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), what's 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 don't 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 : $$p(t<T)=\cdots$$