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According to the mortgage bankers association, 11% of u. S. Mortgages were delinquent last year. A delinquent mortgage is one that has missed at least one payment but has not yet gone to foreclosure. A random sample of eight mortgages was selected. What is the probability that less than three of these mortgages are delinquent?.

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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