To find the proportion of male babies less than 2 months old who weigh more than 12 pounds, we will first convert the weight into a z-score using the given mean and standard deviation.
The formula for the z-score is:
\[ z = \frac{X - \mu}{\sigma} \]
where:
- \(X\) is the value (12 pounds in this case),
- \(\mu\) is the mean (12.3 pounds),
- \(\sigma\) is the standard deviation (4.7 pounds).
Substituting the values:
\[ z = \frac{12 - 12.3}{4.7} = \frac{-0.3}{4.7} \approx -0.0638 \]
Next, we will look up the z-score in the standard normal distribution table or use a calculator to find the proportion of values less than this z-score.
Using the standard normal distribution table, or a calculator:
- The cumulative probability (area to the left of \(z = -0.0638\)) is approximately \(0.4756\).
To find the proportion of babies weighing more than 12 pounds, we calculate:
\[ P(X > 12) = 1 - P(X \leq 12) \]
Thus,
\[ P(X > 12) = 1 - 0.4756 \approx 0.5244 \]
So the proportion of babies who weigh more than 12 pounds is approximately:
\[ \boxed{0.5244} \]