Question

In one region. the Sep. energy consumption levels for a single-family are to be normally distributed with a mean of 1,050 kWh and standard deviation of 218 kWh. If 50 different homes are randomly selected, find probability that their mean energy consumption is greater than 1,075 kWh.

Answer 1

The problem asks about the probability of the (sample) mean, so you have that the sample mean \(\bar{X}\) is normally distributed with mean the same as the population mean \(\mu =1050\)
and with standard deviation being \(\dfrac{\sigma}{\sqrt{n}}\). So, when you find the z-score, you need to do the following:
\[ Z=\frac{\bar{X}-\mu}{\sigma/\sqrt{n}}\]
So the question asks to find \(P(\bar{X}> 1075)\)
, and so by standardizing you get:
\[P\left( \frac{\bar{X}-1050}{218/\sqrt{50}}>\frac{1075-1050}{218/\sqrt{50}}\right) =P(Z>0.811)\] and then use a standard normal table to fnd that probability, or a calculator/software.