Mathematics, 19.06.2020 22:57 christopherluckey7

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 2 or fewer will withdraw. If required, round your answer to four decimal places. (b) Compute the probability that exactly 4 will withdraw. If required, round your answer to four decimal places. (c) Compute the probability that more than 3 will withdraw. If required, round your answer to four decimal places. (d) Compute the expected number of withdrawals.

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Answer from: Lilyy1k

a) 0.206 = 20.6% probability that 2 or fewer will withdraw.

b) 0.2182 = 21.82% probability that exactly 4 will withdraw.

c) 0.5886 = 58.86% probability that more than 3 will withdraw.

d) The expected number of withdrawals is 4.

Step-by-step explanation:

For each student, there are only two possible outcomes. Either they withdraw without completing the introductory statistics course, or they do not. Each student is independent of each other. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

20% of its students withdraw without completing the introductory statistics course.

This means that p = 0.2

Assume that 20 students registered for the course.

This means that n = 20

(a) Compute the probability that 2 or fewer will withdraw.

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

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{20,0}.(0.2)^{0}.(0.8)^{20} = 0.0115$

$P(X = 1) = C_{20,1}.(0.2)^{1}.(0.8)^{19} = 0.0576$

$P(X = 2) = C_{20,2}.(0.2)^{2}.(0.8)^{18} = 0.1369$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0115 + 0.0576 + 0.1369 = 0.206$

0.206 = 20.6% probability that 2 or fewer will withdraw.

(b) Compute the probability that exactly 4 will withdraw.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 4) = C_{20,4}.(0.2)^{4}.(0.8)^{16} = 0.2182$

0.2182 = 21.82% probability that exactly 4 will withdraw.

Either less than 3 withdraw, or more than 3 withdraw. The sum of the probabilities of these events is 1. So

$P(X \leq 3) + P(X 3) = 1$

We want P(X > 3). So

$P(X 3) = 1 - P(X \leq 3)$

In which

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

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{20,0}.(0.2)^{0}.(0.8)^{20} = 0.0115$

$P(X = 1) = C_{20,1}.(0.2)^{1}.(0.8)^{19} = 0.0576$

$P(X = 2) = C_{20,2}.(0.2)^{2}.(0.8)^{18} = 0.1369$

$P(X = 3) = C_{20,3}.(0.2)^{3}.(0.8)^{17} = 0.2054$

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0115 + 0.0576 + 0.1369 + 0.2054 = 0.4114$

$P(X 3) = 1 - P(X \leq 3) = 1 - 0.4114 = 0.5886$

0.5886 = 58.86% probability that more than 3 will withdraw.

(d) Compute the expected number of withdrawals.

The expected value of the binomial distribution is:

E(X) = np

In this question:

E(X) = 20*0.2 = 4

The expected number of withdrawals is 4.

Answer from: Quest

Uneed to add them up 27+18+23=68
Hi everybody! can some one me? ! : ) 30 !

Answer from: Quest

if a = 1, and b = 2, then the answer is 47.99999999. if you round that to the nearest whole number, the answer is 48.

Square root of a^6b times, cube root of a^2b^7

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