Question

A soft drink machine outputs a mean of 27 ounces per cup. the machine's output is normally distributed with a standard deviation of 3 ounces. what is the probability of filling a cup between 24 and 32 ounces? round your answer to four decimal places.

Answer 1

The probability will be the area under the normal distribution $N(27,3^2)$ from 24 to 32.

We normalize these so we can use the standard normal distribution

$z_1 = \dfrac{24 - 27}{3} = -1$

$z_2 = \dfrac{32-27}{3}=\dfrac 5 3$

Typically we'd just use a computer to tell us the integral of the standard normal from -1 to 5/3.  In the old days we'd use the z table.  It tells us the normal integral from 0 to positive z.

We know the integral of the Gaussian from -1 to 1 standard deviation is 68%; from the z table we see the probability from 0 to 1 is half that, 0.34134.  That will also be the integral from -1 to 0. 

The z table says the integral from 0 to 5/3, 1.67, is 0.45254.

So our total probability, integral from -1 to 5/3, is

 $p =0.34134+0.45254=0.79388$

That's really from z=-1 to z=+1.67 so likely a bit off in the last couple digits.