The lengths of a particular snake are approximately normally distributed with a given mean Mu = 15 in. and standard deviation Sigma = 0.8 in. What percentage of the snakes are longer than 16.6 in.?
0.3%
2.5%
3.5%
5%
To find the percentage of snakes longer than 16.6 inches, we can use the standard normal distribution.
First, we need to calculate the z-score for 16.6 inches using the formula:
z = (x - μ) / σ
where x is the value we are interested in (16.6 inches), μ is the mean (15 inches), and σ is the standard deviation (0.8 inches).
z = (16.6 - 15) / 0.8
z = 1.8 / 0.8
z = 2.25
Next, we can use a z-table or a calculator to find the percentage of values greater than 2.25.
From the z-table, we find that the percentage is approximately 1 - 0.9878 = 0.0122, which is approximately 1.22%.
Therefore, the percentage of snakes longer than 16.6 inches is approximately 1.22%.
None of the given options (0.3%, 2.5%, 3.5%, 5%) match this value, so none of the options are correct.