2. Suppose that 75% of all residents of Wellsburg, West Virginia, are at risk of heart disease. Let X equal the number, out of sixteen of these residents who are selected randomly, that are at risk for heart disease. Find the probability that X is (a) at least four, (b) exactly four.

a) 99.9996% probability that X is at least four.

b) 0.0034% probability that X is exactly four.

Step-by-step explanation:

For each resident, there are only two possible outcomes. Either they are at risk of a heart disease, or they are not. The individuals are chosen at random, which means that the probability of a person being at risk is independent from other people. So we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The 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.

75% of all residents of Wellsburg, West Virginia, are at risk of heart disease.

This means that

Sixteen of these residents

This means that

Find the probability that X is

(a) at least four

Either X is lower than four, or it is at least four. The sum of the probabilities of these events is 1. So

We want . So

In which

So

99.9996% probability that X is at least four.

(b) exactly four.

This is P(X = 4).

0.0034% probability that X is exactly four.