Binomial probability distribution

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 18 couples. What is the value of the mean? What is the standard deviation?

Answer 1

mean = N*p = 18*0.5 = 9

 

standard deviation = sqrt ( N * p * (1-p))

 

                             = sqrt ( 18 * 1/2 * 1/2 ) 

 

                             = sqrt( 18/4)

 

                             = sqrt ( 9/2)

 

                            = sqrt ( ( 9 * 2) / (2*2) ) <---- rationalizes the denominator

 

                               = sqrt ( 18/4)

 

                              = sqrt ( 2 * 9 / 4 ) =

 

                             = sqrt ( 2) * sqrt(9) / sqrt(4)

 

                               = 3 * sqrt(2)/ 2

 

                               = 3/2 * sqrt (2) = 2.1213204364....