A Caribbean resort has a nightly limbo contest on the beach. Participants must be less than 64 inches tall. The distribution of heights of adult American men is approximately normal with mean 69 inches and standard deviation 2.5 inches.  b. How tall would a man have to be for his height to be in the top 15% of this distribution of adult American men? (Express in inches)

To determine the height that places a man in the top 15% of the distribution of adult American men, we need to find the corresponding z-score for the 85th percentile (since the top 15% corresponds to the bottom 85%).

Here’s the process:

### Step 1: Find the z-score for the 85th percentile
We can look up or use a calculator to find the z-score corresponding to 0.85 in a standard normal distribution. The z-score for 85% is approximately **1.036**.

### Step 2: Use the z-score formula
We can now use the z-score formula to convert this into a height:

\[
z = \frac{x - \mu}{\sigma}
\]

Where:
- \( z = 1.036 \) (z-score for the 85th percentile),
- \( \mu = 69 \) (mean height of adult American men),
- \( \sigma = 2.5 \) (standard deviation of the height).

Rearrange the formula to solve for \( x \) (the height):

\[
x = z \cdot \sigma + \mu
\]

Substituting the values:

\[
x = 1.036 \cdot 2.5 + 69
\]

\[
x = 2.59 + 69 = 71.59 \, \text{inches}
\]

### Conclusion:
A man would have to be approximately **71.6 inches tall** (rounded to one decimal place) to be in the top 15% of the height distribution.

Would you like more details on any part of this? 

Here are 5 related questions:
1. What is the probability of a man being shorter than 64 inches in this distribution?
2. How would the height cutoff change if the standard deviation were larger?
3. How tall would a man have to be to be in the top 10%?
4. What height corresponds to the bottom 5% of this distribution?
5. How would this calculation change for women, given a different mean and standard deviation?

**Tip:** Percentile ranks can always be converted into z-scores when dealing with a normal distribution!

A Caribbean resort has a nightly limbo contest on the beach. Participants must be less than 64 inches tall. The distribution of heights of adult American men is approximately normal with mean 69 inches and standard deviation 2.5 inches. How tall would a man have to be for his height to be in the top 15% of this distribution of adult American men? (Express in inches)

Learn how to determine the height that places a man in the top 15% of the distribution of adult American men using normal distribution and z-score calculation. This problem involves key mathematical concepts, including standard deviation and percentile ranking, suitable for students in Grades 9-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation