Chapter 2.2, Problem 42E

(a)

To determine

Density curve to model the amount of time after an hour at which a request is received by the web server including scales on both axes.

Answer to Problem 42E

  $f\left(x\right)\approx 0.00027778$

When

  $0\le x\le 3600\text{s}$

Explanation of Solution

Given information:

The distribution has been modeled according to uniform distribution on the interval $0\le x\le 3600$ .

Such that

  $a=0$

And

  $b=3600$ .

Calculations:

With uniform distribution, the density curve is reciprocal to the difference of the boundaries.

On the interval between the boundaries,

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

With

  $0\le x\le 3600$ .

  

(b)

To determine

Proportion of requests received within the first 300 seconds after the hour.

Answer to Problem 42E

About 0.0833 of the requests are received within first 300 seconds after the hour.

Explanation of Solution

Given information:

Time for receiving requests,

  $0\le x\le 300\text{s}$

Calculations:

The area underneath the density curve for all the values preceding 300 will be the probability that the time of receiving requests is less than 300 seconds.

Note that

Area underneath the density curve will be the rectangle.

With

Width, $\text{W}=300-0=300$

And

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

Then

  $\begin{array}{l}P\left(0<x<300\right)=\text{Area}\text{of}\text{rectangle}\\ =\text{W}\times \text{H}\\ =300\times 0.00027778\\ \approx 0.0833\end{array}$

Therefore,

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

  

(c)

To determine

Interquartile range of the distribution.

Answer to Problem 42E

Interquartile range of the distribution is 1800 seconds.

Explanation of Solution

Given information:

The distribution has been modeled according to uniform distribution on the interval $0\le x\le 3600$ .

Such that

  $a=0$

And

  $b=3600$ .

Calculations:

Interquartile range is the difference between the 1^st and 3^rd quartile.

The property of 1^st quartile reveals that 25% of the data values are below it.

The property of 3^rd quartile reveals that 75% of the data values are below it.

Although, the distribution is uniform between $0\le x\le 3600$ , the 1^st quartile has 25% of the data values below it, which means 25% of the distribution will be the one − fourth of the interval.

Then

  $Q_1=3600\times \frac{1}{4}=900$

Although, the distribution is uniform between $0\le x\le 3600$ , the 3^rd quartile has 75% of the data values below it, which means 75% of the distribution will be the three − fourth of the interval.

Then

  $Q_3=3600\times \frac{3}{4}=2700$

The difference between the 1^st and 3^rd quartile gives the interquartile range.

Thus,

  $IQR=Q_3-Q_1=1800$

The units of both interquartile range and data values will be the same.

Therefore,

The interquartile range is 1800 seconds.