Problem
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 23 couples. Complete parts (a) through (c) below. a. Find the mean and the standard deviation for the numbers of girls in groups of 23 births. The value of the mean is $\mu=11.5$. (Type an integer or a decimal. Do not round.) The value of the standard deviation is $\sigma=2.4$. (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 $\square$ girls or fewer are significantly low. (Round to one decimal place as needed.)
Solution
Step 1 ：Given that the probability of a girl being born is 0.5 and there are 23 couples, this is a binomial distribution problem.
Step 2 ：The mean of a binomial distribution is given by \(np\), where \(n\) is the number of trials (couples) and \(p\) is the probability of success (having a girl).
Step 3 ：Substituting the given values, we get \(mean = np = 23 * 0.5 = 11.5\).
Step 4 ：The standard deviation of a binomial distribution is given by \(\sqrt{np(1-p)}\).
Step 5 ：Substituting the given values, we get \(std\_dev = \sqrt{23 * 0.5 * (1-0.5)} = 2.4\) (rounded to one decimal place).
Step 6 ：Final Answer: The mean number of girls in groups of 23 births is \(\boxed{11.5}\) and the standard deviation is \(\boxed{2.4}\).
