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:
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:
Then, consulting the <em>cumulative standard normal table</em> for z = 1.75, we have that the corresponding cumulative probability is .
Thus
Then, the probability is
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).
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:
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.