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 22 couples. Complete parts (a) through (b) below

a.Find the mean and the standard deviation for the numbers of girls in groups of 22 births.

The value of the mean is =_______. (Type an integer or a decimal. Do not round.)

The value of the standard deviation is =________. (Round to one decimal place as needed.)

b.Use the range rule of thumb to find the values separating results that are significantly low or significantly high.

-Values of _________girls or fewer are significantly low. (Round to one decimal place as needed.)

-Values of__________girls or greater are significantly high. (Round to one decimal place as needed.)

Answer 1

Given that,

p = 0.5

q = 1 - p = 1 - 0.5 = 0.5

n = 22

Using binomial distribution,

a) Mean = = n * p = 22 * 0.5 = 11

Standard deviation = = n * p * q = 22 * 0.5 * 0.5 = 2.3

b) Using range rule thumb,

significantly low = - 2 =

significantly low = 11 - 2 * 2.3  

significantly low = 11 - 4.6 = 6.4

Values of 6.4 girls or fewer are significantly low

significantly high =   + 2

significantly high = 11 + 2 * 2.3

significantly high = 11 + 4.6 = 15.6

Values of 15.6 girls or greater are significantly high