Answer:
(a)
(b)
(c) X=4.975 percent
Explanation:
(a) Find the z-value that corresponds to 5.40 percent
.
Hence the net interest margin of 5.40 percent is 2.5 standard deviation above the mean.
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 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.
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.
