Question

A soft drink machine is regulated so that it discharges an average of 8 ounces per cup. If the amountof drink per cup is normally distributed with a standard deviation of 0.5 ounce, find the probability acup contains less than 8.8 ounces.

Answer 1

To determine the probability that a cup contains less than 8.8 ounces, we calculate the z-score and use the standard normal distribution, resulting in approximately a 94.52% chance.

Step-by-step explanation:

To find the probability that a cup contains less than 8.8 ounces, when the machine discharges an average of 8 ounces with a standard deviation of 0.5 ounce, we will use the properties of the normal distribution. First, we calculate the z-score for 8.8 ounces using the formula:

Z = (X - μ) / σ

where X is the value in question (8.8 ounces), μ is the mean (8 ounces), and σ is the standard deviation (0.5 ounces). Plugging the values we get:

Z = (8.8 - 8) / 0.5

Z = 1.6

Next, we look up the z-score of 1.6 in a standard normal distribution table or use a calculator to find the cumulative probability.

The probability corresponding to a z-score of 1.6 is approximately 0.9452.

This means there is about a 94.52% chance that a cup contains less than 8.8 ounces.

Answer 2

`\begin{gathered} \mu=8 \\ \sigma=0.5 \end{gathered}`
`\begin{gathered} P(Z<(X-\mu)/(\sigma)) \\ P(Z<(8.8-8)/(0.5)) \\ P(Z<1.6) \end{gathered}`

Using the z-table:

`P(X<8.8)\approx0.9452`