Question

a university found that 40% of its students withdraw without completing the introductory statistics course. assume that 20 students registered for the course. a. compute the probability that 2 or fewer will withdraw (to 4 decimals). b. compute the probability that exactly 4 will withdraw (to 4 decimals). c. compute the probability that more than 3 will withdraw (to 4 decimals). d. compute the expected number of withdrawals.

Explanation:

Step1: Identify the distribution

This is a binomial distribution problem. Let \(n = 20\) (number of students) and \(p=0.4\) (probability of withdrawal). The binomial probability formula is \(P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}\), where \(C(n,k)=\frac{n!}{k!(n - k)!}\).

Step2: Calculate part a

We need \(P(X\leq2)=P(X = 0)+P(X = 1)+P(X = 2)\)
For \(k = 0\):
\[

$$\begin{align*} C(20,0)&=\frac{20!}{0!(20 - 0)!}=1\\ P(X = 0)&=C(20,0)\times(0.4)^{0}\times(0.6)^{20}\\ &=1\times1\times(0.6)^{20}\approx0.00003656 \end{align*}$$

\]
For \(k = 1\):
\[

$$\begin{align*} C(20,1)&=\frac{20!}{1!(20 - 1)!}=20\\ P(X = 1)&=C(20,1)\times(0.4)^{1}\times(0.6)^{19}\\ &=20\times0.4\times(0.6)^{19}\approx0.00030467 \end{align*}$$

\]
For \(k = 2\):
\[

$$\begin{align*} C(20,2)&=\frac{20!}{2!(20 - 2)!}=\frac{20\times19}{2\times 1}=190\\ P(X = 2)&=C(20,2)\times(0.4)^{2}\times(0.6)^{18}\\ &=190\times0.16\times(0.6)^{18}\approx0.00137102 \end{align*}$$

\]
\(P(X\leq2)=0.00003656 + 0.00030467+0.00137102\approx0.0017\)

Step3: Calculate part b

For \(k = 4\):
\[

$$\begin{align*} C(20,4)&=\frac{20!}{4!(20 - 4)!}=\frac{20\times19\times18\times17}{4\times3\times2\times1}=4845\\ P(X = 4)&=C(20,4)\times(0.4)^{4}\times(0.6)^{16}\\ &=4845\times0.0256\times(0.6)^{16}\approx0.1244 \end{align*}$$

\]

Step4: Calculate part c

\(P(X>3)=1 - P(X\leq3)=1-(P(X = 0)+P(X = 1)+P(X = 2)+P(X = 3))\)
For \(k = 3\):
\[

$$\begin{align*} C(20,3)&=\frac{20!}{3!(20 - 3)!}=\frac{20\times19\times18}{3\times2\times1}=1140\\ P(X = 3)&=C(20,3)\times(0.4)^{3}\times(0.6)^{17}\\ &=1140\times0.064\times(0.6)^{17}\approx0.0350 \end{align*}$$

\]
\(P(X\leq3)=0.00003656+ 0.00030467+0.00137102 + 0.0350\approx0.0367\)
\(P(X>3)=1 - 0.0367 = 0.9633\)

Step5: Calculate part d

The expected value of a binomial distribution is \(E(X)=n\times p\). Here, \(n = 20\) and \(p = 0.4\), so \(E(X)=20\times0.4 = 8\)

Answer:

a. 0.0017
b. 0.1244
c. 0.9633
d. 8