Problem
A factory fills bottles of water. The volume of water in the individual bottles is normally distributed with a mean of 16.8 ounces and a standard deviation of 0.1 ounces. Using the 68-95-99. 7\% rule, approximately what percentage of bottles will be filled with more than 17 ounces? Enter the answer in the box. $\%$
Solution
Step 1 ：Calculate the z-score for 17 ounces: \(z = \frac{17 - 16.8}{0.1} = 2\)
Step 2 ：Using the 68-95-99.7% rule, find the percentage of bottles in the upper tail beyond 2 standard deviations: \(100 - 95 = 5\%\) and since it's the upper tail, divide by 2: \(\frac{5}{2} = 2.5\%\)
Step 3 ：\(\boxed{2.5\%}\) of the bottles will be filled with more than 17 ounces.
