Final answer:
To find the probability, we need to use the standard normal distribution and the z-score formula. (a) The probability that a randomly selected utility bill is less than $88 is approximately 0.2119. (b) The probability that a randomly selected utility bill is between $120 and $130 is approximately 0.0684.
Step-by-step explanation:
To find the probability, we need to use the standard normal distribution and the z-score formula.
(a) To find the probability that a randomly selected utility bill is less than $88, we need to find the z-score for $88 using the formula: z = (x - mean) / standard deviation. Plugging in the values, we get: z = (88 - 100) / 15 = -0.8. Looking up this z-score in a standard normal distribution table, we find that the probability is approximately 0.2119.
(b) To find the probability that a randomly selected utility bill is between $120 and $130, we need to find the z-scores for $120 and $130, and then find the difference between their probabilities. The z-score for $120 is (120 - 100) / 15 = 1.33, and the z-score for $130 is (130 - 100) / 15 = 2. The difference between their probabilities can be found by subtracting the probability corresponding to the z-score of 1.33 from the probability corresponding to the z-score of 2. Looking up these z-scores in a standard normal distribution table, we find that the probability for the z-score of 1.33 is approximately 0.9088, and the probability for the z-score of 2 is approximately 0.9772. The difference between these probabilities is approximately 0.9772 - 0.9088 = 0.0684.
(c) To find the probability that a randomly selected utility bill is more than $140, we need to find the z-score for $140 using the formula: z = (140 - 100) / 15 = 2.67. Looking up this z-score in a standard normal distribution table, we find that the probability is approximately 0.9963.