Given:

\(\displaystyle{x}=\text{Ounces of cup}={8}\)

\(\displaystyle\sigma=\text{Standard deviation}={0.03}\)

The top 1% has the property that 1% of the data values are larger than the cutoff value and thus and thus \(\displaystyle{100}\%-{1}\%={99}\%\) of the data values are smaller than the cutoff value.

We will use the standard normal table in the appendix to determine the z-score corresponding to an area of 99% or 0.99 or the closest area.

In the standard normal table in the appendix, we note that the value 0.99 lies closest to 0.9901.

The value .9901 is mentioned in the row starting with 2.3 and in the column starting with .03, which corresponds to a z-score of 2.33.

\(\displaystyle{z}={2.33}\)

However, we also know that the z-score is the value x decreased by the mean, divided by the standard deviation.

\(\displaystyle{z}={\frac{{{x}-\mu}}{{\sigma}}}\)

\(\displaystyle\mu={x}-{z}\sigma\) Solve to \(\displaystyle\mu\)

\(\displaystyle={8}-{\left({2.33}\right)}{\left({0.03}\right)}\) Substitute

\(\displaystyle={7.9301}\)

Thus the mean amount of coffee to be dispensed is 7.9301 ounces, which would then imply that the cup of 8 ounces will overflow about 1% of the time.

\(\displaystyle{x}=\text{Ounces of cup}={8}\)

\(\displaystyle\sigma=\text{Standard deviation}={0.03}\)

The top 1% has the property that 1% of the data values are larger than the cutoff value and thus and thus \(\displaystyle{100}\%-{1}\%={99}\%\) of the data values are smaller than the cutoff value.

We will use the standard normal table in the appendix to determine the z-score corresponding to an area of 99% or 0.99 or the closest area.

In the standard normal table in the appendix, we note that the value 0.99 lies closest to 0.9901.

The value .9901 is mentioned in the row starting with 2.3 and in the column starting with .03, which corresponds to a z-score of 2.33.

\(\displaystyle{z}={2.33}\)

However, we also know that the z-score is the value x decreased by the mean, divided by the standard deviation.

\(\displaystyle{z}={\frac{{{x}-\mu}}{{\sigma}}}\)

\(\displaystyle\mu={x}-{z}\sigma\) Solve to \(\displaystyle\mu\)

\(\displaystyle={8}-{\left({2.33}\right)}{\left({0.03}\right)}\) Substitute

\(\displaystyle={7.9301}\)

Thus the mean amount of coffee to be dispensed is 7.9301 ounces, which would then imply that the cup of 8 ounces will overflow about 1% of the time.