Questions: In a survey of 18-year-old males, the mean weight was 156.6 pounds with a standard deviation of 46.8 pounds. Assume the distribution can be approximated by a normal distribution. (a) What weight represents the 99th percentile? (b) What weight represents the 37th percentile? (c) What weight represents the first quartile? (a) pounds (Round to one decimal place as needed.)

Solution Steps

To find the weight that represents the 99th percentile, we use the inverse of the cumulative distribution function (CDF) for a normal distribution. Given the mean weight \( \mu = 156.6 \) pounds and the standard deviation \( \sigma = 46.8 \) pounds, we calculate:

\[ x_{0.99} = \mu + z_{0.99} \cdot \sigma \]

Using the z-score for the 99th percentile, we find:

\[ x_{0.99} \approx 265.5 \text{ pounds} \]

Next, we calculate the weight that represents the 37th percentile. Again, we use the inverse CDF:

\[ x_{0.37} = \mu + z_{0.37} \cdot \sigma \]

Calculating this gives us:

\[ x_{0.37} \approx 141.1 \text{ pounds} \]

Finally, we find the weight that represents the first quartile (25th percentile):

\[ x_{0.25} = \mu + z_{0.25} \cdot \sigma \]

This results in:

\[ x_{0.25} \approx 125.0 \text{ pounds} \]

Final Answer