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 webserver sends back

a response. For one particular web server, the time (in seconds) after the start of an hour at which a request is received can be modeled by a uniform distribution on the interval from $0$ to $3600$ seconds.

a. Draw a density curve to model the amount of time after an hour at which a request is received by the webserver. Be sure to include scales on both axes.

b. About what proportion of requests are received within the first $5$ minutes ($300$ seconds) after the hour?

c. Find the interquartile range of this distribution.

On interval $0\le x\le 3600$, the distribution was modeled using a uniform distribution.

Such that

$a=0$

And

$b=3600$

Time for receiving requests,

$0\le x\le 300s$

A density curve is always on or above the horizontal axis.

The density curve for a uniform distribution is proportional to the difference between the boundaries.

In the space between the two limits,

$f\left(x\right)=\frac{1}{b-a}=\frac{1}{3600-0}=\frac{1}{3600}\approx 0.00027778$

With

$0\le x\le 3600$

The likelihood that the time it takes to receive requests is less than $300$ seconds will be represented by the area beneath the density curve for all values before $300$

Note that

The rectangle will be the area beneath the density curve.

With

Width,$W=300-0=300$

And

Height, $H=f\left(x\right)=0.00027778$

Then

Therefore,

About $0.0833$of the requests are received within the first $300$seconds ($5$minutes) after the hour.

The difference between the first and third quartiles is the interquartile range.

The $1st$ quartile's characteristic demonstrates that $25\%$ of the data values are below it.

The $3rd$ quartile's characteristic demonstrates that $75\%$ of the data values are below it.

Despite the fact that the distribution is uniform between $0$ and $3600,$ the $1st$ quartile contains $25\%$ of the data values below it, implying that $25\%$ of the distribution will be one-fourth of the interval.

Then

$Q_3=3600\times \frac{3}{4}=2700$

The interquartile range is the difference between the first and third quartiles.

Thus,

$IQR=Q_3-Q_1=1800$

Both the interquartile range and the data values will be expressed in the same units.

Therefore,

The interquartile range is $1800$seconds.