Question

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).

Answer 1

Answer:

$z=-0.674$

And if we solve for $\mu$ we got

$\mu=12 +0.674*0.4=12.270$

So then the mean is $'mu = 12.270$ for this case.

Step-by-step explanation:

Previous concepts

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:

$X \sim N(\mu,\sqrt{0.16}= 0.4)$  

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

$P(X>12)=0.85$   (a)

$P(X$   (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:

$P(X$  

$P(z$

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

$z=-0.674$

And if we solve for $\mu$ we got

$\mu=12 +0.674*0.4=12.270$

So then the mean is $'mu = 12.270$ for this case.