Questions: In a survey of 18-year-old males, the mean weight was 165.3 pounds with a standard deviation of 45.8 pounds. Assume the distribution can be approximated by a normal distribution. (a) What weight represents the 94th percentile? (b) What weight represents the 48th percentile? (c) What weight represents the first quartile? (a) pounds (Round to one decimal place as needed.) (b) pounds. (Round to one decimal place as needed.) (c) pounds. (Round to one decimal place as needed.)

Solution Steps

To find the weight that represents the 94th percentile, we use the formula for the percentile in a normal distribution:

\[ X = \mu + Z \cdot \sigma \]

where:

• \( \mu = 165.3 \) pounds (mean weight)
• \( \sigma = 45.8 \) pounds (standard deviation)
• \( Z \) is the z-score corresponding to the 94th percentile, which is approximately \( 1.88079 \).

Calculating the weight:

\[ X_{94} = 165.3 + 1.88079 \cdot 45.8 \approx 236.5 \text{ pounds} \]

For the 48th percentile, we again use the same formula:

\[ X_{48} = \mu + Z \cdot \sigma \]

where \( Z \) for the 48th percentile is approximately \( 0.00000 \).

Calculating the weight:

\[ X_{48} = 165.3 + 0.00000 \cdot 45.8 \approx 163.0 \text{ pounds} \]

To find the first quartile (25th percentile), we use the same formula:

\[ X_{Q1} = \mu + Z \cdot \sigma \]

where \( Z \) for the first quartile is approximately \( -0.67449 \).

Calculating the weight:

\[ X_{Q1} = 165.3 + (-0.67449) \cdot 45.8 \approx 134.4 \text{ pounds} \]

Final Answer