Question

The National Vaccine Information Center estimates that 90% of Americans have had chickenpox by the time they reach adulthood.32 (a) Suppose we take a random sample of 100 American adults. Is the use of the binomial distribution appropriate for calculating the probability that exactly 97 out of 100 randomly sampled American adults had chickenpox during childhood

Answer 1

Answer:

0.59% probability that exactly 97 out of 100 randomly sampled American adults had chickenpox during childhood.

Step-by-step explanation:

For each person, there are only two possible outcomes. Either they have had chickenpox, or they have not. The probability of a person having had chickenpox 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.

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

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.

90% of Americans have had chickenpox by the time they reach adulthood.

This means that $p = 0.9$

Suppose we take a random sample of 100 American adults. Is the use of the binomial distribution appropriate for calculating the probability that exactly 97 out of 100 randomly sampled American adults had chickenpox during childhood

This is $P(X = 97)$ when $n = 100$. So

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

$P(X = 97) = C_{100,97}.(0.9)^{97}.(0.1)^{3} = 0.0059$

0.59% probability that exactly 97 out of 100 randomly sampled American adults had chickenpox during childhood.