Answer:
(a)
(b)
(c) X=4.975 percent
Explanation:
(a) Identify the z-value that represents 5.40 percent
.
Thus, a net interest margin of 5.40 percent stands at 2.5 standard deviations above the average.
From the standard normal distribution table, the area to the left of 2.5 is 0.9938. Hence, the likelihood of a randomly selected U.S. bank achieving a net interest margin greater than 5.40 percent is 1-0.9938=0.0062
(b) The z-value corresponding to 4.40 percent is The net interest margin of 4.40 percent is situated at 0.5 standard deviation above the average.
According to the normal distribution table, the area to the left of 0.5 is 0.6915
Thus, the probability of a randomly chosen U.S. bank having a net interest margin below 4.40 percent equals 0.6915
(c) The z-value indicating 95% is 1.65
Substituting 1.65 into the equation enables us to find X.
For a bank that wishes for its net interest margin to fall below that of 95 percent of all U.S. banks, it should aim for a net interest margin of 4.975 percent.
