To solve this problem, we can use the properties of the normal distribution with a mean $\mu = 100$ and a standard deviation $\sigma = 13$.

1. Convert each of the values to a Z-score using the formula: $Z = \frac{X - \mu}{\sigma}$
2. Find the probabilities based on the Z-scores using a standard normal distribution table or calculator.

Let's break down each part:

Part (a): Probability that a randomly selected utility bill is less than $69

1. Calculate the Z-score for $X = 69$: $Z = \frac{69 - 100}{13} = \frac{-31}{13} \approx -2.3846$
2. Look up this Z-score in the standard normal distribution table, or use a calculator for more precision.

The probability corresponding to $Z = -2.3846$ is approximately 0.0086 (rounded to four decimal places).

Part (b): Probability that a randomly selected utility bill is between $88 and $110

1. Calculate the Z-score for $X = 88$: $Z_{88} = \frac{88 - 100}{13} = \frac{-12}{13} \approx -0.9231$
2. Calculate the Z-score for $X = 110$: $Z_{110} = \frac{110 - 100}{13} = \frac{10}{13} \approx 0.7692$
3. Find the probabilities for these Z-scores:
   • For $Z = -0.9231$, the probability is approximately 0.1788.
   • For $Z = 0.7692$, the probability is approximately 0.7794.
4. The probability of being between 88 and 110 is the difference between these probabilities: $0.7794 - 0.1788 = 0.6006$

Part (c): Probability that a randomly selected utility bill is more than $140

1. Calculate the Z-score for $X = 140$: $Z = \frac{140 - 100}{13} = \frac{40}{13} \approx 3.0769$
2. Find the probability for $Z = 3.0769$, which is approximately 0.9991.
3. Since we want the probability of being more than 140, we calculate: $1 - 0.9991 = 0.0009$

Summary of Answers

• (a) $P(X < 69) \approx 0.0086$
• (b) $P(88 < X < 110) \approx 0.6006$
• (c) $P(X > 140) \approx 0.0009$

Would you like further details on any of these steps or have any questions?

Here are five related questions to deepen your understanding:

1. How would the probabilities change if the standard deviation were larger or smaller?
2. What would be the probability if we looked for utility bills between $95 and $105?
3. Can you explain why the probability for bills greater than $140 is so small?
4. How would you interpret a Z-score of 0?
5. What is the total probability for all values below the mean of $100?

Tip: Remember, in normal distributions, about 68% of values lie within one standard deviation, 95% within two, and 99.7% within three standard deviations. This is helpful for quick estimations.