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

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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:

[tex]E(X) = np[/tex]

The standard deviation of the binomial distribution is:

[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]

Normal probability distribution

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

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

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

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 [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].

In this problem, we have that:

[tex]n = 300, p = 0.04[/tex]

So

[tex]\mu = E(X) = np = 300*0.04 = 12[/tex]

[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{300*0.04*0.96} = 3.39[/tex]

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

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{18 - 12}{3.39}[/tex]

[tex]Z = 1.77[/tex]

[tex]Z = 1.77[/tex] has a pvalue of 0.9616

1 - 0.9616 = 0.0384

So the correct answer is:

D. 0.0384