The lengths of a particular snake are approximately normally distributed with a given mean mu is equal to 15 in. and standard deviation sigma is equal to 0 point 8 in. What percentage of the snakes are longer than 16 point 6 in.?

Convert 16.6 inches to a z-score: (16.6 − 15) ÷ 0.8 = 2. A length of 16.6 inches is therefore 2 standard deviations above the mean. In a normal distribution, about 2.3% lie above that point, which is closest to choice A, 2.5 percent.

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