Question

A machine fills containers with a particular product. Assume the filling weights are normally distributed with a variance of 0.81 ounces2. If 85% of the containers hold greater than 20 ounces, find the machine's mean filling weight (in ounces).

a. 19.0640
b. 19.1576
c. 20.8424
d. None of the answers is correct
e. 20.9360

Answer 1

`P(X >20) =0.85`

And by the complement rule we know
`P(X<20) =0.15`

We need to find a z score value that accumulates 0.15 of the area on the left and 0.85 of the area on the right and we got:

`z = -1.036`

Since
`P(Z<-1.036)=0.15` and now using the z score formula we have this:

`-1.03643 = (20 -\mu)/(0.9)`

And solving for the mean we got:

`\mu = 20 + 1.03643 *0.9 = 20.93`

So the best anwer for this case would be:

e. 20.9360

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 weights of a population, and for this case we know the distribution for X is given by:

`X \sim N(\mu,√(0.81)=0.9)`

Where
`\mu` the mean and
`\sigma=0.9`

And the best way to solve this problem is using the normal standard distribution and the z score given by:

`z=(x-\mu)/(\sigma)`

We know the following condition:

`P(X >20) =0.85`

And by the complement rule we know
`P(X<20) =0.15`

We need to find a z score value that accumulates 0.15 of the area on the left and 0.85 of the area on the right and we got:

`z = -1.036`

Since
`P(Z<-1.036)=0.15` and now using the z score formula we have this:

`-1.03643 = (20 -\mu)/(0.9)`

And solving for the mean we got:

`\mu = 20 + 1.03643 *0.9 = 20.93`

So the best anwer for this case would be:

e. 20.9360