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 38% probability that exactly 18 of them say job applicants should follow up within two weeks.

How to find that a given condition can be modelled by binomial distribution?

Binomial distributions consists of n independent Bernoulli trials.

Bernoulli trials are those trials that end up randomly either on success (with probability p) or on failures( with probability 1- p = q (say))

The probability that out of n trials, there'd be x successes is given by

$P(X =x) = \: ^nC_xp^x(1-p)^{n-x}$

Binomial probability distribution  

$P(X =x) = \: ^nC_xp^x(1-p)^{n-x}$

The parameters are:

n is the number of trials.

x is the number of successes.

p is the probability of success on a single trial.

In this problem:

62% say job applicants should follow up within two weeks, p = 0.62

25 managers are selected, n = 25

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

P( X > 18) = 1 -  ( X = 18)

= 1 - 0.62

= 0.38

38 % probability that exactly 18 of them say job applicants should follow up within two weeks.

Learn more about binomial distribution here:

brainly.com/question/13609688

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