assume that when human resource managers are randomly selected, 54% say job applicants should follow up within two weeks. if 14 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 0.0027 (round to four decimal places as needed.)

Question

Accepted Answer

# Explanation:
## Step1: Identify the binomial probability formula
The binomial probability formula is $P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}$, where $n$ is the number of trials, $k$ is the number of successes, $p$ is the probability of success in a single trial, and $C(n,k)=\frac{n!}{k!(n - k)!}$

Here, $n = 14$, $p=0.54$, and we want to find $P(X\lt3)=P(X = 0)+P(X = 1)+P(X = 2)$

## Step2: Calculate $P(X = 0)$
$$
\begin{align*}
C(14,0)&=\frac{14!}{0!(14-0)!}=1\
P(X = 0)&=C(14,0)\times(0.54)^{0}\times(1 - 0.54)^{14-0}\
&=1\times1\times(0.46)^{14}\
\end{align*}
$$
Using a calculator, $(0.46)^{14}\approx0.000007$

## Step3: Calculate $P(X = 1)$
$$
\begin{align*}
C(14,1)&=\frac{14!}{1!(14 - 1)!}=\frac{14!}{1!13!}=14\
P(X = 1)&=C(14,1)\times(0.54)^{1}\times(0.46)^{13}\
&=14\times0.54\times(0.46)^{13}\
\end{align*}
$$
Using a calculator, $14\times0.54\times(0.46)^{13}\approx0.0002$

## Step4: Calculate $P(X = 2)$
$$
\begin{align*}
C(14,2)&=\frac{14!}{2!(14-2)!}=\frac{14\times13}{2\times 1}=91\
P(X = 2)&=C(14,2)\times(0.54)^{2}\times(0.46)^{12}\
&=91\times(0.54)^{2}\times(0.46)^{12}\
\end{align*}
$$
Using a calculator, $91\times(0.54)^{2}\times(0.46)^{12}\approx0.0025$

## Step5: Sum up the probabilities
$P(X\lt3)=P(X = 0)+P(X = 1)+P(X = 2)\approx0.000007 + 0.0002+0.0025=0.0027$

# Answer:
$0.0027$

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Step1: Identify the binomial probability formula

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

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

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

Step5: Sum up the probabilities

Answer: