Question

Let X represent the full height of a certain species of tree. Assume that X has a normal distribution with a mean of 137.1 ft and a standard deviation of 3.2 ft. A tree of this type grows in my backyard, and it stands 132.3 feet tall. Find the probability that the height of a randomly selected tree is as tall as mine or shorter. 0.0668 My neighbor also has a tree of this type growing in her backyard, but hers stands 143.5 feet tall. Find the probability that the full height of a randomly selected tree is at least as tall as hers.

Answer 1

Answer:

0.0668 = 6.68% probability that the height of a randomly selected tree is as tall as mine or shorter.

0.0228 = 2.28% probability that the full height of a randomly selected tree is at least as tall as hers.

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

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

$Z = \frac{X - \mu}{\sigma}$

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:

$\mu = 137.1, \sigma = 3.2$

A tree of this type grows in my backyard, and it stands 132.3 feet tall. Find the probability that the height of a randomly selected tree is as tall as mine or shorter.

This is the pvalue of Z when X = 132.3. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{132.3 - 137.1}{3.2}$

$Z = -1.5$

$Z = -1.5$ has a pvalue of 0.0668

0.0668 = 6.68% 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 143.5 feet tall. Find the probability that the full height of a randomly selected tree is at least as tall as hers.

This is 1 subtracted by the pvalue of Z when X = 143.5. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{143.5 - 137.1}{3.2}$

$Z = 2$

$Z = 2$ has a pvalue of 0.9772

1 - 0.9772 = 0.0228

0.0228 = 2.28% probability that the full height of a randomly selected tree is at least as tall as hers.