Chapter 2.2, Problem 41E

(a)

To determine

Density curve to model the amount of time that Sally has to wait for the bus including scales on both axes.

Answer to Problem 41E

  $f\left(x\right)=0.1$

When

  $0\le x\le 10$

Explanation of Solution

Given information:

The distribution has been modeled according to uniform distribution on the interval $0\le x\le 10\min$ .

Such that

  $a=0$

And

  $b=10$ .

Calculations:

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

On the interval between the boundaries,

  $f\left(x\right)=\frac{1}{b-a}=\frac{1}{10-0}=\frac{1}{10}=0.1$

With

  $0\le x\le 10$ .

  

(b)

To determine

Percentage of days that Sally has to wait between 2.5 and 5.3 minutes for the bus.

Answer to Problem 41E

Sally has to wait for the bus for about 28% of days.

Explanation of Solution

Given information:

Waiting time,

  $2.5\le x\le 5.3\min$

Calculations:

The area underneath the density curve between $2.5\le x\le 5.3\min$ , will be the probability that the waiting time is between $2.5\le x\le 5.3\min$ .

Note that

Area underneath the density curve will be the rectangle.

With

Width, $\text{W}=5.3-2.5=2.8$

And

Height, $\text{H}=f\left(x\right)=0.1$

Then

  $\begin{array}{l}P\left(2.5<x<5.3\right)=\text{Area}\text{of}\text{rectangle}\\ =\text{W}\times \text{H}\\ =2.8\times 0.1\\ =0.28\\ =28\%\end{array}$

Therefore,

Sally has to wait for the bus between 2.5 and 5.3 minutes for about 28% of the days.

  

(c)

To determine

70^th percentile of Sally’s wait times.

Answer to Problem 41E

70^th percentile of Sally’s wait times will be 7 minutes.

Explanation of Solution

Given information:

The distribution has been modeled according to uniform distribution on the interval $0\le x\le 10\min$ .

Such that

  $a=0$

And

  $b=10$ .

Calculations:

According to the property for the 70^th percentile, 70% of the data values should be smaller than the 70^th percentile.

Let

xbe the 70^th percentile.

The area underneath the density curve between 0 and x , will be the probability that the time is between 0 and x .

Note that

Area underneath the density curve will be the rectangle.

With

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

And

Height, $\text{H}=f\left(x\right)=0.1$

Then

  $\begin{array}{l}P\left(0\text{and}x\right)=\text{Area}\text{of}\text{rectangle}\\ =\text{W}\times \text{H}\\ =x\times 0.1\\ =0.1x\end{array}$

We know that

xis the 70^th percentile.

Then

The probability has to be equal to 70% or 0.7.

  $0.1x=0.7$

Divide the above equation by 0.1.

That becomes

  $x=\frac{0.7}{0.1}=7$

Therefore,

The 70^th percentile will be the 7 minutes.