## The proportion of American births that result in a birth defect is approximately 1/33 according to the Centers for Disease Control and Preve

The proportion of American births that result in a birth defect is approximately 1/33 according to the Centers for Disease Control and Prevention (CDC). A local hospital randomly selects five births and lets the random variable X count the number resulting in a defect. Assume the births are independent.

1. Identify an appropriate probability model for X.

a. Uniform distribution with mean 2.5.

b. Poisson distribution with mean 5/33.

c. binomial distribution with n = 5 and p = 1/33.

d. binomial distribution with n = 5 and p = 32/33.

e. Normal distribution with mean 5 and variance 1/33.

## Answers ( )

Answer:c. binomial distribution with n = 5 and p = 1/33.

Step-by-step explanation:For each birth, there are only two possible outcomes. Either it results in a defect, or it does not. The probability of a birth resulting in a defect is independent of other births. So we use the binomial probability distrbution to solve this question.

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

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

And p is the probability of X happening.

The proportion of American births that result in a birth defect is approximately 1/33 according to the Centers for Disease Control and Prevention (CDC).This means that the probability of a birth resulting in a defect is

A local hospital randomly selects five births.This means that

So the correct answer is:c. binomial distribution with n = 5 and p = 1/33.