Question

The probability of a student successfully completing an assignment before the due date is 0.65. If ten students are selected at random, what is the probability that at least 8 but at most 10 students will complete their assignments before the due date

Answer 1

Using the binomial distribution, it is found that there is a 0.2617 = 26.17% probability that at least 8 but at most 10 students will complete their assignments before the due date.

For each student, there are only two possible outcomes, either they complete the assignment, or they do not. The probability of a student completing the assignment is independent of any other student, hence the binomial distribution is used to solve this question.

What is the binomial distribution formula?

The formula is:

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

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

The parameters are:

• x is the number of successes.

• n is the number of trials.

• p is the probability of a success on a single trial.

In this problem:

• The probability of a student successfully completing an assignment before the due date is 0.65, hence p = 0.65.

• Ten students are selected at random, hence n = 10.

The probability is:

$P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10)$

In which:

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

$P(X = 8) = C_{10,8}.(0.65)^{8}.(0.35)^{2} = 0.1757$

$P(X = 9) = C_{10,9}.(0.65)^{9}.(0.35)^{1} = 0.0725$

$P(X = 10) = C_{10,10}.(0.65)^{10}.(0.35)^{0} = 0.0135$

Then:

$P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) = 0.1757 + 0.0725 + 0.0135 = 0.2617$

0.2617 = 26.17% probability that at least 8 but at most 10 students will complete their assignments before the due date.

More can be learned about the binomial distribution at brainly.com/question/14424710