The lengths of a particular snake are approximately normally distributed with a given mean = 15 in. and standard deviation = 0.8

in. What percentage of the snakes are longer than 16.6 in.?

a. 0.3%

b. 2.5%

c. 3.5%

d. 5%

Let x be a random variable representing the length of the snakes, then

P (x > 16.6) = 1 - P (x < 16.6) = 1 - P[z < (16.6 - 15) / 0.8] = 1 - P (z < 2) = 1 - 0.97725 = 0.02275

Therefore, the percentage of snakes longer than 16.6 in. is 0.02275 * 100 = 2.275%