A brewery produces cans of beer that are supposed to contain exactly 12 ounces. But owing to the inevitable variation in the filling equipment, the amount of beer in each can is actually a random variable with a normal distribution. It has a mean of 12 ounces and a standard deviation of 0.30 ounce.
If you bought a six-pack of their beer what is the probability that you are going to actually get less than or equal to a total of 72 ounces of beer in your six-pack?

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".

Solutio to the problem

Let X the random variable that represent the amount of beer in each can of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(12,0.3)[/tex]

Where [tex]\mu=12[/tex] and [tex]\sigma=0.3[/tex]

For this case we select 6 cans and we are interested in the probability that the total would be less or equal than 72 ounces. So we need to find a distribution for the total.

The definition of sample mean is given by:

[tex] \bar X = \frac{\sum_{i=1}^n X_i}{n} = \frac{T}{n}[/tex]

If we solve for the total T we got:

[tex] T= n \bar X[/tex]

For this case then the expected value and variance are given by:

[tex] E(T) = n E(\bar X) =n \mu[/tex]

[tex] Var(T) = n^2 Var(\bar X)= n^2 \frac{\sigma^2}{n}= n \sigma^2[/tex]

And the deviation is just:

[tex] Sd(T) = \sqrt{n} \sigma[/tex]

So then the distribution for the total would be also normal and given by:

[tex] T \sim N (\mu = 6*12=72 , \sigma= \sqrt{6} *0.3=0.735)[/tex]

And we want this probability:

[tex] P(T\leq 72)[/tex]

And we can use the z score formula given by:

[tex] z = \frac{x-\mu}{\sigma}[/tex]

[tex] P(T \leq 72) = P(Z< \frac{72-72}{0.735}) = P(Z<0) = 0.5[/tex]