Chapter 2.2, Problem 44E

(a)

To determine

Height of the density curve.

Answer to Problem 44E

Height of the density curve is 0.2.

Explanation of Solution

Given information:

The distribution has been modeled according to uniform distribution on the interval $-1\le x\le 4$ .

Such that

  $a=-1$

And

  $b=4$ .

Calculations:

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

On the interval between the boundaries,

  $f\left(x\right)=\frac{1}{b-a}=\frac{1}{4-\left(-1\right)}=\frac{1}{4+1}=\frac{1}{5}=0.2$

With

  $-1\le x\le 4$ .

  $f\left(x\right)$ represents the height of the density curve.

Thus, the height of the density curve is 0.2.

  

(b)

To determine

Proportion of time that Mr. Shrager dismisses the class within 1 minute of its scheduled end time.

Answer to Problem 44E

Mr. Sharger dismisses the class about 0.40 of the time within 1 minute of its scheduled end time.

Explanation of Solution

Given information:

Class dismissal time,

  $-1<X<1$

Calculations:

The area underneath the density curve between $-1<X<1$ , will be the probability that the dismissal time is between $-1<X<1$ .

Note that

Area underneath the density curve will be the rectangle.

With

Width, $\text{W}=1-\left(-1\right)=1+1=2$

And

Height, $\text{H}=f\left(x\right)=0.2$

Then

  $\begin{array}{l}P\left(-1<X<1\right)=\text{Area}\text{of}\text{rectangle}\\ =\text{W}\times \text{H}\\ =2\times 0.2\\ =0.4\end{array}$

Therefore,

Mr. Sharger dismisses the class about 0.40 of the time within 1 minute of its scheduled end time.

  

(c)

To determine

20^th percentile of the distribution to be calculated and interpreted.

Answer to Problem 44E

20^th percentile of the distribution is 0 minutes.

Explanation of Solution

Given information:

The distribution has been modeled according to uniform distribution on the interval $-1\le x\le 4$ .

Such that

  $a=-1$

And

  $b=4$ .

Calculations:

According to the property for the 20^th percentile, 20% of the data values should be smaller than the 20^th percentile.

Let

xbe the 20^th percentile.

The area underneath the density curve between-1 and x , will be the probability that the time is between the lower boundary and x .

Note that

Area underneath the density curve will be the rectangle.

With

Width, $\text{W}=x-\left(-1\right)=x+1$

And

Height, $\text{H}=f\left(x\right)=0.2$

Then

  $\begin{array}{l}P\left(X<x\right)=P\left(-1<X<x\right)\text{=Area}\text{of}\text{rectangle}\\ =\text{W}\times \text{H}\\ =\left(x+1\right)\times 0.2\\ =0.2x+0.2\end{array}$

We know that

xis the 20^th percentile.

Then

The probability has to be equal to 20% or 0.2.

  $0.2x+0.2=0.2$

Subtract 0.2 from both sides.

  $0.2x=0$

Divide the above equation by 0.2.

That becomes

  $x=\frac{0}{0.2}=0$

Therefore,

The 20^th percentile of the distribution will be 0 minutes, which means that Mr. Shrager dismisses the class about 20% early of the time.