
#1
Dec2512, 08:24 PM

P: 49

The Information Systems Audit and Control Association surveyed office workers to learn about the anticipated usage of office computers for holiday shopping. Assume that the number of hours a worker spends doing holiday shopping on an office computer follows an exponential distribution.
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 5.8,6.2,6.6, or 7 hours? b) Using the mean time from part a), whats 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 fr holiday shopping between 4 and 8 hours? I just dont know how to calculate the mean. Once i get that I'll be able to do parts b) and c). Any help with part a) will be very helpful.thanks! 



#2
Dec2612, 04:23 AM

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Do you know the general form of the exponential distribution?




#3
Dec2612, 08:49 AM

P: 49

f(x)= 1/a (e)^ x/a
where a is the mean? 



#4
Dec2812, 03:07 AM

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exponential distribution word problem
Good. In terms of time, you may prefer: $$p(t)=\frac{1}{\tau}e^{t/\tau}$$... where ##\tau## is the mean.
Can you turn that into an expression for the probability that the time is less than some specified value T : $$p(t<T)=\cdots$$ 


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