question archive 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
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
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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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Answer:
P(1100 <X<1225) = 0.1971
Step-by-step explanation
Given that mean = 1050 and standard deviation = 218
we have to find P(1100 <X<1225)
= ?P(x−μ/σ?<z<y−μ/σ?)
=P(1100−1050/218?<z<1225−1050/218 )
=P(0.23<z<0.80) =P(z<0.8)−P(z<0.23)
=0.7881−0.5910
.......................using standard normal distribution table =0.1971?
