Respuesta :
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:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
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 [tex]p = 0.9[/tex].
- The professor randomly selected 10 exams, hence [tex]n = 10[/tex].
Item a:
The probability is:
[tex]P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10)[/tex]
In which:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 8) = C_{10,8}.(0.9)^{8}.(0.1)^{2} = 0.1937[/tex]
[tex]P(X = 9) = C_{10,9}.(0.9)^{9}.(0.1)^{1} = 0.3874[/tex]
[tex]P(X = 10) = C_{10,10}.(0.9)^{10}.(0.1)^{0} = 0.3487[/tex]
Then:
[tex]P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) = 0.1937 + 0.3874 + 0.3487 = 0.9298[/tex]
0.9298 = 92.98% probability that at least 8 of them passed.
Item b:
The probability is:
[tex]P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)[/tex]
Using the binomial formula, as in item a, to find each probability, then adding them, it is found that:
[tex]P(X < 5) = 0.0001[/tex]
Hence:
0.0001 = 0.01% probability that fewer than 5 passed.
You can learn more about the the binomial distribution at https://brainly.com/question/24863377
