question archive Suppose that the time required to complete a 1040R tax form is normally distributed with a mean of 100 minutes and a standard deviation of 10 minutes
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Suppose that the time required to complete a 1040R tax form is normally distributed with a mean of 100 minutes and a standard deviation of 10 minutes. What proportion of 1040R tax forms will be completed in at most 93 minutes? Round your answer to at least four decimal places.
Answer:
µ = 100 ; ? = 10
X = 93
Z = (X - µ) / ?
= (93 - 100) / 10 = -0.7
P(Z > -0.7) = 1 - P(Z ? -0.7) = 1 - 0.2419 = 0.7581 = 75.81%