exponential distribution word problem

salma17

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!

Simon Bridge

Do you know the general form of the exponential distribution?

salma17

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

Simon Bridge

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 : $$p(t<T)=\cdots$$