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.

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03-04-2020
Mathematics

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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.

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opudodennis opudodennis · Answer 1 · 2020-04-12T04:04:57+02:00

Answer:

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:

[tex]P(1100<X<1225)=P(\frac{1100-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{1225-\mu}{\sigma})\\\\=P(\frac{1100-1050}{218}<Z<\frac{1225-1050}{218})\\\\=P(0.2294<Z<0.8028)\\\\=P(Z<0.8028)-P(Z<0.2994)\\\\=0.7881-0.5910\\\\=0.1971[/tex]

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