To solve this problem, we need to find the 45th percentile of a normally distributed dataset with a mean ($\mu$) of 1050 kWh and a standard deviation ($\sigma$) of 218 kWh.

Solution Steps

1. Identify the Z-score: For a normal distribution, the 45th percentile corresponds to a cumulative probability of 0.45. We can find the Z-score for this percentile using standard normal distribution tables or statistical software.

   For the 45th percentile, the Z-score is approximately $-0.1257$ (since the 50th percentile is 0 in a standard normal distribution, the 45th percentile is slightly below the mean).

2. Apply the Z-score formula: The formula to find a specific percentile $X$ in a normal distribution is: $X = \mu + Z \cdot \sigma$ where:

   • $\mu = 1050$
   • $\sigma = 218$
   • $Z = -0.1257$

3. Calculate $X$: Substitute the values into the formula: $X = 1050 + (-0.1257) \cdot 218$ $X \approx 1050 - 27.93$ $X \approx 1022.07$

4. Choose the closest answer: The closest answer to 1022.07 is D. 1021.7.

Answer

The 45th percentile of the energy consumption distribution is approximately 1021.7 kWh (Option D).

Would you like more details on any step, or do you have any questions?

Here are some related questions to consider:

1. How do you calculate the Z-score for a given percentile?
2. What does the 45th percentile represent in the context of energy consumption?
3. How would you calculate the 95th percentile of the same distribution?
4. What is the significance of standard deviation in a normal distribution?
5. How does the mean shift affect the percentile values in a normal distribution?

Tip: Percentiles in a normal distribution can be quickly estimated using Z-scores for common percentiles, which saves time in manual calculations.