Step 1 :We are given that the utility bills 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 between $84 and $90.
Step 2 :To solve this, we can use the Z-score formula which is \(Z = \frac{X - \mu}{\sigma}\), where \(X\) is the value from the dataset, \(\mu\) is the mean and \(\sigma\) is the standard deviation. The Z-score tells us how many standard deviations an element is from the mean.
Step 3 :We can calculate the Z-scores for $84 and $90, and then find the area under the normal distribution curve between these two Z-scores. This area represents the probability that a randomly selected utility bill is between $84 and $90.
Step 4 :Calculating the Z-scores for $84 and $90, we get \(Z_{84} = \frac{84 - 100}{13} = -1.2307692307692308\) and \(Z_{90} = \frac{90 - 100}{13} = -0.7692307692307693\) respectively.
Step 5 :The probability that a randomly selected utility bill is between $84 and $90 is the area under the normal distribution curve between these two Z-scores, which is approximately 0.11167357040283893.
Step 6 :Final Answer: The probability that a randomly selected utility bill is between $84 and $90 is approximately \(\boxed{0.1117}\).