Question

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 24 couples. complete parts (a) through (c) below.
a. find the mean and the standard deviation for the numbers of girls in groups of 24 births.
the value of the mean is $mu=square$.
(type an integer or a decimal. do not round.)

Explanation:

Step1: Identify the distribution

This is a binomial distribution problem where \(n = 24\) (number of trials, i.e., number of couples) and \(p=0.5\) (probability of a girl).

Step2: Calculate the mean

The mean of a binomial distribution is given by \(\mu = np\). Substitute \(n = 24\) and \(p = 0.5\) into the formula: \(\mu=24\times0.5\).

Step3: Calculate the standard - deviation

The standard deviation of a binomial distribution is \(\sigma=\sqrt{np(1 - p)}\). Substitute \(n = 24\) and \(p = 0.5\) into the formula: \(\sigma=\sqrt{24\times0.5\times(1 - 0.5)}=\sqrt{24\times0.5\times0.5}=\sqrt{6}\).

Answer:

The value of the mean is \(\mu = 12\)