a vending machine dispenses coffee into a sixteen - ounce cup. the amount of coffee dispensed into the cup…

a vending machine dispenses coffee into a sixteen - ounce cup. the amount of coffee dispensed into the cup is normally distributed with a standard deviation of 0.07 ounce. you can allow the cup to overfill 2% of the time. what amount should you set as the mean amount of coffee to be dispensed? click to view page 1 of the table. click to view page 2 of the table. ounces (round to two decimal places as needed.)

a vending machine dispenses coffee into a sixteen - ounce cup. the amount of coffee dispensed into the cup is normally distributed with a standard deviation of 0.07 ounce. you can allow the cup to overfill 2% of the time. what amount should you set as the mean amount of coffee to be dispensed? click to view page 1 of the table. click to view page 2 of the table. ounces (round to two decimal places as needed.)

Answer

Explanation:

Step1: Find the z - score

We know that we allow over - filling 2% of the time. So the area to the right of the z - score is 0.02. The area to the left of the z - score is $1 - 0.02=0.98$. Looking up in the standard normal distribution table, the z - score corresponding to an area of 0.98 is approximately $z = 2.05$.

Step2: Use the z - score formula

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $x = 16$ (the capacity of the cup), $\sigma=0.07$ (standard deviation), and $\mu$ is the mean we want to find. We can re - arrange the formula to solve for $\mu$: $\mu=x - z\sigma$.

Step3: Substitute values

Substitute $x = 16$, $z = 2.05$, and $\sigma = 0.07$ into the formula: $\mu=16-2.05\times0.07$. $\mu=16 - 0.1435$. $\mu = 15.86$ (rounded to two decimal places).

Answer:

$15.86$