Let T denote the time it takes for a computer to shut down S

Let T denote the time it takes for a computer to shut down. Suppose T follows an Exponential distribution with mean 0.80 minutes. A computer lab has 6 independent computers that must all be shut down at the end of the day.

a) What is the probability that at most 2 computers take longer than 1 minute to shut down on a given day?

b) What is the probability that at least 2 computers take longer than 2 minutes to shut down on a given day?

Solution

Let T denote the time it takes for a computer to shut down. Suppose T follows an Exponential distribution with mean 0.80 minutes. A computer lab has 6 independent computers that must all be shut down at the end of the day.

a) What is the probability that at most 2 computers take longer than 1 minute to shut down on a given day?

For the exponential distribution:

P(X >= x) = e^(-x/mean)

Since the mean is .8

P(X >=x) = e^(-x/.8) = e^(-1.25x)

The probability that a computer takes longer than 1 minute = e^(-1.25*1) = e^-1.25 = .286505

The probability that at most 2 computers out of 6 computers which are independent taking longer than 1 minute to shut down is binomial

For a binomial distribution, P(X=x) = ncx*p^x*(1-p)^(n-x)

n = 6, p = .286505, at most 2 would be 0, 1, or 2

P(X=0) = 6c0*.286505^0*(1-.286505)^(6-0) = .131931

P(X=1) = 6c1*.286505^1*(1-.286505)^(6-1) = .317861

P(X=2) = 6c2*.286505^2*(1-.286505)^(6-2) = .319095

Add up the probabilities: .131931 + .317861 + .319095 = .768887

Answer: The probability that at most 2 computers out of 6 take longer than 1 minute to shut down is 0.7689

(alternatively you can use a calculator with the binomcdf function, binomcdf(6, .286505, 2) = 0.7689

(b) What is the probability that at least 2 computers take longer than 2 minutes to shut down on a given day?

At least 2 is 2 or more

P(X is at least 2) = 1