Assume that different groups of couples use a particular method of gender selection and each couple gives birth to one baby. This method is designed to increase the likelihood that each baby will be a girl, but assume that the method has no effect, so the probability of a girl is 0.5 . Assume that the groups consist of 18 couples. Complete parts (a) through (c) below.
a. Find the mean and the standard deviation for the numbers of girls in groups of 18 births.
The value of the mean is $\mu=$
(Type an integer or a decimal. Do not round.)
The value of the standard deviation is $\sigma=$
(Round to one decimal place as needed.)
b. Use the range rule of thumb to find the values separating results that are significantly low or significantly high.
Values of girls or fewer are significantly low.
(Round to one decimal place as needed.)
Values of $\square$ girls or greater are significantly high.
(Round to one decimal place as needed.)
c. Is the result of 16 girls a result that is significantly high? What does it suggest about the effectiveness of the method?
The result significantly high, because 16 girls is girls. A result of 16 girls would suggest that the method
(Round to one decimal place as needed.)
Answer
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.
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.
