Question

In a study, 43% of adults questioned reported that their health was excellent. A researcher wishes to study the health of people living close to a nuclear power plant. Among 13 adults randomly selected from this area, only 3 reported that their health was excellent. Find the probability that when 13 adults are randomly selected, at most 3 are in excellent health.

Answer 1

Answer:

13.44% probabilith that at most 3 are in excellent health.

Step-by-step explanation:

For each adult, there are only two possible outcomes. Either they are in excellent health, or they are not. The probability of an adult being in excellent health 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.

43% of adults questioned reported that their health was excellent.

This means that $p = 0.43$

13 adults randomly selected

This means that $n = 13$

Find the probability that when 13 adults are randomly selected, at most 3 are in excellent health.

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

In which:

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

$P(X = 0) = C_{13,0}.(0.42)^{0}.(0.58)^{13} = 0.0008$

$P(X = 1) = C_{13,1}.(0.42)^{1}.(0.58)^{12} = 0.0079$

$P(X = 2) = C_{13,2}.(0.42)^{2}.(0.58)^{11} = 0.0344$

$P(X = 3) = C_{13,3}.(0.42)^{3}.(0.58)^{10} = 0.0913$

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0008 + 0.0079 + 0.0344 + 0.0913 = 0.1344$

13.44% probabilith that at most 3 are in excellent health.