91Ó°ÊÓ

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 $\stackrel{¯}{x}$.

Let us consider the mean of the random variable

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

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 $\mathbf{μ}$and variance $\frac{{\mathbf{σ}}^{\mathbf{2}}}{\mathbf{n}}$.

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

$$\begin{gathered} E\left(\stackrel{¯}{x}\right)=\mathrm{μ} \\ =1800 \end{gathered}$$

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

Let us consider the variance of the random sample,

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

So, form the central limit theorem,

$\begin{array}{rcl}Var\left(\stackrel{¯}{x}\right)& =& \frac{{\mathrm{σ}}^2}{n}\\ & =& \frac{1080000}{60}\\ & =& 18000\\ & & \end{array}$

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 $\stackrel{¯}{x}$is between 1700 and 1900. That is,

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

Thus, the required probability is 0.5468.

Step 5: Calculate the probability

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

$\begin{array}{rcl}\mathrm{Pr}\left(\stackrel{¯}{x}>2000\right)& =& \mathrm{Pr}\left[\frac{\stackrel{¯}{x}-\mathrm{μ}}{\sqrt{Var\left(\stackrel{¯}{x}\right)}}>\frac{2000-1800}{\sqrt{18000}}\right]\\ & =& \mathrm{Pr}\left[Z>1.49\right]\\ & =& 1-\mathrm{Pr}\left[Zâ\copyright ½1.49\right]\\ & =& 1-0.93183\\ & =& 0.0681\\ & & \end{array}$

Thus, the required probability is 0.0681.