In one region, the September energy consumption levels for single -family homes are found to be normally distributed with a mean of 1050 kWh and a standard deviation of 218 kWh. For a randomly selected home, find the probability that the September energy consumption level is between 1100 kWh and 1225 kWh.

To find the probability that the September energy consumption level is between 1100 kWh and 1225 kWh, we can standardize the values and use the standard normal distribution (Z-distribution).

### Given Data:
- Mean, \( \mu = 1050 \) kWh
- Standard deviation, \( \sigma = 218 \) kWh
- Lower bound, \( X_1 = 1100 \) kWh
- Upper bound, \( X_2 = 1225 \) kWh

We will calculate the Z-scores for both values and then find the area between these Z-scores.

### Step 1: Calculate the Z-scores
The Z-score formula is:
\[
Z = \frac{X - \mu}{\sigma}
\]
For \( X_1 = 1100 \) kWh:
\[
Z_1 = \frac{1100 - 1050}{218} = \frac{50}{218} \approx 0.2294
\]
For \( X_2 = 1225 \) kWh:
\[
Z_2 = \frac{1225 - 1050}{218} = \frac{175}{218} \approx 0.8028
\]

### Step 2: Find the probabilities
Using the standard normal distribution table or a calculator:

- The probability corresponding to \( Z_1 = 0.2294 \) is approximately 0.5910.
- The probability corresponding to \( Z_2 = 0.8028 \) is approximately 0.7888.

### Step 3: Calculate the probability between the two Z-scores
The probability that the energy consumption is between 1100 kWh and 1225 kWh is the difference between the two probabilities:
\[
P(1100 \leq X \leq 1225) = P(Z_2) - P(Z_1) = 0.7888 - 0.5910 = 0.1978
\]

### Final Answer:
The probability that the energy consumption level is between 1100 kWh and 1225 kWh is approximately **0.1978**, or **19.78%**.

Would you like more details on any of these steps or further explanations?

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Here are 5 related questions:
1. How do you interpret the result of a Z-score?
2. What is the probability that energy consumption exceeds 1225 kWh?
3. How do you find the probability that energy consumption is less than 1100 kWh?
4. Can you calculate the probability for consumption between 1000 kWh and 1300 kWh?
5. How would the result change if the standard deviation were higher?

**Tip:** When using a Z-table, remember that it gives the cumulative probability up to a Z-score. Subtract to find probabilities between two Z-scores.

In one region, the September energy consumption levels for single-family homes are found to be normally distributed with a mean of 1050 kWh and a standard deviation of 218 kWh. For a randomly selected home, find the probability that the September energy consumption level is between 1100 kWh and 1225 kWh.

Find the probability that the September energy consumption for a single-family home falls between 1100 kWh and 1225 kWh, assuming a normal distribution with a mean of 1050 kWh and a standard deviation of 218 kWh. Learn how to calculate Z-scores and use the standard normal distribution to solve this statistics problem. Suitable for Grades 10-12.

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