Question

Recently, FHA mortgages, which are insured by the federal government, accounted for 28% of all home-purchase mortgages that were approved. A random sample of 150 mortgage applications was selected. What is the probability that 48 or more from this sample were insured by the FHA?

Answer 1

15.87% probability that 48 or more from this sample were insured by the FHA

Explanation:

I am going to use the normal approximation to the binomial to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

`E(X) = np`

The standard deviation of the binomial distribution is:

`√(V(X)) = √(np(1-p))`

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that
`\mu = E(X)`,
`\sigma = √(V(X))`.

In this problem, we have that:

`p = 0.28, n = 150`

So

`\mu = E(X) = np = 150*0.28 = 42`

`\sigma = √(V(X)) = √(np(1-p)) = √(150*0.28*0.72) = 5.5`

What is the probability that 48 or more from this sample were insured by the FHA?

Using continuity correction, this is
`P(X \geq 48 - 0.5) = P(X \geq 47.5)`, which is 1 subtracted by the pvalue of Z when X = 47.5. So

`Z = (X - \mu)/(\sigma)`

`Z = (47.5 - 42)/(5.5)`

`Z = 1`

`Z = 1` has a pvalue of 0.8413.

1 - 0.8413 = 0.1587

15.87% probability that 48 or more from this sample were insured by the FHA