a university found that 40% of its students withdraw without completing the introductory statistics course…

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