Question

When grading an exam, 90% of a professor's 50 students passed. If the professor randomly selected 10 exams, what is the probability that a.) at least 8 of them passed? b.) fewer than 5 passed?

Answer 1

Using the binomial distribution, it is found that there is a:

a) 0.9298 = 92.98% probability that at least 8 of them passed.

b) 0.0001 = 0.01% probability that fewer than 5 passed.

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

What is the binomial probability 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:

• 90% of the students passed, hence $p = 0.9$.

• The professor randomly selected 10 exams, hence $n = 10$.

Item a:

The probability is:

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

In which:

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

$P(X = 8) = C_{10,8}.(0.9)^{8}.(0.1)^{2} = 0.1937$

$P(X = 9) = C_{10,9}.(0.9)^{9}.(0.1)^{1} = 0.3874$

$P(X = 10) = C_{10,10}.(0.9)^{10}.(0.1)^{0} = 0.3487$

Then:

$P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) = 0.1937 + 0.3874 + 0.3487 = 0.9298$

0.9298 = 92.98% probability that at least 8 of them passed.

Item b:

The probability is:

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

Using the binomial formula, as in item a, to find each probability, then adding them, it is found that:

$P(X < 5) = 0.0001$

Hence:

0.0001 = 0.01% probability that fewer than 5 passed.

You can learn more about the the binomial distribution at brainly.com/question/24863377