To find the proportion of male babies less than 2 months old that weigh more than 14 pounds, we can use the properties of the normal distribution.
Given:
- Mean (\( \mu \)) = 12.5 pounds
- Standard deviation (\( \sigma \)) = 5.1 pounds
We can standardize the weight of 14 pounds to find the corresponding \( z \)-score using the formula:
\[ z = \frac{(X - \mu)}{\sigma} \]
where \( X \) is the value we are interested in (14 pounds).
Substituting the values:
\[ z = \frac{(14 - 12.5)}{5.1} = \frac{1.5}{5.1} \approx 0.2941 \]
Next, we can use the \( z \)-score to find the proportion of babies that weigh more than 14 pounds. We look up the \( z \)-score in the standard normal distribution table or use a calculator that provides the cumulative distribution function (CDF).
A \( z \)-score of approximately 0.2941 typically corresponds to a cumulative probability of about 0.6151 (this value may vary slightly depending on the table or calculator used).
To find the proportion of babies that weigh more than 14 pounds, we subtract the cumulative probability from 1:
\[ P(X > 14) = 1 - P(Z < 0.2941) \approx 1 - 0.6151 = 0.3849 \]
Thus, the proportion of male babies less than 2 months old that weigh more than 14 pounds is approximately 0.3849, or 38.49%.