Each newborn baby has a probability of approximately 0.49 of being female and 0.51 of being male. for a family with four children, let x = the number of children who are girls.

I believe this problem has 3 questions:

a. Explain why the three conditions are satisfied for X to have the binomial distribution.

b. Identify n and p for the binomial distribution.

c. Find the probability that the family has two girls and two boys.

Answers:

a. First because there are only 2 possible outcomes for each birth: male or female. Hence a binomial distribution. 
Second, because the probability of giving out a girl is constant: 0.49 for each birth. 
Third, the probability of a giving out a girl does not depend on whether or not there is already a boy or a girl in the family. 

b. The n is the total number of children, so n = 4

While the p is the success of being a girls, so P = 0.49

c. We use the binomial probability equation:

P (X) = nCx * p^x * q^(n-x)

P(X=2) = 4!/(2!2!) * 0.49^2 * 0.51^2 = 0.3747

nothjing

step-by-step explanation:

the functions have the same rate of change i think. i narrowed it down to this answer and "the rate of change of tour 1 is greater" but i got it wrong using that answer so im assuming its the answer above. (this is a question i also answered from the iready site if anyone needs a reference)