Step 1 :We are given that the monthly utility bills in a city are normally distributed, 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 less than \$70.
Step 2 :To solve this problem, we need to use the concept of Z-score in statistics. The Z-score is a measure of how many standard deviations an element is from the mean. We can calculate the Z-score using the formula: \(Z = \frac{X - \mu}{\sigma}\), where \(X\) is the value we are looking for, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.
Step 3 :In this case, we want to find the probability that a randomly selected utility bill is less than \$70. So, \(X = 70\), \(\mu = 100\), and \(\sigma = 13\).
Step 4 :Substituting these values into the Z-score formula, we get \(Z = \frac{70 - 100}{13} = -2.3076923076923075\).
Step 5 :After calculating the Z-score, we can use a Z-table to find the probability associated with this Z-score. The Z-table gives us the probability that the value is less than \(X\) (in this case, less than \$70).
Step 6 :Looking up the Z-score of -2.3076923076923075 in the Z-table, we find that the probability is approximately 0.0105.
Step 7 :Final Answer: The probability that a randomly selected utility bill is less than \$70 is approximately \(\boxed{0.0105}\).