https://file.aiquickdraw.com/mathbot/file/1726890011615hj120adb.jpg

The problem is asking to find the percentage of noonday asphalt temperatures either **less than 90°F** or **more than 114°F** given a **normal distribution** with a mean of **102°F** and a standard deviation of **16°F**.

### Steps:
1. **Convert the temperatures to z-scores** using the formula:
   \[
   z = \frac{x - \mu}{\sigma}
   \]
   where:
   - \( x \) is the temperature,
   - \( \mu = 102 \) (mean),
   - \( \sigma = 16 \) (standard deviation).

2. **Calculate the z-score for 90°F:**
   \[
   z = \frac{90 - 102}{16} = \frac{-12}{16} = -0.75
   \]

3. **Calculate the z-score for 114°F:**
   \[
   z = \frac{114 - 102}{16} = \frac{12}{16} = 0.75
   \]

4. **Find the probability using the standard normal distribution:**
   - For **\( z = -0.75 \)**, the cumulative probability from z-tables is approximately **0.2266**.
   - For **\( z = 0.75 \)**, the cumulative probability is approximately **0.7734**.

5. The percentage of temperatures **between 90°F and 114°F** is the difference:
   \[
   P(90 < T < 114) = 0.7734 - 0.2266 = 0.5468
   \]
   or **54.68%**.

6. To find the percentage **outside** this range:
   \[
   P(T < 90 \text{ or } T > 114) = 1 - 0.5468 = 0.4532
   \]
   So, the percentage is **45.32%**.

Therefore, the correct answer should be approximately **45.32%**, not 54.67% as shown.

Would you like more details or further clarifications on this process?

### Related Questions:
1. How can I better interpret z-scores in different contexts?
2. How does changing the mean or standard deviation affect the probabilities?
3. What is the significance of cumulative probability in normal distributions?
4. How do I use the standard normal table effectively?
5. What if the temperatures were modeled using a different distribution?

**Tip**: Always double-check the z-score table for accuracy when calculating probabilities.

Find the percentage of noonday temperatures of the asphalt less than 90°F or more than 114°F, given a normal distribution with a mean of 102°F and a standard deviation of 16°F.

Learn how to calculate the percentage of noonday asphalt temperatures either less than 90°F or more than 114°F using a normal distribution with a mean of 102°F and a standard deviation of 16°F. This guide explains step-by-step how to use Z-scores and standard normal distribution tables.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation