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. If 50 different homes are randomly selected, find the probability that their mean energy consumption level for September is greater than 1075 kWh.
Solution :
Given that ,
mean = = 1050
standard deviation = = 218
n = 50
_{} = 1050 and
_{} = / n = 218 / 50 = 30.8299
P( > 1075) = 1 - P( < 1075)
= 1 - P(( - _{} ) / _{} < (1075 - 1050) / 30.8299)
= 1 - P(z < 0.81)
= 1 - 0.791
= 0.209
Probability = 0.209
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