Question

Because of the relatively high interest rates, most consumers attempt to pay off their credit card bills promptly. However, this is not always possible. An analysis of the amount of interest paid monthly by a bank's Visa cardholders reveals that the amount is normally distributed with a mean of 27 dollars and a standard deviation of 9 dollars.
A. What proportion of the bank's Visa cardholders pay more than 29 dollars in interest?
B. What proportion of the bank's Visa cardholders pay more than 35 dollars in interest?
C. What proportion of the bank's Visa cardholders pay less than 14 dollars in interest?
D. What interest payment is exceeded by only 18% of the bank's Visa cardholders?

Answer 1

Answer:

a. 0.4129 = 41.29% of the bank's Visa cardholders pay more than 29 dollars in interest.

b. 0.1867 = 18.67% of the bank's Visa cardholders pay more than 35 dollars in interest.

c. 0.0742 = 7.42% of the bank's Visa cardholders pay less than 14 dollars in interest.

d. An interest payment of $35.2 is exceeded by only 18% of the bank's Visa cardholders.

Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Mean of 27 dollars and a standard deviation of 9 dollars.

This means that $\mu = 27, \sigma = 9$

A. What proportion of the bank's Visa cardholders pay more than 29 dollars in interest?

This is 1 subtracted by the p-value of Z when X = 29, so:

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

$Z = \frac{29 - 27}{9}$

$Z = 0.22$

$Z = 0.22$ has a p-value of 0.5871.

1 - 0.5871 = 0.4129

0.4129 = 41.29% of the bank's Visa cardholders pay more than 29 dollars in interest.

B. What proportion of the bank's Visa cardholders pay more than 35 dollars in interest?

This is 1 subtracted by the p-value of Z when X = 35, so:

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

$Z = \frac{35 - 27}{9}$

$Z = 0.89$

$Z = 0.89$ has a p-value of 0.8133.

1 - 0.8133 = 0.1867

0.1867 = 18.67% of the bank's Visa cardholders pay more than 35 dollars in interest.

C. What proportion of the bank's Visa cardholders pay less than 14 dollars in interest?

This is the p-value of Z when X = 14. So

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

$Z = \frac{14 - 27}{9}$

$Z = -1.445$

$Z = -1.445$ has a p-value of 0.0742.

0.0742 = 7.42% of the bank's Visa cardholders pay less than 14 dollars in interest.

D. What interest payment is exceeded by only 18% of the bank's Visa cardholders?

This is the 100 - 18 = 82nd percentile, which is X when Z has a p-value of 0.82, so X when Z = 0.915.

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

$0.915 = \frac{X - 27}{9}$

$X - 27 = 0.915*9$

$X = 35.2$

An interest payment of $35.2 is exceeded by only 18% of the bank's Visa cardholders.