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

 

[blue_raver22]

a) To calculate the mean (μ) of an exponential distribution, we can use the formula μ = 1/λ, where λ is the rate parameter. In this case, since we are given the probability of a worker spending 5 hours or less on holiday shopping (P(x ≤ 5) = 0.53), we can set up the following equation:

0.53 = 1 - e^(-λ * 5)

Solving for λ, we get λ ≈ 0.126.

Therefore, the mean time spent using an office computer for holiday shopping is closest to 6.6 hours (1/0.126 ≈ 6.6).

b) Using the mean time of 6.6 hours, we can calculate the probability of a worker spending more than 10 hours on holiday shopping by using the formula P(x > 10) = 1 - e^(-λ * 10). Plugging in our value of λ = 0.126, we get P(x > 10) ≈ 0.06 or 6%.

c) To calculate the probability of a worker spending between 4 and 8 hours on holiday shopping, we can use the formula P(4 < x < 8) = e^(-λ * 4) - e^(-λ * 8). Plugging in our value of λ = 0.126, we get P(4 < x < 8) ≈ 0.36 or 36%.