Find the Value of a Normal Random Variable
-The amount of soda a dispensing machine pours into a 12 ounce can of soda follows a normal distribution with a standard deviation of 0.12 ounce. Every can that has more than 12.30 ounces of soda poured into it causes a spill and the can needs to go through a special cleaning process before it can be sold. What is the mean amount of soda the machine should dispense if the company wants to limit the percentage that need to be cleaned because of spillage to 3%
A) 12.0744 oz
B) 12.5256 oz
C) 12.0396 oz
D) 12.5604 oz

Question

Quizplus · Accepted Answer

To limit spillage to 3%, the amount of soda that causes a spill (12.30 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 $\mu$. The Z-score corresponding to the 97th percentile is approximately 1.88. Plugging in the values, $1.88 = \frac{12.30 - \mu}{0.12}$, solving for $\mu$ gives $\mu \approx 12.0744$ ounces.