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Sagot :
Using the binomial distribution, it is found that there is a:
A. 0.1075 = 10.75% probability that exactly 8 of the students had type AB blood.
B. 0.2467 = 24.67% probability that at least 8 of the students had type AB blood.
For each student, there are only two possible outcomes, either they have type AB blood, or they do not. The probability of a student having type AB blood 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:
[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:
- About 12% of the US population has type AB blood, hence [tex]p = 0.12[/tex].
- A sample of 50 students is taken, hence [tex]n = 50[/tex].
Item a:
The probability is P(X = 8), hence:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 8) = C_{50,8}.(0.12)^{8}.(0.88)^{42} = 0.1075[/tex]
0.1075 = 10.75% probability that exactly 8 of the students had type AB blood.
Item b:
The probability is:
[tex]P(X \geq 8) = 1 - P(X < 8)[/tex]
In which:
[tex]P(X < 8) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7)[/tex]
Using a calculator to find each probability, we have that:
[tex]P(X < 8) = 0.7533[/tex]
Then:
[tex]P(X \geq 8) = 1 - P(X < 8) = 1 - 0.7533 = 0.2467[/tex]
0.2467 = 24.67% probability that at least 8 of the students had type AB blood.
You can learn more about the binomial distribution at https://brainly.com/question/24863377
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