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 30 couples. complete parts (a) through (c) below.
a. find the mean and the standard deviation for the numbers of girls in groups of 30 births.
the value of the mean is \\( \mu= \\) \\( \square \\).
(type an integer or a decimal. do not round.)

Explanation:

Step1: Recall the formula for the mean of a binomial distribution

The formula for the mean $\mu$ of a binomial distribution is $\mu = np$, where $n$ is the number of trials and $p$ is the probability of success on a single trial.
Here, $n = 30$ (number of couples / births) and $p=0.5$ (probability of a girl).
So, $\mu=np$.

Step2: Calculate the mean

Substitute $n = 30$ and $p = 0.5$ into the formula:
$\mu=30\times0.5$.

Answer:

$15$