Answer:
The 99% confidence interval estimate of the percentage of girls born is (74.37%, 85.63%).
Usually, 50% of the babies are girls. This confidence interval gives values considerably higher than that, so the method to increase the probability of conceiving a girl appears to be very effective.
Step-by-step explanation:
Confidence Interval for the proportion:
In a sample with a number n of people surveyed with a probability of a success of
, and a confidence level of
, we have the following confidence interval of proportions.

In which
z is the zscore that has a pvalue of
.
For this problem, we have that:

99% confidence level
So
, z is the value of Z that has a pvalue of
, so
.
The lower limit of this interval is:

The upper limit of this interval is:

For the percentage:
Multiplying the proportions by 100.
The 99% confidence interval estimate of the percentage of girls born is (74.37%, 85.63%).
Usually, 50% of the babies are girls. This confidence interval gives values considerably higher than that, so the method to increase the probability of conceiving a girl appears to be very effective.