Answer:
The probability of filling a cup between 18 and 33 ounces is 0.93719.
Step-by-step explanation:
The variable here is the machine's output which is <em>normally distributed</em>.
The normal distribution is defined by <em>two parameters</em>, namely, the mean and the standard deviation.
The population mean for this <em>normal distribution</em> is
ounces per cup, and a population standard deviation of
ounces (per cup).
To solve this question, we can use the <em>standard normal distribution</em> (which has a mean = 0, and a standard deviation = 1). All we have to do is to "transform" those raw scores into z-scores, and then consult a standard normal table (available in every Statistics book or on the Internet).
Roughly speaking, the <em>z-scores</em> are values that tell us the distance from the mean in standard deviations units. For <em>positive</em> values of it, the value is <em>above</em> the mean, and <em>negative</em> values tell us that the value is <em>below</em> the mean.
We can obtain the z-scores using the next formula:
[1]
Where
<em>z</em> is the z-score.
<em>x</em> is the raw score.
is the population mean.
is the population standard deviation.
The probability of filling a cup between 18 and 33 ounces
Having all the information above, we can proceed as follows:
<em>First Step: obtaining the z-score for 18 to find the corresponding probability of this value in this normal distribution.</em>
Using the formula [1], we have:
![\\ z = \frac{18 - 25}{4}](https://tex.z-dn.net/?f=%20%5C%5C%20z%20%3D%20%5Cfrac%7B18%20-%2025%7D%7B4%7D)
This tells us that the value is <em>below</em> the mean and is -1.75 standard units from it.
Most of the standard normal tables are made for positive values of <em>z</em>. However, since the normal distribution is <em>symmetrical</em> around the mean, we, fortunately, can obtain the corresponding probability considering that:
![\\ P(z](https://tex.z-dn.net/?f=%20%5C%5C%20P%28z%3C-1.75%29%20%3D%201%20-%20P%28z%3C1.75%29%20%3D%20P%28z%3E1.75%29)
Then, consulting the <em>cumulative standard normal table</em> for z = 1.75, we have that the corresponding cumulative probability is
.
Thus
![\\ P(z1.75)](https://tex.z-dn.net/?f=%20%5C%5C%20P%28z%3C-1.75%29%20%3D%201%20-%200.95994%20%3D%20P%28z%3E1.75%29)
![\\ P(z1.75)](https://tex.z-dn.net/?f=%20%5C%5C%20P%28z%3C-1.75%29%20%3D%200.04006%20%3D%20P%28z%3E1.75%29)
Then, the probability is ![\\ P(z](https://tex.z-dn.net/?f=%20%5C%5C%20P%28z%3C-1.75%29%20%3D%200.04006)
We have to remember that this is the standardized value for x = 18, and for this normal distribution (mean = 25, standard deviation = 4), it has the same cumulative probability [P(x<18) = 0.04006].
<em>Second Step: obtaining the z-score for 33 to find the corresponding probability of this value in this normal distribution.</em>
We can proceed in the same way to obtain the z-score and the associated probability with the raw score x = 33. We can see that this value is above the mean (positive).
![\\ z = \frac{33 - 25}{4}](https://tex.z-dn.net/?f=%20%5C%5C%20z%20%3D%20%5Cfrac%7B33%20-%2025%7D%7B4%7D)
![\\ z = \frac{8}{4}](https://tex.z-dn.net/?f=%20%5C%5C%20z%20%3D%20%5Cfrac%7B8%7D%7B4%7D)
![\\ z = 2](https://tex.z-dn.net/?f=%20%5C%5C%20z%20%3D%202)
The raw score is two (2) <em>standard deviations above the mean</em>. The corresponding cumulative probability is (consulting the standard normal table):
Then the cumulative probability for
. We have to remember that this is the standardized value for x = 33, and for this normal distribution, it has the same cumulative probability [P(x<33) = 0.97725].
<em>Third step: subtract both values and we obtain "the probability of filling a cup between 18 and 33 ounces"</em>, that is:
![\\ P(18 < x < 33) = 0.97725 - 0.04006](https://tex.z-dn.net/?f=%20%5C%5C%20P%2818%20%3C%20x%20%3C%2033%29%20%3D%200.97725%20-%200.04006)
![\\ P(18 < x < 33) = 0.93719](https://tex.z-dn.net/?f=%20%5C%5C%20P%2818%20%3C%20x%20%3C%2033%29%20%3D%200.93719)
Thus, <em>the probability of filling a cup between 18 and 33 ounces </em>is 0.93719.
We can see the area that represents this in the graph below.