Tree heights: Cherry trees in a certain orchard have heights that are normally distributed with mean μ = 112 inches and standard deviation σ = 10 inches. Use the TI-84 PLUS calculator to answer the following. Round the answers to at least two decimals. (a) Find the 22nd percentile of the tree heights. (b) Find the 80th percentile of the tree heights. (c) Find the first quartile of the tree heights. (d) An agricultural scientist wants to study the tallest 1% of the trees to determine whether they have a certain gene that allows them to grow taller. To do this, she needs to study all the trees above a certain height. What height is this? Part: 0 / 40 of 4 Parts Complete Part 1 of 4 Find the 22nd percentile of the tree heights. The 22nd percentile of tree heights is inches.

Explained below.

Step-by-step explanation:

Let X represents the tree heights.

It is provided that X follows a normal distribution with mean μ = 112 inches and standard deviation σ = 10 inches.

(a)

Compute the 22nd percentile of the tree heights as follows:

The z-score for the corresponding probability is, z = -0.77.

Compute the value of x as follows:

Thus, the 22nd percentile of the tree heights is 104.3 inches.

(b)

Compute the 80th percentile of the tree heights as follows:

The z-score for the corresponding probability is, z = 0.84.

Compute the value of x as follows:

Thus, the 80th percentile of the tree heights is 120.4 inches.

(c)

Compute the first quartile of the tree heights as follows:

The z-score for the corresponding probability is, z = -0.67.

Compute the value of x as follows:

Thus, the first quartile of the tree heights is 113.3 inches.

(d)

The tallest 1% of the trees, i.e. P (X > x) = 0.01.

That is, P (X < x) = 0.99.

⇒ P (Z < z) = 0.99

The corresponding z-value is, z = 2.33.

Compute the value of x as follows:

Thus, the tallest 1% of the trees are more than 121.9 inches tall.

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