Step 1 :Given that the number of trials (n) is 18 and the probability of success (p) is 0.5, we can calculate the mean and standard deviation of the binomial distribution.
Step 2 :The mean (\(\mu\)) of a binomial distribution is given by np, so \(\mu = np = 18 \times 0.5 = 9.0\).
Step 3 :The standard deviation (\(\sigma\)) is given by \(\sqrt{np(1-p)}\), so \(\sigma = \sqrt{18 \times 0.5 \times (1-0.5)} = 2.1\).
Step 4 :According to the range rule of thumb, most values should fall within 2 standard deviations of the mean. So, values that are significantly low are those that are less than \(\mu - 2\sigma\), and values that are significantly high are those that are greater than \(\mu + 2\sigma\).
Step 5 :Calculating these values, we find that values of 4.8 girls or fewer are significantly low, and values of 13.2 girls or greater are significantly high.
Step 6 :Comparing the result of 16 girls with this range, we find that 16 is greater than 13.2, so the result of 16 girls is significantly high.
Step 7 :Final Answer: The mean number of girls in groups of 18 births is \(\boxed{9.0}\). The standard deviation is \(\boxed{2.1}\). Values of \(\boxed{4.8}\) girls or fewer are significantly low. Values of \(\boxed{13.2}\) girls or greater are significantly high. The result of 16 girls is significantly high, suggesting that the method may be effective.