In one region, the September energy consumption levels for single-family homes are found to be normally distributed with a mean
1 answer:
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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<em>Additional comment</em>
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.
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