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 μ = 12.
(type an integer or a decimal. do not round.)
the value of the standard deviation is σ =
(round to one decimal place as needed.)

Explanation:

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

The mean $\mu$ of a binomial distribution is given by $\mu = np$, where $n$ is the number of trials and $p$ is the probability of success on each trial. Here, $n = 24$ (number of couples/birth - events) and $p=0.5$ (probability of having a girl). So, $\mu=np = 24\times0.5=12$.

Step2: Recall the formula for the standard deviation of a binomial distribution

The standard deviation $\sigma$ of a binomial distribution is given by $\sigma=\sqrt{np(1 - p)}$. Substitute $n = 24$ and $p = 0.5$ into the formula:
\[

$$\begin{align*} \sigma&=\sqrt{24\times0.5\times(1 - 0.5)}\\ &=\sqrt{24\times0.5\times0.5}\\ &=\sqrt{6}\\ &\approx2.4 \end{align*}$$

\]

Answer:

The value of the standard deviation is $\sigma = 2.4$