The mean weight of all newborn babies in the US is 7 pounds with a standard deviation of 1.8 pounds. The weights of newborn babies have a normal distribution. Newborn babies who are in the lightest $16 \%$ of birth weights will require an incubator. What is the minimum weight allowed before an incubator is used?

Step 1 ：The problem is asking for the weight that corresponds to the 16th percentile of the normal distribution of newborn baby weights in the US. The mean weight of all newborn babies in the US is 7 pounds with a standard deviation of 1.8 pounds.

Step 2 ：We can use the properties of the normal distribution to solve this problem. The formula for a z-score is \(Z = \frac{X - \mu}{\sigma}\), where \(X\) is the value from the dataset, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.

Step 3 ：We need to find the \(X\) that corresponds to the z-score of the 16th percentile. We can use a z-table or a function that gives the z-score for a given percentile. The z-score for the 16th percentile is approximately -0.994457883209753.

Step 4 ：Substituting the z-score, the mean, and the standard deviation into the z-score formula, we can solve for \(X\). The calculation is as follows: \(X = Z \cdot \sigma + \mu = -0.994457883209753 \cdot 1.8 + 7\).

Step 5 ：The result of the calculation is approximately 5.21. This is the minimum weight for a newborn baby to require an incubator.

Step 6 ：Final Answer: The minimum weight allowed before an incubator is used is approximately \(\boxed{5.21}\) pounds.