Question

A soft drink-dispensing machine uses paper cups that hold a maximum of 12 ounces. The machine is set to dispense a mean of 10 ounces of the drink. Because of an imprecise process, the amount dispensed actually varies; it is normally distributed with a standard deviation of 1 ounce. If we simulated the filling process, and had the random number .564, how many ounces would the cup contain? Give your answer to 3 decimal places.

Answer 1

The cup would contain 10.16 ounces.

Explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (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 pvalue, we get the probability that the value of the measure is greater than X.

Mean of 10 ounces, standard deviation of 1 ounce:

This means that
`\mu = 10, \sigma = 1`

If we simulated the filling process, and had the random number .564, how many ounces would the cup contain?

This means that we have to find X when Z has a pvalue of 0.564. So X when Z = 0.16. So

`Z = (X - \mu)/(\sigma)`

`0.16 = (X - 10)/(1)`

`X - 10 = 0.16*1`

`X = 10.16`

The cup would contain 10.16 ounces.