Question

Not Answered 4.Not Answered 5.Not Answered 6.Not Answered Question Workspace Check My Work (2 remaining) A university found that 20% of its students withdraw without completing the introductory statistics course. Assume that 20 students registered for the course. If you compute the binomial probabilities manually, make sure to carry at least four decimal digits in your calculations. Compute the probability that 2 or fewer will withdraw (to 4 decimals).

Answer 1

Answer:

0.2060 = 20.60% probability that 2 or fewer will withdraw

Step-by-step explanation:

For each student, there are only two possible outcomes. Either they withdraw, or they do not. The probability of a student withdrawing is independent from other students. 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 $n = 20$

Compute the probability that 2 or fewer will withdraw (to 4 decimals).

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

In which

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

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

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

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

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0115 + 0.0576 + 0.1369 = 0.2060$

0.2060 = 20.60% probability that 2 or fewer will withdraw