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.
a. What percent of adult American males could enter this contest? 
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) 


We are working with a normal distribution where the mean height (\(\mu\)) of adult American males is 69 inches, and the standard deviation (\(\sigma\)) is 2.5 inches.

### Part a: What percent of adult American males could enter the contest?
The height requirement is less than 64 inches. To find the percentage of adult males shorter than 64 inches, we need to calculate the z-score.

The z-score is given by the formula:

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

Where:
- \(X = 64\) (the height threshold),
- \(\mu = 69\) (mean height),
- \(\sigma = 2.5\) (standard deviation).

Let's calculate the z-score:

\[
z = \frac{64 - 69}{2.5} = \frac{-5}{2.5} = -2
\]

Using the standard normal distribution table, a z-score of -2 corresponds to a percentile of approximately **2.28%**. Therefore, about 2.28% of adult American males could enter the contest.

### Part b: What height would place a man in the top 15% of this distribution?
We need to find the height corresponding to the top 15% of the distribution. This is equivalent to finding the height at the 85th percentile, since the top 15% corresponds to 100% - 15% = 85%.

From the z-score table, the z-score corresponding to the 85th percentile is approximately **1.04**.

Now, we use the z-score formula in reverse to find the height:

\[
X = \mu + z \times \sigma
\]

Substituting the values:

\[
X = 69 + 1.04 \times 2.5 = 69 + 2.6 = 71.6 \, \text{inches}
\]

So, a man would need to be approximately **71.6 inches** tall to be in the top 15% of the height distribution for adult American men.

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Do you want any details or further clarification on this solution? 

Here are five related questions:
1. How would the percentage change if the standard deviation were larger?
2. What height corresponds to the top 10% of the distribution?
3. What is the probability that a randomly selected man is taller than 72 inches?
4. What height would place a man in the bottom 5% of this distribution?
5. How would the percentage of eligible participants change if the height limit was raised to 66 inches?

**Tip**: Understanding z-scores is essential for working with normal distributions, as they allow you to standardize different values and compare them directly across distributions.

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. a. What percent of adult American males could enter this contest? 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)

Solve a normal distribution problem related to heights of adult American men with a mean of 69 inches and a standard deviation of 2.5 inches. Learn how to calculate the percentage of men eligible for a limbo contest and find the height in the top 15%. Perfect for students in Grades 10-12.

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