Question

A university found that 20% of its students withdraw without completing the introductory statistics course. Assume that 20 students registered for the course. (Round your answers to four decimal places.) (a) Compute the probability that 2 or fewer will withdraw. (b) Compute the probability that exactly 4 will withdraw. (c) Compute the probability that more than 3 will withdraw. (d) Compute the expected number of withdrawals.

Answer 1

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;

$P(X =r) = \binom{n}{r} \times p^{r} \times (1-p)^{n-r};x=0,1,2,3,......$

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 $\leq$ 2)

P(X $\leq$ 2) =  P(X = 0) + P(X = 1) + P(X = 2)

=  $\binom{20}{0} \times 0.20^{0} \times (1-0.20)^{20-0}+ \binom{20}{1} \times 0.20^{1} \times (1-0.20)^{20-1}+ \binom{20}{2} \times 0.20^{2} \times (1-0.20)^{20-2}$

=  $1 \times1 \times 0.80^{20}+ 20 \times 0.20^{1} \times 0.80^{19}+ 190\times 0.20^{2} \times 0.80^{18}$

=  0.2061

(b) Probability that exactly 4 will withdraw is given by = P(X = 4)

                      P(X = 4) =  $\binom{20}{4} \times 0.20^{4} \times (1-0.20)^{20-4}$

                                 =  $4845\times 0.20^{4} \times 0.80^{16}$

                                 =  0.2182

(c) Probability that more than 3 will withdraw is given by = P(X > 3)

P(X > 3) =  1 - P(X $\leq$ 3) = 1 - P(X = 0) - P(X = 1) - P(X = 2) - P(X = 3)

=  $1-(\binom{20}{0} \times 0.20^{0} \times (1-0.20)^{20-0}+ \binom{20}{1} \times 0.20^{1} \times (1-0.20)^{20-1}+ \binom{20}{2} \times 0.20^{2} \times (1-0.20)^{20-2}+\binom{20}{3} \times 0.20^{3} \times (1-0.20)^{20-3})$

=  $1-(1 \times1 \times 0.80^{20}+ 20 \times 0.20^{1} \times 0.80^{19}+ 190\times 0.20^{2} \times 0.80^{18}+1140\times 0.20^{3} \times 0.80^{17})$

=  1 - 0.4114 = 0.5886

(d) The expected number of withdrawals is given by;

                        E(X)  =  $n\times p$

                                 =  $20 \times 0.20$ = 4 withdrawals