Respuesta :
Using the binomial distribution, it is found that there is a 0.4295 = 42.95% probability that at least 4 of them say job applicants should follow up within two weeks.
For each manager, there are only two possible outcomes. Either they say job applicants should follow up within two weeks, or they do not say it. The opinion of a manager is independent of any other manager, which means that 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:
- n is the number of trials.
- x is the number of successes.
- p is the probability of a success on a single trial.
In this problem:
- 41% say job applicants should follow up within two weeks, thus [tex]p = 0.41[/tex]
- 8 managers are selected, thus [tex]n = 8[/tex].
The probability that at least 4 of them say job applicants should follow up within two weeks is [tex]P(X \geq 4)[/tex], which is given by:
[tex]P(X \geq 4) = 1 - P(X < 4)[/tex]
In which:
[tex]P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)[/tex]
Then:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{8,0}.(0.41)^{0}.(0.59)^{8} = 0.0147[/tex]
[tex]P(X = 1) = C_{8,1}.(0.41)^{1}.(0.59)^{7} = 0.0816[/tex]
[tex]P(X = 2) = C_{8,2}.(0.41)^{2}.(0.59)^{6} = 0.1985[/tex]
[tex]P(X = 3) = C_{8,3}.(0.41)^{3}.(0.59)^{5} = 0.2759[/tex]
[tex]P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0147 + 0.0816 + 0.1985 + 0.2759 = 0.5707[/tex]
[tex]P(X \geq 4) = 1 - P(X < 4) = 1 - 0.5705 = 0.4295[/tex]
0.4295 = 42.95% probability that at least 4 of them say job applicants should follow up within two weeks.
A similar problem is given at https://brainly.com/question/25166697
