Assume that when human resource managers are randomly selected, 43% say job applicants should follow up within two weeks. If 6 human resource managers are randomly selected, find the probability that at least 2 of them say job applicants should follow up within two weeks.

Respuesta :

If 6 human resource managers are randomly selected, the probability that at least 2 of them say job applicants should follow up within two weeks is; 81.047%

How to solve Binomial Probability distribution problems?

The binomial probability is defined as the probability of exactly x successes on n repeated trials, with p probability

The general formula of binomial probability distribution is;

P(X = x) = ⁿCₓ * pˣ * (1 - p)⁽ⁿ ⁻ ˣ⁾

where;

p is probability of success

n is number of trials or sample size

x is number of successful trials

Thus,  If 6 human resource managers are randomly selected, the probability that at least 2 of them say job applicants should follow up within two weeks is;

P(X ≥ 2) = 1 - (P(0) + P(1))

P(0) = ⁶C₀ * 0.43⁰ * (1 - 0.43)⁽⁶ ⁻ ⁰⁾ = 0.0343

P(1) = ⁶C₁ * 0.43¹ * (1 - 0.43)⁽⁶ ⁻ ¹⁾ = 0.15524

P(X ≥ 2) = 1 - (0.0343 + 0.15524)

P(X ≥ 2) = 0.81047 = 81.047%

Read more about Binomial probability distribution at; https://brainly.com/question/9325204

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