Using the binomial probability distribution it is found that there is a 0.1402 = 14.02% probability that exactly 20 of them say job applicants should follow up within two weeks.
The binomial distribution is a probability distribution that summarizes the likelihood that a value will take one of two unbiased values below a given set of parameters or assumptions.
Let X be the binomial random variable with parameters n and p, in which "n" represents the number of managers randomly selected and "p" represents the probability that the chosen managers say job applicants should follow up within two weeks.
According to the question,
63% say job applicants should follow up within two weeks therefore p=0.63
30 human resource managers are selected, therefore n = 30
For a binomial distribution, the probability of "m" successes is :
P ( X = m ) = ^nCm (p)^m (1 - p)^n-m = n! / m! (n - m)! (p)^m (1 - p)^n-m
According to the question, we can compute the probability that exactly 20 of the 30 selected applicants say job applicants should follow up within two weeks as :
n = 30, p = 0.63, m = 20
P ( X = m ) = n! / m! (n - m)! (p)^m (1 - p)^n-m
Therefore, P ( X = 6 ) = 30! / 20! ( 30 - 20 )! ( 0.63 )^20 (1 - 0.63)^30 - 20
P ( X = 6 ) = 30! / 20! 10! ( 0.63 )^20 ( 0.37 )^10
= 30045015 * ( 0.63 )^20 ( 0.37 )^10
=0.1402 (rounded to 4 decimal places)
Therefore the probability is 0.1402
Learn more about probability here brainly.com/question/13604758
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