A survey found that women's heights are normally distributed with mean 63.4 in. and standard deviation 2.4 in. The survey also found that men's heights are normally distributed with a mean 68.1 in. and standard deviation 2.8 . Complete parts a through c below. a. Most of the live characters at an amusement park have height requirements with a minimum of 4 ft 9 in . and a maximum of 6 ft 2 in . Find the percentage of women meeting the height requirement. The percentage of women who meet the height requirement is $$ 99.62 \% $$. (Round to two decimal places as needed ) b. Find the percentage of men meeting the height requirement. The percentage of men who meet the height requirement is $$ 98.26 \% $$. (Round to two decimal places as needed) c. If the height requirements are changed to exclude only the tallest $$ 5 \% $$ of men and the shortest $$ 5 \% $$ of women, what are the new height requirements? The new height requirements are at least $$ \square $$ in and at most $$ \square $$ in]. (Round to one decimal place as needed)

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To solve part c, we need to find the heights that correspond to the 5th percentile and the 95th percentile for both women and men, given their respective normal distributions.

### For Women:
- Mean ($$ \mu $$) = 63.4 inches
- Standard Deviation ($$ \sigma $$) = 2.4 inches

**Finding the 5th percentile:**
To find the height corresponding to the 5th percentile, we can use the z-score formula:
$$
z = \frac{X - \mu}{\sigma}
$$
For the 5th percentile, the z-score is approximately -1.645 (from z-tables).

Now we can rearrange the formula to solve for $$ X $$:
$$
X = \mu + z \cdot \sigma
$$
$$
X = 63.4 + (-1.645) \cdot 2.4
$$
$$
X \approx 63.4 - 3.948 = 59.452 \text{ inches}
$$

**Finding the 95th percentile:**
For the 95th percentile, the z-score is approximately 1.645.

Using the same formula:
$$
X = 63.4 + (1.645) \cdot 2.4
$$
$$
X \approx 63.4 + 3.948 = 67.348 \text{ inches}
$$

### For Men:
- Mean ($$ \mu $$) = 68.1 inches
- Standard Deviation ($$ \sigma $$) = 2.8 inches

**Finding the 5th percentile:**
Using the z-score of -1.645:
$$
X = 68.1 + (-1.645) \cdot 2.8
$$
$$
X \approx 68.1 - 4.606 = 63.494 \text{ inches}
$$

**Finding the 95th percentile:**
Using the z-score of 1.645:
$$
X = 68.1 + (1.645) \cdot 2.8
$$
$$
X \approx 68.1 + 4.606 = 72.706 \text{ inches}
$$

### Summary of Results:
- The new height requirements for women are at least $$ 59.5 $$ inches and at most $$ 67.3 $$ inches.
- The new height requirements for men are at least $$ 63.5 $$ inches and at most $$ 72.7 $$ inches.

### Final Answer:
The new height requirements are at least $$ 59.5 $$ inches and at most $$ 72.7 $$ inches.
The new height requirements are at least 59.5 in and at most 72.7 in.

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