A machine fills containers with a particular product. Assume the filling weights are normally distributed with a variance of 0.16 ounces2. If 85% of the containers hold greater than 12 ounces, find the machine's mean filling weight (in ounces).

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(\mu,\sqrt{0.16}= 0.4)[/tex]

For this part we want to find a value a, such that we satisfy this condition:

[tex]P(X>12)=0.85[/tex] (a)

[tex]P(X<12)=0.25[/tex] (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.25 of the area on the left and 0.85 of the area on the right it's z=-0.674. On this case P(Z<-0.674)=0.25 and P(z>-0.674)=0.85

If we use condition (b) from previous we have this:

[tex]P(X<12)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.25[/tex]

[tex]P(z<\frac{12-\mu}{\sigma})=0.25[/tex]

But we know which value of z satisfy the previous equation so then we can do this:

[tex]z=-0.674<\frac{12-\mu}{0.4}[/tex]

And if we solve for [tex]\mu[/tex] we got

[tex]\mu=12 +0.674*0.4=12.270[/tex]

So then the mean is [tex] 'mu = 12.270[/tex] for this case.