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) 


Given the problem, we are working with a normal distribution for the heights of adult American men. Let's break down the steps to solve each part of the question:

### Part (a): What percent of adult American males could enter this contest?

We are asked to find the percentage of men who are less than 64 inches tall. The height distribution is normal with:
- **Mean (μ)** = 69 inches
- **Standard deviation (σ)** = 2.5 inches

We want to find the proportion of the population with heights below 64 inches.

First, we calculate the **z-score** for 64 inches using the formula:

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

where \( x \) is the value of interest (64 inches), \( \mu \) is the mean (69 inches), and \( \sigma \) is the standard deviation (2.5 inches).

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

Using the z-table (or a normal distribution calculator), we look up the value corresponding to \( z = -2 \). The area to the left of \( z = -2 \) represents the percentage of men who are shorter than 64 inches.

From the z-table:
\[
P(z < -2) \approx 0.0228
\]

So, approximately **2.28%** of adult American men are shorter than 64 inches and would qualify for the limbo contest.

### Part (b): How tall would a man have to be for his height to be in the top 15% of this distribution?

We are looking for the height that corresponds to the **85th percentile** (since the top 15% means 100% - 15% = 85%).

To find the corresponding z-score for the 85th percentile, we use a z-table or a normal distribution calculator, which gives us:
\[
P(z < z_{85}) = 0.85 \quad \text{(z-score for 85th percentile)} \quad z_{85} \approx 1.036
\]

Now, we convert this z-score back to a height using the formula:

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

Substitute \( z = 1.036 \), \( \mu = 69 \) inches, and \( \sigma = 2.5 \) inches:

\[
x = 1.036 \cdot 2.5 + 69 = 2.59 + 69 = 71.59 \text{ inches}
\]

So, a man would have to be approximately **71.59 inches tall** to be in the top 15% of adult American men.

---

### Summary:
a. **2.28%** of adult American men are less than 64 inches tall.
b. A man needs to be **about 71.59 inches tall** to be in the top 15% of adult American men.

Would you like more details or have any questions about the solution? Here are 5 related questions to deepen your understanding:

1. What is the z-score for a man who is 67 inches tall?
2. If the contest height limit was increased to 66 inches, what percentage of men could participate?
3. How would the percentage change if the standard deviation was larger, say 3 inches?
4. What height corresponds to the top 5% of this distribution?
5. What is the probability that a randomly selected man is taller than 72 inches?

**Tip**: When working with normal distributions, z-scores help translate raw values into positions relative to the mean, making it easy to calculate probabilities.

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).

Explore how to calculate the percentage of adult American men eligible for a limbo contest at a Caribbean resort based on height (less than 64 inches). Learn how to find the height required to be in the top 15% of this normal distribution with a mean of 69 inches and a standard deviation of 2.5 inches. Suitable for high school students learning normal distribution and z-scores.

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