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?
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%.