Question

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.

Answer 1

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

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 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) = (n!)/(x!(n-x)!)`

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
`p = (1)/(33)`

A local hospital randomly selects five births.

This means that
`n = 5`

So the correct answer is:

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