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skelet666 [1.2K]
2 years ago
8

When grading an exam, 90% of a professor's 50 students passed. If the professor randomly selected 10 exams, what is the probabil

ity that a.) at least 8 of them passed? b.) fewer than 5 passed?
Mathematics
1 answer:
Damm [24]2 years ago
3 0

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.

<h3>What is the binomial probability distribution formula?</h3>

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

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