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 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 μ = 11.5
(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 binomial distribution formula for standard - deviation

For a binomial distribution $X\sim B(n,p)$, the standard - deviation is $\sigma=\sqrt{np(1 - p)}$, where $n$ is the number of trials and $p$ is the probability of success in a single trial.

Step2: Identify values of $n$ and $p$

We are given that $n = 23$ (number of couples or births) and $p=0.5$ (probability of having a girl).

Step3: Calculate the standard - deviation

Substitute $n = 23$ and $p = 0.5$ into the formula $\sigma=\sqrt{np(1 - p)}$.
$\sigma=\sqrt{23\times0.5\times(1 - 0.5)}=\sqrt{23\times0.5\times0.5}=\sqrt{5.75}\approx2.4$.

Answer:

$2.4$