Question

Based on past experience, a bank believes that 4% of the people who receive loans will not make payments on time. The bank has recently approved 300 loans. 6% of these clients did not make timely payments. What is the probability that over 6% will not make timely payments?A. 0.0721B. 0.9616C. 0.9279D. 0.0384

Answer 1

Answer:

D. 0.0384

Step-by-step explanation:

For each loan, there are only two possible outcomes. Either the client makes timely payments, or he does not. The probability of a client making a timely payment is independent from other clients. So we use the binomial probability distribution to solve this question.

However, our sample is big. So i am going to aproximate this binomial distribution to the normal.

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:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

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:

$Z = \frac{X - \mu}{\sigma}$

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 $\mu = E(X)$, $\sigma = \sqrt{V(X)}$.

In this problem, we have that:

$n = 300, p = 0.04$

So

$\mu = E(X) = np = 300*0.04 = 12$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{300*0.04*0.96} = 3.39$

What is the probability that over 6% will not make timely payments?

This is 1 subtracted by the pvalue of Z when X = 0.06*300 = 18. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{18 - 12}{3.39}$

$Z = 1.77$

$Z = 1.77$ has a pvalue of 0.9616

1 - 0.9616 = 0.0384

So the correct answer is:

D. 0.0384