The length of fully grown male alligators has a normal distribution. Alligators that are one standard deviation less than the mean are 3.8 m in length. Alligators that are on standard deviation longer than the mean are 4.2m in length. What is the probability that an alligator will be longer than 4.0m in length?

Answer 1

To solve this problem, we need to calculate the z-scores for the lengths of 3.8m, 4.0m, and 4.2m using the mean and standard deviation provided.

Let x be the mean length of fully grown male alligators.
Given:
x - 3.8 = 1σ
x + 4.2 = 1σ

Adding the two equations:
x - 3.8 + x + 4.2 = 2σ
2x + 0.4 = 2σ
2x = 2σ - 0.4
x = σ - 0.2

Now, we substitute the mean back into the original equations:
σ - 0.2 - 3.8 = 1σ
σ - 4 = 1σ
-4 = 0σ
σ = 4

So, the mean (μ) = 4, and the standard deviation (σ) = 4.

The z-score for 4.0m is:
z = (4 - 4) / 4 = 0

Using a standard normal distribution table, we find that the probability of an alligator being longer than 4.0m in length is approximately 50%.

Therefore, the probability that an alligator will be longer than 4.0m in length is 0.5 or 50%.