Question

Assume that when Human Resource managers are randomly selected, 62% say job applicants should follow up within two weeks. If 25 human resource managers are randomly selected, find the probability that exactly 18 of them say job applicants should follow up within two weeks.

Answer 1

Using the binomial distribution, it is found that there is a 0.1008 = 10.08% probability that exactly 18 of them say job applicants should follow up within two weeks.

What is the binomial distribution formula?

The formula is:

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

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

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.

The values of the parameters are given as follows:

n = 25, p = 0.62.

The probability that exactly 18 of them say job applicants should follow up within two weeks is P(X = 18), hence:

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

$P(X = 18) = C_{25,18}.(0.62)^{18}.(0.38)^{7} = 0.1008$

0.1008 = 10.08% probability that exactly 18 of them say job applicants should follow up within two weeks.

More can be learned about the binomial distribution at brainly.com/question/24863377

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