Question 1064873
In one region, the September energy consumption levels for single -family homes are found to be normally distributed with a mean of 1060 kWh and a standard deviation of 233 kWh. If 44 different homes are randomly selected, find the probability that their mean energy consumption level for September is greater than 1045 kWh.
-----------------
z(1045) = (1045-1060)/(233/sqrt(44)) = -15*sqrt(44)/233 = -0.4270
====
P(x > 1045) = P(z > -0.4270) = 0.6653
----------------
Cheers,
Stan H.
------------