Net interest margin—often referred to as spread—is the difference between the rate banks pay on deposits and the rate they charge for loans. Suppose that the net interest margins for all U.S. banks are normally distributed with a mean of 4.15 percent and a standard deviation of .5 percent. (a) Find the probability that a randomly selected U.S. bank will have a net interest margin that exceeds 5.40 percent. (Round answer to 4 decimal places.) P (b) Find the probability that a randomly selected U.S. bank will have a net interest margin less than 4.40 percent. (Round answer to 4 decimal places.) P (c) A bank wants its net interest margin to be less than the net interest margins of 95 percent of all U.S. banks. Where should the bank’s net interest margin be set? (Round the z value to 3 decimal places. Round answer to 4 decimal places.)

The area to the left of 2.5 from the standard normal distribution table is 0.9938.The probability that a randomly selected U.S. bank will have a net interest margin that exceeds 5.40 percent is 1-0.9938=0.0062

(b) The z-value that corresponds to 4.40 percent is [tex]Z=\frac{4.40-4.15}{0.5}=0.5[/tex]The net interest margin of 4.40 percent is 0.5 standard deviation above the mean.

Using the normal distribution table, the area under the curve to the left of 0.5 is 0.6915

Therefore the probability that a randomly selected U.S. bank will have a net interest margin less than 4.40 percent is 0.6915

(c) The z-value that corresponds to 95% which is 1.65

We substitute the 1.65 into the formula and solve for X.[tex]1.65=\frac{X-4.15}{0.5}[/tex]

[tex]1.65\times 0.5=X-4.15[/tex][tex]0.825=X-4.15[/tex]

[tex]0.825+4.15=X[/tex]

[tex]4.975=X[/tex]

A bank that wants its net interest margin to be less than the net interest margins of 95 percent of all U.S. banks should set its net interest margin to 4.975 percent.