Respuesta :
Answer:
(a) Probability that 2 or fewer will withdraw is 0.2061.
(b) Probability that exactly 4 will withdraw is 0.2182.
(c) Probability that more than 3 will withdraw is 0.5886.
(d) The expected number of withdrawals is 4.
Step-by-step explanation:
We are given that a university found that 20% of its students withdraw without completing the introductory statistics course.
Assume that 20 students registered for the course.
The above situation can be represented through binomial distribution;
[tex]P(X =r) = \binom{n}{r} \times p^{r} \times (1-p)^{n-r};x=0,1,2,3,......[/tex]
where, n = number of trials (samples) taken = 20 students
r = number of success
p = probability of success which in our question is probability
that students withdraw without completing the introductory
statistics course, i.e; p = 20%
Let X = Number of students withdraw without completing the introductory statistics course
So, X ~ Binom(n = 20 , p = 0.20)
(a) Probability that 2 or fewer will withdraw is given by = P(X [tex]\leq[/tex] 2)
P(X [tex]\leq[/tex] 2) = P(X = 0) + P(X = 1) + P(X = 2)
= [tex]\binom{20}{0} \times 0.20^{0} \times (1-0.20)^{20-0}+ \binom{20}{1} \times 0.20^{1} \times (1-0.20)^{20-1}+ \binom{20}{2} \times 0.20^{2} \times (1-0.20)^{20-2}[/tex]
= [tex]1 \times1 \times 0.80^{20}+ 20 \times 0.20^{1} \times 0.80^{19}+ 190\times 0.20^{2} \times 0.80^{18}[/tex]
= 0.2061
(b) Probability that exactly 4 will withdraw is given by = P(X = 4)
P(X = 4) = [tex]\binom{20}{4} \times 0.20^{4} \times (1-0.20)^{20-4}[/tex]
= [tex]4845\times 0.20^{4} \times 0.80^{16}[/tex]
= 0.2182
(c) Probability that more than 3 will withdraw is given by = P(X > 3)
P(X > 3) = 1 - P(X [tex]\leq[/tex] 3) = 1 - P(X = 0) - P(X = 1) - P(X = 2) - P(X = 3)
= [tex]1-(\binom{20}{0} \times 0.20^{0} \times (1-0.20)^{20-0}+ \binom{20}{1} \times 0.20^{1} \times (1-0.20)^{20-1}+ \binom{20}{2} \times 0.20^{2} \times (1-0.20)^{20-2}+\binom{20}{3} \times 0.20^{3} \times (1-0.20)^{20-3})[/tex]
= [tex]1-(1 \times1 \times 0.80^{20}+ 20 \times 0.20^{1} \times 0.80^{19}+ 190\times 0.20^{2} \times 0.80^{18}+1140\times 0.20^{3} \times 0.80^{17})[/tex]
= 1 - 0.4114 = 0.5886
(d) The expected number of withdrawals is given by;
E(X) = [tex]n\times p[/tex]
= [tex]20 \times 0.20[/tex] = 4 withdrawals
