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 $\mu=$
(Type an integer or a decimal. Do not round.)

Step 1 ：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.

Step 2 ：We are asked to find the mean and the standard deviation for the numbers of girls in groups of 23 births.

Step 3 ：The mean of a binomial distribution is given by the formula \(\mu = np\), where n is the number of trials (in this case, the number of couples, which is 23) and p is the probability of success (in this case, the probability of having a girl, which is 0.5). So, we need to calculate \(23 \times 0.5\) to find the mean.

Step 4 ：The standard deviation of a binomial distribution is given by the formula \(\sigma = \sqrt{np(1-p)}\), where n is the number of trials, p is the probability of success, and sqrt is the square root function. So, we need to calculate \(\sqrt{23 \times 0.5 \times (1 - 0.5)}\) to find the standard deviation.

Step 5 ：By calculating, we find that the mean is 11.5 and the standard deviation is approximately 2.4.

Step 6 ：Final Answer: The mean number of girls in groups of 23 births is \(\boxed{11.5}\) and the standard deviation is \(\boxed{2.4}\) (rounded to one decimal place).