Question

assume that when human resource managers are randomly selected, 53% say job applicants should follow up within two weeks. if 12 human resource managers are randomly selected, find the probability that fewer than 3 of them say job applicants should follow up within two weeks. the probability is (round to four decimal places as needed.)

Explanation:

Step1: Identify the binomial distribution parameters

This is a binomial - distribution problem. Let \(n = 12\) (the number of trials, i.e., the number of human - resource managers selected), \(p=0.53\) (the probability that a single human - resource manager says job applicants should follow up within two weeks), and \(q = 1 - p=1 - 0.53 = 0.47\). We want to find \(P(X\lt3)=P(X = 0)+P(X = 1)+P(X = 2)\).
The binomial probability formula is \(P(X = k)=C(n,k)\times p^{k}\times q^{n - k}\), where \(C(n,k)=\frac{n!}{k!(n - k)!}\).

Step2: Calculate \(P(X = 0)\)

$$
LATEXBLOCK0
$$

Step3: Calculate \(P(X = 1)\)

$$
LATEXBLOCK1
$$

Step4: Calculate \(P(X = 2)\)

$$
LATEXBLOCK2
$$

Step5: Calculate \(P(X\lt3)\)

$$
LATEXBLOCK3
$$

Answer:

0.0090