Question

Alligators captured in Florida are found to have a mean length of 2 meters and a standard deviation of 0.35 meters. The lengths of alligators are believed to be approximately normally distributed. What percent of alligators have lengths greater than 2.2 meters?

Answer 1

28.43% of alligators have lengths greater than 2.2 meters

Explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

`\mu = 2, \sigma = 0.35`

What percent of alligators have lengths greater than 2.2 meters?

This is 1 subtracted by the pvalue of Z when X = 2.2. So

`Z = (X - \mu)/(\sigma)`

`Z = (2.2 - 2)/(0.35)`

`Z = 0.57`

`Z = 0.57` has a pvalue of 0.7157

1 - 0.7157 = 0.2843

28.43% of alligators have lengths greater than 2.2 meters