Question

According to the CDC, the rate of Cesarean births in the United States in 2013 was about 33%. Suppose a random sample of 200 births is selected. Let X be the number of Cesarean births out of all 200 births. What are the values of the parameters for the binomial random variable X?
n=______
p=________

Answer 1

Answer:

n = 200

p = 0.33.

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem we have that:

33% of the births are Cesarean, so $p = 0.33$

Sample of 200 births, which eans that, $n = 200$