Q:

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. your answer should be a decimal rounded to the fourth decimal place.

Accepted Solution

A:
We first calculate the z-score corresponding to x = 1075 kWh. Given the mean of 1050 kWh, SD of 218 kWh, and sample size of n = 50, the formula for z is:
z = (x - mean) / (SD/sqrt(n)) = (1075 - 1050) / (218/sqrt(50)) = 0.81
From a z-table, the probability that z > 0.81 is 0.2090. Therefore, the probability that the mean of the 50 households is > 1075 kWh is 0.2090.