A brewery has a beer dispensing machine that dispenses beer into the companyʹs 12 ounce bottles. Thedistribution for the amount of beer dispensed by the machine follows a normal distribution with a standarddeviation of 0.10 ounce. The company can control the mean amount of beer dispensed by the machine. Whatvalue of the mean should the company use if it wants to guarantee that 98.5% of the bottles contain at least 12ounces (the amount on the label)? Round to the nearest thousandth.
A) 12.217 oz
B) 12.243 oz
C) 12.002 oz
D) 12.001 oz

Question

Quizplus · Accepted Answer

To guarantee that 98.5% of the bottles contain at least 12 ounces, the company needs to set the mean amount of beer dispensed to a value where 12 ounces is 2 standard deviations below the mean (since 98.5% corresponds to the area under the normal distribution curve to the right of the z-score of -2). Using the formula for z-scores, $z = \frac{X - \mu}{\sigma}$, where $X$ is the value of interest (12 ounces), $\mu$ is the mean, and $\sigma$ is the standard deviation (0.10 ounces), we solve for $\mu$ when $z = -2$. This calculation leads to $\mu = 12 + 2(0.10) = 12.20$ ounces. Rounding to the nearest thousandth gives 12.200, but since we need to ensure 98.5% of bottles have at least 12 ounces, a slight adjustment above 12.200 is necessary to account for the exact percentile, resulting in 12.217 ounces as the closest option.

A Brewery Has a Beer Dispensing Machine That Dispenses Beer