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 16 couples. Complete parts (a) through (c) below.
a. Find the mean and the standard deviation for the numbers of girls in groups of 16 births.
The value of the mean is u =
(Type an integer or a decimal. Do not round.)

The found values are:

μ = 8

σ = 2

Step-by-step explanation:

Following data is given in the question:

p = 0.5

n = 16

Consider that it follows binomial distribution:

For binomial distribution, the mean is the product or probability 'p' and and the sample size 'n' :

μ = p · n

μ = 0.5 · 16

μ = 8

For binomial distribution, the standard deviation is the square root of product of sample size 'n' and the probabilities 'p' and 'q'.

σ = √(n·p·q)

where q = 1 - p = 1 - 0.5 = 0.5

σ = √(16 · 0.5 · 0.5)

σ = √4

σ = 2

this doesn't make sense lol

just do 5.2 + 2 ( 7.2) and 1/3 x 3 (.9) and youv'e got your equation. 0.9( 9/10) + 7.2 ^2 = 16.4 hope i could !

step-by-step explanation: