Question

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?.

Answer 1

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:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

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:

• 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:

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$

In which:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{8,0}.(0.11)^{0}.(0.89)^{8} = 0.3937$

$P(X = 1) = C_{8,1}.(0.11)^{1}.(0.89)^{7} = 0.3892$

$P(X = 2) = C_{8,2}.(0.11)^{2}.(0.89)^{6} = 0.1684$

Then:

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.3937 + 0.3892 + 0.1684 = 0.9513$

0.9513 = 95.13% probability that less than three of these mortgages are delinquent.

You can learn more about the binomial distribution at brainly.com/question/24863377