Respuesta :
Using the binomial distribution, it is found that there is a 0.2415 = 24.15% probability that at least 4 of the people selected will always take medicine as prescribed.
For each person, there are only two possible outcomes, either they take their medicines as prescribed, or they do not. The probability of a person taking their medicine as prescribed is independent of any other person, hence the binomial distribution is used to solve this question.
Binomial probability distribution
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
The parameters are:
- x is the number of successes.
- n is the number of trials.
- p is the probability of a success on a single trial.
In this problem:
- The probability that a person will always take medicine as prescribed is 0.54, hence [tex]p = 0.54[/tex].
- 5 people are selected at random from the population, hence [tex]n = 5[/tex].
The probability that at least 4 of the people selected will always take medicine as prescribed is:
[tex]P(X \geq 4) = P(X = 4) + P(X = 5)[/tex]
In which:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 4) = C_{5,4}.(0.54)^{4}.(0.46)^{1} = 0.1956[/tex]
[tex]P(X = 5) = C_{5,5}.(0.54)^{5}.(0.46)^{0} = 0.0459[/tex]
Then:
[tex]P(X \geq 4) = P(X = 4) + P(X = 5) = 0.1956 + 0.0459 = 0.2415[/tex]
0.2415 = 24.15% probability that at least 4 of the people selected will always take medicine as prescribed.
To learn more about the binomial distribution, you can take a look at https://brainly.com/question/24863377
