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Thepotemich [5.8K]

3 years ago

8

A house is being purchased at the price of $138,000.00. The 30-year mortgage has a 10% down payment at an interest rate of 4.875

% and a PMI payment of $25.88 each month for 77 months. The yearly taxes are $2400.00 and the insurance is $750.00 per year, which is to be placed into an escrow account. What is the total cost of the loan? Round your answer to the nearest one hundred dollars. Enter a number, such as $123,500.00.

1 answer:

Mama L [17]3 years ago

3 0

Answer:

$238,600

Step-by-step explanation:

The amount being borrowed is 100% -10% = 90% of the purchase price, or ...

0.90 × $138,000 = $124,200.

Then the monthly payment can be computed using the amortization formula:

A = P(r/12)/(1 -(1 +r/12)^(-12t))

A = 124200(.04875/12)/(1 -(1 +.04875/12)^(-12·30)) ≈ 657.28

and the total repaid is ...

(360 months) × ($657.28/month) = $236,619.58*

The cost of mortgage insurance is considered part of the cost of the loan. That amount is ...

(77 months) × ($25.88/month) = $1992.76

so the total amount paid for the loan is ...

$236,619.58 +1992.76 = $238,612.34

The cost of the loan is about $238,600.

_____

* In figuring the total repayment cost, we used the full-precision value for the loan payment. The actual repayment amount will depend on the rounded value of the loan payment and the way the final payment is made. When rounded to the nearest $100, these details become unimportant.

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Step-by-step explanation:

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Answer:

a) There are going to be at least 15 items both priced correctly, and incorrectly, which means that we can use the normal approximation to the binomial to solve the exercise in this supposed problem.

b) 20.36% probability that exactly one of the five items is priced incorrectly by the scanner

c) 22.62% probability that at least one of the five items is priced incorrectly by the scanner

Step-by-step explanation:

For each item, there are only two possible outcomes. Either it is priced correctly, or it is not. The probability of an item being priced incorrectly is independent of other items. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

And p is the probability of X happening.

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This means that $n = 5$

One of 20 is priced incorrectly.

This means that $p = \frac{1}{20} = 0.05$

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This is P(X = 1).

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$P(X = 1) = C_{5,1}.(0.05)^{1}.(0.95)^{4} = 0.2036$

20.36% probability that exactly one of the five items is priced incorrectly by the scanner

c) What is the probability that at least one of the five items is priced incorrectly by the scanner?

Either no items are priced incorrectly, or at least one is. The sum of the probabilities of these events is decimal 1. So

$P(X = 0) + P(X \geq 1) = 0$

We want $P(X \geq 1)$ . So

$P(X \geq 1) = 1 - P(X = 0)$

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$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.7738 = 0.2262$

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