Cherry trees in a certain orchard have heights that are normally distributed with mean u = 113 inches and standard deviation o = 11 inches. Round the answers to at least four decimal places.

Question

(a) What proportion of trees are more than 131 inches tall?

(b) What proportion of trees are less than 98 inches tall?

(c) What is the probability that a randomly chosen tree is between 90 and 105 inches tall?

Answer 1

To solve these problems, we will use the properties of the normal distribution and the z-score formula.

The z-score is calculated using the formula: \[ z = \frac{(X - \mu)}{\sigma} \] where:

• \(X\) is the value,
• \(\mu\) is the mean (113 inches),
• \(\sigma\) is the standard deviation (11 inches).

We will also use the standard normal distribution table (z-table) or a calculator to find the proportions.

(a) Proportion of trees more than 131 inches tall

1. Calculate the z-score for \(X = 131\): \[ z = \frac{(131 - 113)}{11} = \frac{18}{11} \approx 1.6364 \]
2. Find the area to the right of \(z = 1.6364\):
   • Using a z-table or calculator, the proportion of trees less than 131 inches tall (P(Z < 1.6364)) is approximately \(0.9495\).
   • Therefore, the proportion of trees more than 131 inches tall is: \[ P(Z > 1.6364) = 1 - P(Z < 1.6364) = 1 - 0.9495 = 0.0505 \]

(b) Proportion of trees less than 98 inches tall

1. Calculate the z-score for \(X = 98\): \[ z = \frac{(98 - 113)}{11} = \frac{-15}{11} \approx -1.3636 \]
2. Find the area to the left of \(z = -1.3636\):
   • Using a z-table or calculator, the proportion of trees less than 98 inches tall (P(Z < -1.3636)) is approximately \(0.0869\).

(c) Probability that a randomly chosen tree is between 90 and 105 inches tall

1. Calculate the z-score for \(X = 90\): \[ z_{90} = \frac{(90 - 113)}{11} = \frac{-23}{11} \approx -2.0909 \]
2. Calculate the z-score for \(X = 105\): \[ z_{105} = \frac{(105 - 113)}{11} = \frac{-8}{11} \approx -0.7273 \]
3. Find the area between these z-scores:
   • For \(z_{90}\), the area (P(Z < -2.0909)) is approximately \(0.0183\).
   • For \(z_{105}\), the area (P(Z < -0.7273)) is approximately \(0.2333\).
4. The probability that a tree is between 90 and 105 inches tall: \[ P(90 < X < 105) = P(Z < -0.7273) - P(Z < -2.0909) = 0.2333 - 0.0183 = 0.2150 \]

Summary of Answers

(a) Proportion of trees more than 131 inches tall: 0.0505

(b) Proportion of trees less than 98 inches tall: 0.0869

(c) Probability that a randomly chosen tree is between 90 and 105 inches tall: 0.2150