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The Information Systems Audit and Control Association surveyed office workers to learn about the anticipated usage of office co...Asked by Jenna
The Information Systems Audit and Control Association surveyed office workers to learn about the anticipated usage of office computers for personal holiday shopping. Assume that the number of hours a worker spends doing holiday shopping on an office computer follows an exponential distribution.
Round your answers to four decimal places.
a.)The study reported that there is a .53 probability that a worker uses the office computer for holiday shopping 5 hours or less. Is the mean time spent using an office computer for holiday shopping closest to 5.5, 6, 6.6, or 7 hours?
The answer is 6.6.
b.)Using the mean time from part (a), what is the probability that a worker uses the office computer for holiday shopping more than 10 hours?
c.) What is the probability that a worker uses the office computer for holiday shopping between four and eight hours?
Round your answers to four decimal places.
a.)The study reported that there is a .53 probability that a worker uses the office computer for holiday shopping 5 hours or less. Is the mean time spent using an office computer for holiday shopping closest to 5.5, 6, 6.6, or 7 hours?
The answer is 6.6.
b.)Using the mean time from part (a), what is the probability that a worker uses the office computer for holiday shopping more than 10 hours?
c.) What is the probability that a worker uses the office computer for holiday shopping between four and eight hours?
Answers
Answered by
Nonetheless
If the mean = 6.6, then the probability density function is:
f(x) = (1/6.6)e^(-x/6.6)
P(x>10) = 1- int_{0}^{10} ((1/6.6)e^(-x/6.6))dx
= e^(-10/6.6)
~ 0.2198
P(4<x<8) = P(x>4) - P(x>8)
= e^(-4/6.6) - e^(-8/6.6)
~ 0.2479
f(x) = (1/6.6)e^(-x/6.6)
P(x>10) = 1- int_{0}^{10} ((1/6.6)e^(-x/6.6))dx
= e^(-10/6.6)
~ 0.2198
P(4<x<8) = P(x>4) - P(x>8)
= e^(-4/6.6) - e^(-8/6.6)
~ 0.2479
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