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probability, normal distribution

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 of218 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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Answer #1
Given X~Normal(mean=1050, s=218)

SO the probability is
P(1100<X<1225)
=P((1100-1050)/219 <(X-mean)/s <(1225-1050)/218)
=P( 0.23<Z< 0.8)
= 0.1972 (check standard normal table)
answered by: Alyss
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