Statistics problem...?

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 = number of children who are girls.





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.





*Notes:


• Each of n trials has two possible outcomes. The outcome of interest is called a "success" and the other outcome is called a "failure."


• Each trial has the same probability of a success. This is denoted by p.


• The probability of a failure is denoted by 1 - p.


• The n trials are independent. That is, the result for one trial does not depend on the results of others.





The binomial random variable X is the number of successes in the n trials.





Probabilities for a Binomial Distribution:


P(x) = (n!) /( x!(n-x)!) * p^x (1 - p)^(n-x), x = 0,1,2,...,n.

Statistics problem...?
a)


1. There are only two possible outcomes with each birth: male or female.


2. The probability of a female birth remains the same: 0.49 for each birth.


3. The probability of a female birth doesn't depend on whether or not there has already been a male or female birth in the family.





b)


n=4 The number of children.


p=0.49 The probability of a female.





c)


P(X=2) = 4!/(2!2!) 0.49^2 0.51^2 = 0.3747
Reply:A) The gender of a child is a Bernoulli trail and since all the children together is considered a sum of n independent and identically distributed Bernoulli trials you have a Binomial distribution.





B) n = 4, p = 0.49





C)


P(X = 0) = P(0 females, 4 males) = 0.06765201


P(X = 1) = P(1 female, 3 males) = 0.25999596


P(X = 2) = P(2 females, 2 males) = 0.37470006


P(X = 3) = P(3 females, 1 male) = 0.24000396


P(X = 4) = P(4 females, 0 males) = 0.05764801