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Sagot :
Using the binomial distribution, it is found that there is a 0.9513 = 95.13% probability that less than three of these mortgages are delinquent.
For each mortgage, there are only two possible outcomes, either they are delinquent or they are not. The probability of a mortgage being delinquent is independent of any other mortgage, 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:
- 11% of mortgages were delinquent last year, hence p = 0.11.
- A random sample of eight mortgages was selected, hence n = 8.
The probability that less than three of these mortgages are delinquent is:
[tex]P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)[/tex]
In which:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{8,0}.(0.11)^{0}.(0.89)^{8} = 0.3937[/tex]
[tex]P(X = 1) = C_{8,1}.(0.11)^{1}.(0.89)^{7} = 0.3892[/tex]
[tex]P(X = 2) = C_{8,2}.(0.11)^{2}.(0.89)^{6} = 0.1684[/tex]
Then:
[tex]P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.3937 + 0.3892 + 0.1684 = 0.9513[/tex]
0.9513 = 95.13% probability that less than three of these mortgages are delinquent.
You can learn more about the binomial distribution at https://brainly.com/question/24863377
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