The weekly demand for a particular automobile manufacturer follows a normal distribution with a mean of 40,000 cars and a standard deviation of 10,000. Below you will find probability and percentile calculations related to the customer purchase amounts. Use this information to answer the following questions.Probability CalculationsP(Sales < 2,000,000) = 0.134, P(Sales < 2,050,000) = 0.339P(Sales < 2,100,000) = 0.609, P(Sales < 2,150,000) = 0.834Percentiles Calculations1st percentile = 1,912,245, 5th percentile = 1,961,38895th percentile = 2,198,612, 99th percentile = 2,247,755
What number of cars, equidistant from the mean, such that 98% of car sales are between these values?
What will be an ideal response?
1,912,245 to 2,247,755
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