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.
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$\boxed{2.5\%}$ of the bottles will be filled with more than 17 ounces.

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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.

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