Question

The fill volume of an automated filling machine used for filling cans of carbonated beverage is normally distributed. Suppose, that the mean of the filling operation can be adjusted easily, but the standard deviation remains at 0.4 fluid ounce. (a) At what value should the mean be set so that 99.9% of all cans exceed 12 fluid ounces

Answer 1

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 $\mu$ and standard deviation $\sigma$.

Hence, random variable Z will be introduced as follows.

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

(a) Set the value $\sigma = 0.1$ and the equation will be written down as follows.

$0.999 = P (X \geq 12) \\= P (Z \geq \frac{12 - \mu}{0.1})\\= 1 -\phi (\frac{12 - \mu}{0.1})\\\phi (\frac{12 - \mu}{0.1}) = 0.001$

According  to the tables,

$\frac{12 - \mu}{0.1} = -3\\\mu = 12.3$

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.