Question

A soft drink machine outputs a mean of 25 ounces per cup. The machine's output is normally distributed with a standard deviation of 4 ounces. What is the probability of filling a cup between 18 and 33 ounces

Answer 1

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 normally distributed.

The normal distribution is defined by two parameters, namely, the mean and the standard deviation.  

The population mean for this normal distribution is $\\ \mu = 25$ ounces per cup, and a population standard deviation of $\\ \sigma = 4$ ounces (per cup).

To solve this question, we can use the standard normal distribution (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 z-scores are values that tell us the distance from the mean in standard deviations units. For positive values of it, the value is above the mean, and negative values tell us that the value is below the mean.

We can obtain the z-scores using the next formula:

$\\ z = \frac{x - \mu}{\sigma}$ [1]

Where

z is the z-score.

x is the raw score.

$\\ \mu$ is the population mean.

$\\ \sigma$ 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:

First Step: obtaining the z-score for 18 to find the corresponding probability of this value in this normal distribution.

Using the formula [1], we have:

$\\ z = \frac{18 - 25}{4}$

$\\ z = \frac{-7}{4}$

$\\ z = -1.75$

This tells us that the value is below the mean and is -1.75 standard units from it.

Most of the standard normal tables are made for positive values of z. However, since the normal distribution is symmetrical around the mean, we, fortunately, can obtain the corresponding probability considering that:

$\\ P(z$

Then, consulting the cumulative standard normal table for z = 1.75, we have that the corresponding cumulative probability is $\\ P(z$.

Thus

$\\ P(z1.75)$

$\\ P(z1.75)$

Then, the probability is $\\ P(z$

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].

Second Step: obtaining the z-score for 33 to find the corresponding probability of this value in this normal distribution.

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).

$\\ z = \frac{33 - 25}{4}$

$\\ z = \frac{8}{4}$

$\\ z = 2$

The raw score is two (2) standard deviations above the mean. The corresponding cumulative probability is (consulting the standard normal table):

$\\ P(z$

Then the cumulative probability for $\\ P(z$. 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].

Third step: subtract both values and we obtain "the probability of filling a cup between 18 and 33 ounces", that is:

$\\ P(18 < x < 33) = 0.97725 - 0.04006$

$\\ P(18 < x < 33) = 0.93719$

Thus, the probability of filling a cup between 18 and 33 ounces is 0.93719.

We can see the area that represents this in the graph below.