Question

Still waiting for the server? How does your web browser get a file from the Internet? Your computer sends a request for the file to a web server, and the web server sends back a response. Let Y = the amount of time (in seconds) after the start of an hour at which a randomly selected request is received by a particular web server. The probability distribution of Y can be modeled by a uniform density curve on the interval from 0 to 3600 seconds. Find the probability that the request is received by this server within the first 5 minutes (300 seconds) after the hour.

Short answer

The probability is$\frac{1}{12}\approx 0.0833$.

Step-by-step solution

Step 1. Given information.

It is given that "Y" is the amount of time (in seconds) after the start of an hour at which a randomly selected request is received by a particular web server.

A uniform density curve on the interval 0 to 3600 seconds can be used to simulate the probability distribution of Y.

Step 2. Find the probability.

$a=0;b=3600$

On the interval between the boundaries (0 somewhere), the density curve of a uniform distribution is the reciprocal of the difference between the boundaries:

$$\begin{gathered} f\left(x\right)=\frac{1}{b-a} \\ =\frac{1}{3600-0} \\ =\frac{1}{3600};0\le x\le 3600 \end{gathered}$$

The region beneath the density curve between 0 and 300 represents the probability that the time is fewer than 300 seconds. This region is then a rectangle with a width of role="math" localid="1654411338356" $300-0=300$and a height of $\frac{1}{3600}$.

The product of the width and height equals the area of a rectangle.

role="math" localid="1654411447841" $$\begin{gathered} P\left(X<300\right)=\left(300-0\right)\times \frac{1}{3600} \\ =\frac{300}{3600} \\ =\frac{1}{12}\approx 0.0833 \end{gathered}$$

As a result, on roughly 28% of the days, Sally must wait between 2.5 and 5.3 minutes for the bus.