assume that when human resource managers are randomly selected, 54% say job applicants should follow up within two weeks. if 13 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.)

Question

Accepted Answer

# Explanation:
## Step1: Identify the binomial formula
The binomial probability formula is $P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}$, where $n = 13$, $p=0.54$, and $C(n,k)=\frac{n!}{k!(n - k)!}$

## Step2: Calculate $P(X = 0)$
$$
\begin{align*}
C(13,0)&=\frac{13!}{0!(13-0)!}=1\
P(X = 0)&=1\times(0.54)^{0}\times(1 - 0.54)^{13}\
&=1\times1\times(0.46)^{13}\
\end{align*}
$$
Using a calculator, $(0.46)^{13}\approx0.000001$

## Step3: Calculate $P(X = 1)$
$$
\begin{align*}
C(13,1)&=\frac{13!}{1!(13 - 1)!}=\frac{13!}{1!12!}=13\
P(X = 1)&=13\times(0.54)^{1}\times(0.46)^{12}\
&=13\times0.54\times(0.46)^{12}\
\end{align*}
$$
$(0.46)^{12}\approx0.000003$, $13\times0.54\times0.000003 = 0.000021$

## Step4: Calculate $P(X = 2)$
$$
\begin{align*}
C(13,2)&=\frac{13!}{2!(13-2)!}=\frac{13\times12\times11!}{2\times1\times11!}=78\
P(X = 2)&=78\times(0.54)^{2}\times(0.46)^{11}\
&=78\times0.2916\times(0.46)^{11}\
\end{align*}
$$
$(0.46)^{11}\approx0.000007$, $78\times0.2916\times0.000007\approx0.000159$

## Step5: Sum the probabilities
$P(X\lt3)=P(X = 0)+P(X = 1)+P(X = 2)$
$P(X\lt3)=0.000001 + 0.000021+0.000159=0.000181\approx0.0002$

# Answer:
$0.0002$

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Question

Explanation:

Step1: Identify the binomial formula

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

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

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

Step5: Sum the probabilities

Answer: