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. Find the mean and the standard deviation for the numbers of girls in groups of 23 births.

Answer :

We will use the binomial distribution to answer this question.

p success = 0.5

q = 0.5

n = 23

For the binomial distribution, the mean is equal to n*p

mean = 0.5*23 = 11.5 number of girls

standard deviation = 2.40

[tex]\begin{gathered} sd\text{ = }\sqrt{np(1-p)} \\ \text{ =}\sqrt{(23)(0.5)(0.5)} \\ \text{ =2.40} \end{gathered}[/tex]

Other Questions