The amount of soda a dispensing machine pours into a 12 ounce can of soda follows a normal distribution witha standard deviation of 0.06 ounce. Every can that has more than 12.15 ounces of soda poured into it causes aspill and the can needs to go through a special cleaning process before it can be sold. What is the mean amountof soda the machine should dispense if the company wants to limit the percentage that need to be cleanedbecause of spillage to 3%?
A) 12.0372 oz
B) 12.2628 oz
C) 12.0198 oz
D) 12.2802 oz

Question

Quizplus · Accepted Answer

To limit the spillage to 3%, the amount of soda that causes a spill (12.15 ounces) should correspond to the 97th percentile of the normal distribution (since 100% - 3% = 97%). Using the Z-score formula $Z = \frac{X - \mu}{\sigma}$, where $X$ is the value from the distribution, $\mu$ is the mean, and $\sigma$ is the standard deviation, we can solve for the mean $\mu$. Given that the standard deviation $\sigma = 0.06$ ounces and the Z-score corresponding to the 97th percentile is approximately 1.88 (from Z-tables), we can set up the equation $1.88 = \frac{12.15 - \mu}{0.06}$. Solving for $\mu$ gives us approximately 12.0372 ounces, meaning the machine should dispense an average of 12.0372 ounces to limit the spillage to 3%.

The Amount of Soda a Dispensing Machine Pours into a 12