Find the indicated probability. 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.

p=0.1971

Step-by-step explanation:

-Let X be the normally distributed random variable with mean=1050kWh and standard deviation=218kWh.

-We use the z-test to test for the probability of mean consumption being between 1100kWh and 1225kWh as:

Hence, the probability of mean consumption being between 1100 and 1225kWh is 0.1971

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