Question

A company that manufactures and bottles apple juice uses a machine that automatically fills 32-ounce bottles. There is some variation, however, in the amount of liquid dispensed into the bottles. The amount dispensed has been observed to be approximately normally distributed with mean 32 ounces and standard deviation 1 ounce. Determine the proportion of bottles that will have more than 30 ounces dispensed into them. (Round your answer to four decimal places.)

Answer 1

The proportion of bottles that will have more than 30 ounces dispensed into them is 0.9772 = 97.72%.

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 = (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.

The amount dispensed has been observed to be approximately normally distributed with mean 32 ounces and standard deviation 1 ounce.

This means that
`\mu = 32, \sigma = 1`

Determine the proportion of bottles that will have more than 30 ounces dispensed into them.

This is 1 subtracted by the p-value of Z when X = 30, so:

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

`Z = (30 - 32)/(1)`

`Z = -2`

`Z = -2` has a p-value of 0.0228.

1 - 0.0228 = 0.9772

The proportion of bottles that will have more than 30 ounces dispensed into them is 0.9772 = 97.72%.