Question

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.

Answer 1

p=0.1971

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:

`P(1100<X<1225)=P((1100-\mu)/(\sigma)<(X-\mu)/(\sigma)<(1225-\mu)/(\sigma))\\\\=P((1100-1050)/(218)<Z<(1225-1050)/(218))\\\\=P(0.2294<Z<0.8028)\\\\=P(Z<0.8028)-P(Z<0.2994)\\\\=0.7881-0.5910\\\\=0.1971`

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