To determine the percentage of snakes that are longer than 16.6 inches, we can follow these steps:
1. Identify the Given Values:
- Mean length of snakes (\(\mu\)): 15 inches
- Standard deviation of lengths (\(\sigma\)): 0.8 inches
- Length threshold to consider: 16.6 inches
2. Calculate the Z-Score:
The Z-score is a measure of how many standard deviations an element is from the mean. It can be calculated using the formula:
[tex]\[
Z = \frac{X - \mu}{\sigma}
\][/tex]
where \(X\) is the value of the threshold length (16.6 inches).
Plugging in the values, we get:
[tex]\[
Z = \frac{16.6 - 15}{0.8} = \frac{1.6}{0.8} = 2
\][/tex]
3. Determine the Cumulative Probability:
The cumulative probability associated with a Z-score tells us the proportion of values below that Z-score in a standard normal distribution. For a Z-score of 2, the cumulative probability is approximately 0.9772 (or 97.72%).
4. Calculate the Percentage Above the Threshold:
To find the percentage of snakes longer than 16.6 inches, we need to consider the proportion of the distribution above this Z-score. Since the total area under the normal distribution curve is 1 (or 100%), the area above the Z-score of 2 is:
[tex]\[
1 - \text{Cumulative Probability} = 1 - 0.9772 = 0.0228
\][/tex]
Converting this to a percentage:
[tex]\[
0.0228 \times 100\% = 2.28\%
\][/tex]
Therefore, the percentage of snakes longer than 16.6 inches is approximately 2.28%. Among the given options, the closest value is:
[tex]\[ \boxed{2.5\%} \][/tex]
