Question

The National Center for Health Statistics (NCHS) reports that 70%70% of U.S. adults aged 6565 and over have ever received a pneumococcal vaccination. Suppose that you obtain an independent sample of 2020 adults aged 6565 and over who visited the emergency room and the pneumococcal vaccination rate applies to the sample. Determine the probability that exactly 12 members of the sample received a pneumococcal vaccination. Let XX represent the number of successes in a binomial setting with nn trials and probability pp of success in each trial. The probability of obtaining exactly kk successes is given by the formula

Answer 1

Answer:

11.44% probability that exactly 12 members of the sample received a pneumococcal vaccination.

Step-by-step explanation:

For each adult, there are only two possible outcomes. Either they received a pneumococcal vaccination, or they did not. The probability of an adult receiving a pneumococcal vaccination is independent of other adults. So we use the binomial probability distribution 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.

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

And p is the probability of X happening.

70% of U.S. adults aged 65 and over have ever received a pneumococcal vaccination.

This means that $p = 0.7$

20 adults

This means that $n = 20$

Determine the probability that exactly 12 members of the sample received a pneumococcal vaccination.

This is P(X = 12).

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

$P(X = 12) = C_{20,12}.(0.7)^{12}.(0.3)^{8} = 0.1144$

11.44% probability that exactly 12 members of the sample received a pneumococcal vaccination.