To determine the percentage of snakes that are longer than 16.6 inches, we can follow these detailed steps:
1. Understand the Parameters of the Distribution:
The lengths of the snakes are normally distributed with a mean \(\mu = 15\) inches and a standard deviation \(\sigma = 0.8\) inches.
2. Calculate the Z-Score:
The Z-score helps us determine how many standard deviations a particular value (in this case, 16.6 inches) is from the mean.
The formula for the Z-score is:
[tex]\[
Z = \frac{X - \mu}{\sigma}
\][/tex]
where \(X\) is the value we are interested in (16.6 inches).
Plugging in the values:
[tex]\[
Z = \frac{16.6 - 15}{0.8} = \frac{1.6}{0.8} = 2
\][/tex]
3. Determine the Cumulative Probability:
The Z-score tells us the number of standard deviations above the mean the value lies. We use the standard normal distribution to find the cumulative probability up to this Z-score.
The cumulative probability for \(Z = 2\) from standard normal distribution tables or software is approximately 0.9772. This represents the probability that a snake is shorter than or equal to 16.6 inches.
4. Calculate the Complement to Find the Required Probability:
We are interested in the percentage of snakes that are longer than 16.6 inches. This is the complement of the cumulative probability we found.
[tex]\[
P(X > 16.6) = 1 - P(X \leq 16.6)
\][/tex]
So, substituting in the cumulative probability:
[tex]\[
P(X > 16.6) = 1 - 0.9772 = 0.0228
\][/tex]
5. Convert the Probability to a Percentage:
To convert this probability to a percentage, we multiply by 100:
[tex]\[
0.0228 \times 100 = 2.28\%
\][/tex]
Therefore, approximately 2.28% of the snakes are longer than 16.6 inches. The closest answer choice to this percentage is 2.5%. Hence, the correct answer is:
[tex]\[
\boxed{2.5\%}
\][/tex]
