A university found that 20% of its students withdraw without completing the introductory statistics course. Assume that 20 students registered for the course.

a. Compute the probability that two or fewer will withdraw.
b. Compute the probability that exactly four will withdraw.
c. Compute the probability that more than three will withdraw.
d. Compute the expected number of withdrawals.

This situation can be modeled with a binomial distribution, where p = probability of “success” (completing the course) equals 80% = 0.8 and the probability of “failure” (withdrawing) equals 0.2.

So, the probability of exactly k withdrawals in 20 cases is given by

[tex]\large P(20;k)=\binom{20}{k}(0.2)^k(0.8)^{20-k}[/tex]

a)

We are looking for

P(0;20)+P(0;1)+P(0;2) =

[tex]\large \binom{20}{0}(0.2)^0(0.8)^{20}+\binom{20}{1}(0.2)^1(0.8)^{19}+\binom{20}{2}(0.2)^2(0.8)^{18}=[/tex]

0.0115292150460685 + 0.0576460752303424 + 0.136909428672063 = 0.206084718948474≅ 0.2061 or 20.61%

b)

Here we want P(20;4)

[tex]\large P(20;4)=\binom{20}{4}(0.2)^4(0.8)^{16}=0.218199402\approx 0.2182=21.82\%[/tex]

c)

Here we need

[tex]\large \sum_{k=4}^{20}P(20;k)=1-\sum_{k=1}^{3}P(20;k)[/tex]

But we already have P(0;20)+P(0;1)+P(0;2) =0.2061 and

[tex]\large \sum_{k=1}^{3}P(20;k)=0.2061+P(20;3)=0.2061+0.205364 \approx 0.4236=42.36\%[/tex]

d)

For a binomial distribution the expectance of “succeses” in n trials is np where p is the probability of “succes”, and the expectance of “failures” is nq, so the expectance for withdrawals in 20 students is 20*0.2 = 4 withdrawals.