Question 341087: Demand for residential electricity at 6:00 P.M. on the first Monday in October in Santa Theresa County is normally distributed with a mean of 4,905 MW (megawatts) and a standard deviation of 355 MW. Due to scheduled maintenance and unexpected system failures in a generating station the utility can supply a maximum of 5,200 MW at that time. What is the probability that the utility
will have to purchase electricity from other utilities or allow brownouts?
So I have tried this:
(5200-4905)/355 = 0.830986
I do not know if I have answered this question correctly. Please review and help me, if you can.
Thank you!!
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Demand for residential electricity at 6:00 P.M. on the first Monday in October in Santa Theresa County is normally distributed with a mean of 4,905 MW (megawatts) and a standard deviation of 355 MW.
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Due to scheduled maintenance and unexpected system failures in a generating station the utility can supply a maximum of 5,200 MW at that time.
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What is the probability that the utility will have to purchase electricity from other utilities or allow brownouts?
So I have tried this:
z(5200) = (5200-4905)/355 = 0.8310
Note: You have found the z-value associated with 5200.
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You now need to find the probability that z is greater than 0.8310
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P(x > 5200) = P(z > 0.8310) = normalcdf(0.8310,100) = 0.2030
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Cheers,
Stan H.
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