According to a government energy agency, the mean monthly household electricity bill in the United States in 2011 was $109.55. Assume the amounts are normally distributed with standard deviation .

To solve these problems, we will use the standard normal distribution (Z-distribution). We have the following information:

• Mean (\(\mu\)) = $109.55
• Standard deviation (\(\sigma\)), which needs to be defined; however, since we do not have the standard deviation value, we cannot compute exact probabilities. Let's denote it as \(\sigma\).

To provide answers to the questions, I'll outline the steps used for calculations.

(a) What proportion of bills are greater than $135?

1. Calculate the Z-score for $135:

\[ Z = \frac{X - \mu}{\sigma} = \frac{135 - 109.55}{\sigma} \]

2. Use the Z-score to find the corresponding probability from the standard normal distribution table (or use a calculator).

3. The proportion of bills greater than $135 is:

\[ P(X > 135) = 1 - P(Z \leq Z) \]

(b) What proportion of bills are between $84 and $143?

1. Find the Z-scores for both amounts:

For $84:

\[ Z_{84} = \frac{84 - 109.55}{\sigma} \]

For $143:

\[ Z_{143} = \frac{143 - 109.55}{\sigma} \]

2. Find the probability for both Z-scores using a standard normal distribution table:

\[ P(84 < X < 143) = P(Z_{143}) - P(Z_{84}) \]

(c) What is the probability that a randomly selected household had a monthly bill less than $119?

1. Calculate the Z-score for $119:

\[ Z = \frac{119 - 109.55}{\sigma} \]

2. Find the probability from the standard normal distribution:

\[ P(X < 119) = P(Z \leq Z) \]

Summary

As we do not have the standard deviation value \(\sigma\), we cannot compute exact numerical answers. If you provide the value for the standard deviation, I can help you compute the exact proportions.

Just to clarify how you would finish this, let’s say \(\sigma = 20\) as an example:

1. For (a), you would compute for $135.
2. For (b), you would compute for $84 and $143.
3. For (c), you would compute for $119.

Each step would involve looking up or calculating the normal probabilities based on actual Z-scores. If you provide the value of \(\sigma\), I can calculate exact answers.

Great! Now that we have the standard deviation (\(\sigma = 21.00\)), we can compute the required probabilities.

(a) Proportion of bills greater than $135:

1. Calculate the Z-score for $135:

\[ Z = \frac{135 - 109.55}{21} = \frac{25.45}{21} \approx 1.2143 \]

2. Find the probability using the Z-table: Using the Z-table, we find \(P(Z \leq 1.2143)\). The corresponding probability is approximately \(0.8888\).

3. Calculate the proportion of bills greater than $135:

\[ P(X > 135) = 1 - P(Z \leq 1.2143) \approx 1 - 0.8888 = 0.1112 \]

(b) Proportion of bills between $84 and $143:

1. Calculate the Z-scores:

   • For $84: \[ Z_{84} = \frac{84 - 109.55}{21} = \frac{-25.55}{21} \approx -1.2167 \]

   • For $143: \[ Z_{143} = \frac{143 - 109.55}{21} = \frac{33.45}{21} \approx 1.5905 \]

2. Find the probabilities:

   • \(P(Z \leq -1.2167) \approx 0.1101\) (from the Z-table).
   • \(P(Z \leq 1.5905) \approx 0.9429\).

3. Calculate the proportion of bills between $84 and $143:

\[ P(84 < X < 143) = P(Z \leq 1.5905) - P(Z \leq -1.2167) \approx 0.9429 - 0.1101 = 0.8328 \]

(c) Probability that a randomly selected household had a monthly bill less than $119:

1. Calculate the Z-score for $119:

\[ Z = \frac{119 - 109.55}{21} = \frac{9.45}{21} \approx 0.4500 \]

2. Find the corresponding probability:

   \(P(Z \leq 0.4500) \approx 0.6736\) (from the Z-table).

Final Answers:

• (a) The proportion of bills greater than $135 is approximately 0.1112.
• (b) The proportion of bills between $84 and $143 is approximately 0.8328.
• (c) The probability that a randomly selected household had a monthly bill less than $119 is approximately 0.6736.

These values can be rounded to at least four decimal places as requested.