Suppose the lengths of bank visits form a normal distribution with a mean length of 7.0 min and a standard deviation of 3.6 min.  The probability that a bank visit selected at random will last more than 16.0 min is most nearly:

To solve this problem, we first want to figure out how many standard deviations a value of 16 minutes is above the mean. We are told that the mean is 7 minutes and the standard deviation of the distribution is 3.6. Therefore:

$\frac{16-7}{3.6}=\frac{9}{3.6}=2.5$

Therefore, 16 minutes is 2.5 standard distributions above the mean.

Since the problem asks for the probability that the bank visit last longer than 16 minutes, we are essentially being asked what is the probability that an event occurs greater than 2.5 standard deviations above the mean on a normal distribution.

To answer this question, we must reference pg. 76 of the FE Reference Handbook, for the table titled Unit Normal Distribution.

The column labeled R(x) shows probabilities above a defined standard deviation on a normal distribution, which is exactly what we are after. We then navigate to x = 2.5 standard deviations and see that the probability is equal to .0062.

Therefore, the probability that a bank visit will last more than 16 minutes is .0062.