Assume that the height of a type of tree has a normal probability distribution with a mean of 55.2 ft and a standard deviation of 5.7 ft.

A tree of this type grows in my backyard and it stands 39.8 feet tall. Find the probability that the height of a randomly selected tree is as tall as mine or shorter.

My neighbor also has a tree of this type growing in her backyard, but hers stands 68.3 feet tall. Find the probability that the full height of a randomly selected tree is at least as tall as hers.

Enter your answers as decimals accurate to 3 decimal places.

Hmm, let me perform some tree-mendous calculations for you!

To find the probability that a randomly selected tree is as tall as yours or shorter, we need to calculate the area under the normal distribution curve up to your tree height.

We can use the Z-score formula to standardize the value first:
Z = (39.8 - 55.2) / 5.7

Calculating this value gives us:
Z ≈ -2.692

Now, we need to find the probability corresponding to this Z-score using a Z-table or calculator. The table shows that the area to the left of -2.692 is 0.0035.

So, the probability that a randomly selected tree is as tall as yours or shorter is approximately 0.0035.

Now, let's leaf through to the second question. To find the probability that a randomly selected tree is at least as tall as your neighbor's tree, we need to calculate the area under the normal distribution curve starting from her tree height.

Let's standardize her tree's height using the Z-score formula:
Z = (68.3 - 55.2) / 5.7

This calculation gives us:
Z ≈ 2.298

Again, we'll refer to a Z-table or use a calculator to find the probability corresponding to this Z-score. The table shows that the area to the left of 2.298 is 0.9899.

However, we want the probability of being "at least as tall," which means we need to subtract this value from 1. So, the probability that a randomly selected tree is at least as tall as your neighbor's is approximately 1 - 0.9899 = 0.0101.

I hope these answers bring some arboreal amusement!