Using the normal distribution, it is found that 2.5% of the snakes are longer than 16.6 in.
Normal Probability Distribution
In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- It measures how many standard deviations the measure is from the mean.
- After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
In this problem, the mean and the standard deviation are respectively, given by [tex]\mu = 15, \sigma = 0.8[/tex].
The proportion of the snakes are longer than 16.6 in is 1 subtracted by the p-value of Z when X = 16.6, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{16.6 - 15}{0.8}[/tex]
[tex]Z = 2[/tex]
[tex]Z = 2[/tex] has a p-value of 0.975.
1 - 0.975 = 0.025 = 2.5% of the snakes are longer than 16.6 in.
More can be learned about the normal distribution at https://brainly.com/question/24663213
