Question

The waiting time for service at a hospital emergency department (in hours) follows a distribution with probability density function f(x) = 0.5exp(-0.5x) for 0 < x. Determine the following:

Answer 1

The waiting time for service at a hospital emergency department follows an exponential distribution. To find the probability P(2 < x < 5), we need to integrate the probability density function (PDF) over the given interval. The approximate probability P(2 < x < 5) is 0.3176.

Step-by-step explanation:

The waiting time for service at a hospital emergency department follows an exponential distribution with the probability density function (PDF) f(x) = 0.5exp(-0.5x) for x > 0. To determine the probability P(2 < x < 5), we need to integrate the PDF over this interval.

To do the integration, we can calculate the definite integral of the PDF from 2 to 5. This will give us the probability of the waiting time being between 2 and 5 hours.

The integral of the PDF f(x) = 0.5exp(-0.5x) from 2 to 5 is:

∫[0.5exp(-0.5x)]dx, evaluated from 2 to 5.

Using the antiderivative of the PDF, which is -exp(-0.5x), we can evaluate the definite integral:

-[exp(-0.5x)] evaluated from 2 to 5.

Plugging in the upper and lower limits:

-[exp(-0.5(5)) - exp(-0.5(2))].

Simplifying the expression:

-[exp(-2.5) - exp(-1)].

Using a calculator, we can find that the probability P(2 < x < 5) is approximately 0.3176.