]0.999 = P (X \geq 12) \\= P (Z \geq \frac{12 - \mu}{0.1})\\= 1 -\phi (\frac{12 - \mu}{0.1})\\\phi (\frac{12 - \mu}{0.1}) = 0.001[/tex]
According to the tables,
[tex]\frac{12 - \mu}{0.1} = -3\\\mu = 12.3[/tex]
Thus, we can conclude that at a value of 12.3 the mean should be set so that 99.9% of all cans exceed 12 fluid ounces.
