a vending machine dispenses coffee into an 8 oz cup. the amount of coffee dispensed into the cup is normally distributed with a standard deviation of 0.03 oz.you can allow the cup to overfill 5 percent of the time. what amont should you set as the mean amount of coffee to be dispensed?

Use z-scores and solve for the mean:

z = (8 - mean)/(0.03)

To find z to plug into the above formula, use a z-table and the percentage given in the problem.

ryahwryhar666

Jsns

To determine the mean amount of coffee to be dispensed, we need to find the value that corresponds to the 95th percentile of the normal distribution. This is the value at which the cup will overfill 5 percent of the time.

To find this value, we can use the z-score formula:

z = (X - μ) / σ

Where:
- X is the value we are trying to find (the mean amount of coffee)
- μ is the mean of the distribution
- σ is the standard deviation of the distribution

We need to find the z-score corresponding to the 95th percentile. Since the normal distribution is symmetric, we can use a standard normal distribution table (also known as a z-table) to find the value of the z-score.

The z-score corresponding to the 95th percentile is approximately 1.645. This means that 95% of the data falls below this value.

Now we can solve for X:

1.645 = (X - μ) / 0.03

Multiply both sides of the equation by 0.03:

0.04935 = X - μ

Add μ to both sides:

X = 0.04935 + μ

Therefore, to allow the cup to overfill 5 percent of the time, you should set the mean amount of coffee to be dispensed to approximately 0.04935 oz.