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.
A) 0.3791
B) 0.2881
C) 0.1971
D) 0.0910

Question

Quizplus · Accepted Answer

To find the probability that the energy consumption is between 1100 kWh and 1225 kWh, we use the Z-score formula: $Z = \frac{X - \mu}{\sigma}$, where $X$ is the value, $\mu$ is the mean, and $\sigma$ is the standard deviation. For 1100 kWh, $Z = \frac{1100 - 1050}{218} \approx 0.229$, and for 1225 kWh, $Z = \frac{1225 - 1050}{218} \approx 0.802$. Using the standard normal distribution table, the probabilities corresponding to these Z-scores are approximately 0.5904 and 0.7875, respectively. The probability of being between these two values is the difference, 0.7875 - 0.5904 = 0.1971.

In One Region, the September Energy Consumption Levels for Single