Solve the problem.
-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. If 50 different homes are randomly selected, find the probability that their mean energy consumption level for September is greater than 1075 kWh.
A)0.2090
B)0.0438
C)0.2910
D)0.4562

Question

Quizplus · Accepted Answer

To find the probability that the mean energy consumption level for September is greater than 1075 kWh for a sample of 50 homes, we use the formula for the z-score of a sample mean: $z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}}$, where $\bar{x}$ is the sample mean, $\mu$ is the population mean, $\sigma$ is the population standard deviation, and $n$ is the sample size. Plugging in the given values, we get $z = \frac{1075 - 1050}{218/\sqrt{50}}$. Calculating this gives a z-score, which we then look up in a standard normal distribution table or use a calculator to find the probability corresponding to this z-score. The result is approximately 0.2090, indicating the probability that the mean energy consumption level for September is greater than 1075 kWh.

Solve the Problem. -In One Region, the September Energy Consumption Levels for Single-Family