Requests to a Web server. In Exercise 4.175 (p. 282) youlearned that Brighton Webs LTD modeled the arrivaltime of requests to a Web server within each hour, using auniform distribution. Specifically, the number of seconds xfrom the start of the hour that the request is made is uniformly

distributed between 0 and 3,600 seconds. In a randomsample of n= 60 Web server requests, letrepresentthe sample mean number of seconds from the start of thehour that the request is made.

1. Find $\mathbf{E}\left(\overline{\mathbf{x}}\right)$and interpret its value.
2. Find $\mathbf{Var}\left(\overline{\mathbf{x}}\right)$.
3. Describe the shape of the sampling distribution of $\overline{\mathbf{x}}$.
4. Find the probability that $\overline{\mathbf{x}}$is between 1,700 and 1,900seconds.
5. Find the probability that $\overline{\mathbf{x}}$exceeds 2,000 seconds.

Referring to exercise 4.175 (p 282), consider that Brighton Webs LTD modeled the arrival time of requests to a web server within each hour.

Here x, the number of seconds from the start of the hour. The request is made from distribution on the interval [0,3600]. A random sample of 60 observation has mean of $\overline{x}$.

Let us consider the mean of the random variable

$\begin{array}{rcl}\mu & =& \frac{a+b}{2}\\ & =& \frac{0+3600}{2}\\ & =& 1800\\ & & \end{array}$

By the central limit theorem, there can be concluded that if the sample size is large then the mean of the sample follows the normal distribution with mean $\mathbf{\mu }$and variance $\frac{{\mathbf{\sigma }}^{\mathbf{2}}}{\mathbf{n}}$.

Therefore, the expected value of $\overline{x}$is,

$$\begin{gathered} E\left(\overline{x}\right)=\mu \\ =1800 \end{gathered}$$

Thus, the mean of the seconds from the start of the hour for the sampling distribution of $\overline{x}$is 1800.

Let us consider the variance of the random sample,

$\begin{array}{rcl}{\sigma }^2& =& \frac{{\left(b-a\right)}^2}{12}\\ & =& \frac{{\left(3600-0\right)}^2}{12}\\ & =& 1080000\\ & & \end{array}$

So, form the central limit theorem,

$\begin{array}{rcl}Var\left(\overline{x}\right)& =& \frac{{\sigma }^2}{n}\\ & =& \frac{1080000}{60}\\ & =& 18000\\ & & \end{array}$

Thus, the variance is 18000.

As there is stated that if the sample size is large then the mean of the sample follows a normal distribution. So, the shape of this sampling distribution with size n=60 is same as a normal distribution.

Consider the probability that $\overline{x}$is between 1700 and 1900. That is,

$\begin{array}{rcl}\mathrm{Pr}\left(1700<\overline{x}<1900\right)& =& \mathrm{Pr}\left[\frac{1700-1800}{\sqrt{18000}}<\frac{\overline{x}-\mu }{\sqrt{Var\left(\overline{x}\right)}}<\frac{1900-1800}{\sqrt{18000}}\right]\\ & =& \mathrm{Pr}\left(-0.75<Z<0.75\right)\\ & =& \mathrm{Pr}\left(Z<0.75\right)-\mathrm{Pr}\left(Z<-0.75\right)\\ & =& 0.2734+0.2734\\ & =& 0.5468\\ & & \end{array}$

Thus, the required probability is 0.5468.

Step 5: Calculate the probability

Consider the probability that $\overline{x}$exceed 2000 seconds, that is,

$\begin{array}{rcl}\mathrm{Pr}\left(\overline{x}>2000\right)& =& \mathrm{Pr}\left[\frac{\overline{x}-\mu }{\sqrt{Var\left(\overline{x}\right)}}>\frac{2000-1800}{\sqrt{18000}}\right]\\ & =& \mathrm{Pr}\left[Z>1.49\right]\\ & =& 1-\mathrm{Pr}\left[Z⩽1.49\right]\\ & =& 1-0.93183\\ & =& 0.0681\\ & & \end{array}$

Thus, the required probability is 0.0681.