Chapter 4, Problem 98SE

Let X = the time it takes a read/write head to locate a desired record on a computer disk memory device once the head has been positioned over the correct track. If the disks rotate once every 25 millisec, a reasonable assumption is that X is uniformly distributed on the interval [0, 25].

a. Compute P(10 ≤ X ≤ 20).

b. Compute P(X ≥ 10).

c. Obtain the cdf F(X).

d. Compute E(X) and σ_X.

To determine

Compute $P\left(10\le X\le 20\right)$.

Answer to Problem 98SE

The value of $P\left(10\le X\le 20\right)$ is 0.4.

Explanation of Solution

Given info:

The time taken to read/write head to locate the record on a computer disk memory device over the correct track follows uniform distribution in $\left[0,25\right]$.

Calculation:

The probability density function of uniform distribution is given below:

$P\left(x;A,B\right)=\{\begin{array}{l}\frac{1}{B-A}A\le x\le B\\ 0\text{otherwise}\end{array}$

The probability density function of uniform distribution for the interval $\left[0,25\right]$ is given below:

$P\left(x;0,25\right)=\{\begin{array}{l}\frac{1}{25}0\le x\le 25\\ 0\text{otherwise}\end{array}$

The value of $P\left(10\le X\le 20\right)$ is obtained as shown below:

$\begin{array}{c}P\left(10\le X\le 20\right)={\displaystyle \underset{10}{\overset{20}{\int }}\frac{1}{25}dx}\\ =\frac{1}{25}{\left[x\right]}_{10}^{20}\\ =\frac{1}{25}\left[20-10\right]\\ =\frac{10}{25}\\ =0.4\end{array}$

Thus, the value of $P\left(10\le X\le 20\right)$ is 0.4.

To determine

Compute $P\left(X\ge 10\right)$.

Answer to Problem 98SE

The value of $P\left(X\ge 10\right)$ is 0.6.

Explanation of Solution

Calculation:

The value of $P\left(X\ge 10\right)$ is obtained as shown below:

$\begin{array}{c}P\left(X\ge 10\right)={\displaystyle \underset{10}{\overset{25}{\int }}\frac{1}{25}dx}\\ =\frac{1}{25}{\left[x\right]}_{10}^{25}\\ =\frac{1}{25}\left[25-10\right]\\ =\frac{15}{25}\\ =0.6\end{array}$

Thus, the value of $P\left(X\ge 10\right)$ is 0.6.

To determine

Obtain cumulative density function of $F\left(X\right)$.

Answer to Problem 98SE

The cumulative density function $F\left(X\right)$ is,

$\underset{\_}{F\left(X\right)=\{\begin{array}{l}0x\le 0\\ \frac{x}{25}0<x\le 25\\ 1x>25\end{array}}$

Explanation of Solution

Calculation:

The cumulative density function of uniform distribution is given below:

$F\left(X\right)={\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx}$

The cumulative density function $F\left(X\right)$ is obtained as shown below:

The random variable X is uniformly distributed over $\left[0,25\right]$

For $\left[0<x\le 25\right]$,

$\begin{array}{c}F\left(X\right)={\displaystyle \underset{0}{\overset{x}{\int }}\frac{1}{25}du}\\ =\frac{1}{25}{\left[u\right]}_0^x\\ =\frac{x}{25}\end{array}$

Since the range of $\left[0,25\right)$, all the values of x lie above zero.

For $\left[x\le 0\right]$,

$F\left(X\right)=0$

Since the range of $\left[0,25\right]$, all the values of x lie below 25.

For $\left[x>25\right]$,

$F\left(X\right)=1$

Thus, the cumulative density function $F\left(X\right)$ is $\underset{\_}{F\left(X\right)=\{\begin{array}{l}0x\le 0\\ \frac{x}{25}0<x\le 25\\ 1x>25\end{array}}$.

To determine

Compute $E\left(X\right)$ and $V\left(X\right)$.

Answer to Problem 98SE

The value of $E\left(X\right)$ is 12.5.

The value of ${\sigma }_X$ is 7.22.

Explanation of Solution

Calculation:

The mean of the uniform distribution is given below:

$E\left(X\right)=\frac{A+B}{2}$

Substitute $A=0$ and $B=25$,

$\begin{array}{c}E\left(X\right)=\frac{0+25}{2}\\ =12.5\end{array}$

Thus, the value of $E\left(X\right)$ is 12.5.

The $V\left(X\right)$ is obtained as shown below:

The variance of the uniform distribution is given below:

$V\left(X\right)=\frac{{\left(B-A\right)}^2}{12}$

Substitute $A=0$ and $B=25$,

$\begin{array}{c}V\left(X\right)=\frac{{\left(25-0\right)}^2}{12}\\ =\frac{{25}^2}{12}\\ =\frac{625}{12}\\ =52.083\end{array}$

The standard deviation is obtained as given below:

$\begin{array}{c}{\sigma }_X=\sqrt{V\left(X\right)}\\ =\sqrt{52.083}\\ =7.22\end{array}$

Thus, the value of ${\sigma }_X$ is 7.22.