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Anni [7]
3 years ago
14

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

nts 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.
Mathematics
1 answer:
Mkey [24]3 years ago
3 0

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 = <u><em>Number of students withdraw without completing the introductory statistics course</em></u>

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}

=  <u>0.2061</u>

(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}

                                 =  <u>0.2182</u>

(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 = <u>0.5886</u>

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

                        E(X)  =  n\times p

                                 =  20 \times 0.20 = 4 withdrawals

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