the heights of a certain type of tree are approximately normally distributed with a mean height $mu = 5$ ft…

Explanation:

Step1: Recall z - score formula

The z - score formula is $z=\frac{x - \mu}{\sigma}$, where $x$ is the value from the data set, $\mu$ is the mean and $\sigma$ is the standard deviation.

Step2: Check option 1

For $x = 5.4$, $z=\frac{5.4 - 5}{0.4}=\frac{0.4}{0.4}=1$. So a tree of height 5.4 ft is 1 standard deviation above the mean, not below.

Step3: Check option 2

For $x = 4.6$, $z=\frac{4.6 - 5}{0.4}=\frac{- 0.4}{0.4}=-1$. So a tree of height 4.6 ft is 1 standard deviation below the mean, not above.

Step4: Check option 3

For $x = 5.8$, $z=\frac{5.8 - 5}{0.4}=\frac{0.8}{0.4}=2$. So a tree of height 5.8 ft is 2 standard deviations above the mean, not 2.5.

Step5: Check option 4

For $x = 6.2$, $z=\frac{6.2 - 5}{0.4}=\frac{1.2}{0.4}=3$. So a tree with a height of 6.2 ft is 3 standard deviations above the mean.

Answer:

A tree with a height of 6.2 ft is 3 standard deviations above the mean.