Problem 3: A survey found that women's heights are normally distributed with mean 63.6 in. and standard deviation 3.1 in. The survey also found that men's heights are normally distributed with mean 68.2 in . and standard deviation 3.8 in. Most of the live characters employed at an amusement park have height requirements of a minimum of 55 in. and a maximum of 63 in. Complete parts (a) and (b) below. a) Find the percentage of men meeting the height requirement. What does the result suggest about the genders of the people who are employed as characters at the amusement park? b) If the height requirements are changed to exclude only the tallest $$ 50 \% $$ of men and the shortest $$ 5 \% $$ of men, what are the new height requirements?

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**(a)**  
Let $$ X $$ denote a man’s height, where  
$$
X \sim N(68.2,\,3.8^2)
$$

The current height requirement is $$55 \leq X \leq 63$$ inches. Compute the corresponding $$z$$-scores:

$$
z_{55} = \frac{55 - 68.2}{3.8} \approx -3.47,
$$
$$
z_{63} = \frac{63 - 68.2}{3.8} \approx -1.37.
$$

The probability that a man’s height falls within this range is

$$
P(55 \leq X \leq 63) = \Phi(-1.37) - \Phi(-3.47).
$$

Using the standard normal distribution table or a calculator, we approximate

$$
\Phi(-1.37) \approx 0.0853 \quad \text{and} \quad \Phi(-3.47) \approx 0.0003.
$$

Thus,

$$
P(55 \leq X \leq 63) \approx 0.0853 - 0.0003 \approx 0.085.
$$

This means that about $$8.5\%$$ of men meet the height requirement. Since only a small fraction of men qualify under these conditions, it suggests that most of the live characters employed at the amusement park (meeting these specific height restrictions) are likely to be women.

---

**(b)**  
Now, suppose the height requirements are changed to exclude only the tallest $$50\%$$ of men and the shortest $$5\%$$ of men. This means the acceptable range is between the $$5^\text{th}$$ percentile and the $$50^\text{th}$$ percentile (the median) of the men's height distribution.

1. **Lower bound (5th percentile):**  
   For the 5th percentile, the corresponding $$z$$-score is approximately $$-1.645$$. Using the formula for conversion, we have

$$
   x = \mu + z \sigma = 68.2 + (-1.645)(3.8).
   $$

Calculate the product:

$$
   1.645 \times 3.8 \approx 6.251.
   $$

Thus, the lower bound is

$$
   x \approx 68.2 - 6.251 \approx 61.95 \text{ inches}.
   $$

2. **Upper bound (50th percentile):**  
   The 50th percentile (median) for a normal distribution is the mean. Hence, the upper bound is

$$
   x = 68.2 \text{ inches}.
   $$

Therefore, the new height requirements for men are approximately

$$
61.95 \text{ inches} \leq X \leq 68.2 \text{ inches}.
$$
**(a)** About $$8.5\%$$ of men meet the height requirement. This suggests that most live characters employed as amusement park characters are likely women. **(b)** The new height requirements for men are approximately $$61.95$$ inches to $$68.2$$ inches.

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