Cherry trees in a certain orchard have heights that are normally distributed with mean U = 112 inches and standard deviation O = 10 inches.

To solve these problems, we can use the properties of the normal distribution along with the z-score formula. The z-score formula is given by:

\[ z = \frac{(X - \mu)}{\sigma} \]

where \(X\) is the value of interest, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.

Given:

• Mean (\(\mu\)) = 112 inches
• Standard deviation (\(\sigma\)) = 10 inches

We will use the standard normal distribution table (z-table) or a calculator to find the z-scores corresponding to the percentiles.

(a) Find the 22nd percentile of the tree heights.

The z-score corresponding to the 22nd percentile (0.22) can be found using a z-table or calculator, which gives a z-score of approximately -0.77.

Now we use the z-score formula to find \(X\):

\[ -0.77 = \frac{(X - 112)}{10} \]

Solving for \(X\):

\[ X - 112 = -0.77 \times 10 \] \[ X - 112 = -7.7 \] \[ X = 112 - 7.7 = 104.3 \text{ inches} \]

(b) Find the 80th percentile of the tree heights.

The z-score corresponding to the 80th percentile (0.80) is approximately 0.84.

Now we use the z-score formula to find \(X\):

\[ 0.84 = \frac{(X - 112)}{10} \]

Solving for \(X\):

\[ X - 112 = 0.84 \times 10 \] \[ X - 112 = 8.4 \] \[ X = 112 + 8.4 = 120.4 \text{ inches} \]

(c) Find the first quartile of the tree heights.

The first quartile (Q1) corresponds to the 25th percentile (0.25). The z-score for the 25th percentile is approximately -0.67.

Now we use the z-score formula to find \(X\):

\[ -0.67 = \frac{(X - 112)}{10} \]

Solving for \(X\):

\[ X - 112 = -0.67 \times 10 \] \[ X - 112 = -6.7 \] \[ X = 112 - 6.7 = 105.3 \text{ inches} \]

(d) Height of the tallest 1% of the trees.

The tallest 1% corresponds to the 99th percentile (0.99). The z-score for the 99th percentile is approximately 2.33.

Now we use the z-score formula to find \(X\):

\[ 2.33 = \frac{(X - 112)}{10} \]

Solving for \(X\):

\[ X - 112 = 2.33 \times 10 \] \[ X - 112 = 23.3 \] \[ X = 112 + 23.3 = 135.3 \text{ inches} \]

Summary of results:

(a) 22nd percentile: 104.3 inches
(b) 80th percentile: 120.4 inches
(c) First quartile: 105.3 inches
(d) Height for the tallest 1%: 135.3 inches