Question

A soft drink machine outputs a mean of 29 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 33 and 35 ounces? Round your answer to four decimal places.

Answer 1

Answer:

0.8919 = 89.19% probability of filling a cup between 33 and 35 ounces.

Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

A soft drink machine outputs a mean of 29 ounces per cup. The machine's output is normally distributed with a standard deviation of 4 ounces.

This means that $\mu = 29, \sigma = 4$

What is the probability of filling a cup between 33 and 35 ounces?

This is the p-value of Z when X = 35 subtracted by the p-value of Z when X = 33.

X = 35

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{35 - 29}{4}$

$Z = 1.5$

$Z = 1.5$ has a p-value of 0.9332.

X = 33

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{33 - 29}{4}$

$Z = 1$

$Z = 1$ has a p-value of 0.0413.

0.9332 - 0.0413 = 0.8919

0.8919 = 89.19% probability of filling a cup between 33 and 35 ounces.