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.

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DeanR
The probability will be the area under the normal distribution [tex]N(27,3^2)[/tex] from 24 to 32.

We normalize these so we can use the standard normal distribution

[tex]z_1 = \dfrac{24 - 27}{3} = -1[/tex]

[tex]z_2 = \dfrac{32-27}{3}=\dfrac 5 3[/tex]

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

 
[tex]p =0.34134+0.45254=0.79388[/tex]

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