The monthly utility bills in a city are normally distributed, with a mean of $\$ 100$ and a standard deviation of $\$ 13$. Find the probability that a randomly selected utility bill is (a) less than $\$ 70$, (b) between $\$ 84$ and $\$ 90$, and (c) more than $\$ 120$. (a) The probability that a randomly selected utility bill is less than $\$ 70$ is 0.0105 . (Round to four decimal places as needed.) (b) The probability that a randomly selected utility bill is between $\$ 84$ and $\$ 90$ is 0.1117 (Round to four decimal places as needed.) (c) The probability that a randomly selected utility bill is more than $\$ 120$ is (Round to four decimal places as needed.)

Step 1 ：We are given a normal distribution with a mean of $100 and a standard deviation of $13. We are asked to find the probability that a randomly selected utility bill is more than $120.

Step 2 ：To solve this, we need to convert the $120 to a z-score, which is a measure of how many standard deviations an element is from the mean. The formula for calculating the z-score is: \(z = \frac{X - \mu}{\sigma}\) where: \(X\) is the value we are interested in, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.

Step 3 ：Substituting the given values into the formula, we get: \(z = \frac{120 - 100}{13} = 1.5384615384615385\)

Step 4 ：After calculating the z-score, we can use a z-table to find the probability that a randomly selected utility bill is more than $120. The probability is approximately 0.061967902836371214

Step 5 ：Rounding to four decimal places as needed, the final answer is: \(\boxed{0.0620}\)