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(Solved): Assume that when human resource managers are randomly selected, 55% say job applicants should follo ...




Assume that when human resource managers are randomly selected, \( 55 \% \) say job applicants should follow up within two we
Assume that when human resource managers are randomly selected, say job applicants should follow up within two weeks. If 10 human resource managers are randomly selected, find the probability that at least 7 of them say job applicants should follow up within two weeks. The probability is (Round to four decimal places as needed.)


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To solve this problem, we can use the binomial probability formula. Let's define the following variables:
n = 10 (number of human resource managers randomly selected) p = 0.55 (probability that a human resource manager says job applicants should follow up within two weeks)
We need to find the probability of at least 7 managers saying job applicants should follow up within two weeks. This includes the probability of exactly 7, 8, 9, or 10 managers saying so.
Using the binomial probability formula, the probability of getting exactly k successes out of n trials is given by:
P(X = k) = C(n, k) * p^k * (1 - p)^(n - k)
where C(n, k) represents the number of combinations of n items taken k at a time, and it can be calculated as:
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