Question

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 greater than 1100 kWh.
Please show work!

Answer 1

Answer:

  0.4093

Step-by-step explanation:

A suitable probability calculator will answer this question with no "work" required, other than to enter the numbers into the calculator. The attachment shows the probability of interest is about 0.4093 that consumption will exceed 1100 kWh.

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Additional comment

All spreadsheets have probability functions built in. Many graphing calculators are similarly capable of giving area from probability, or probability from area. A TI-84 requires two limits. In most cases, an upper limit of about 10 standard deviations from the mean (here, 3000 kWh) will work just fine. It tells me ...

  normalcdf(1100,3000,1050,218) = 0.409295417

This is in agreement with my phone app: 0.4092954170395281.