Chapter 2.2, Problem 43E

(a)

To determine

Height of the density curve.

Answer to Problem 43E

Height of the density curve is 0.25.

Explanation of Solution

Given information:

The distribution has been modeled according to uniform distribution on the interval $1\le x\le 5$ .

Such that

  $a=1$

And

  $b=5$ .

Calculations:

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

On the interval between the boundaries,

  $f\left(x\right)=\frac{1}{b-a}=\frac{1}{5-1}=\frac{1}{4}=0.25$

With

  $1\le x\le 5$ .

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

Thus, the height of the density curve is 0.25.

  

(b)

To determine

Percentage of time that the light flashes more than 3.75 seconds after the subject clicks “Start”.

Answer to Problem 43E

31.75% of the time that the light flashes more than 3.75 seconds.

Explanation of Solution

Given information:

Time until light flashes,

  $3.75<X<5$

Calculations:

The area underneath the density curve between $3.75<X<5$ , will be the probability that the time until light flashes is between $3.75<X<5$ (maximum time is 5 seconds).

Note that

Area underneath the density curve will be the rectangle.

With

Width, $\text{W}=5-3.75=1.25$

And

Height, $\text{H}=f\left(x\right)=0.25$

Then

  $\begin{array}{l}P\left(X>3.75\right)=P\left(3.75<X<5\right)=\text{Area}\text{of}\text{rectangle}\\ =\text{W}\times \text{H}\\ =1.25\times 0.25\\ =0.3125\\ =31.25\%\end{array}$

Therefore,

31.25% of the time that the light flashes more than 3.75 seconds after the subject clicks “Start”.

  

(c)

To determine

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

Answer to Problem 43E

38^th percentile of the distribution is 2.52 seconds.

Explanation of Solution

Given information:

The distribution has been modeled according to uniform distribution on the interval $1\le x\le 5$ .

Such that

  $a=1$

And

  $b=5$ .

Calculations:

According to the property for the 38^th percentile, 38% of the data values should be smaller than the 38^th percentile.

Let

xbe the 38^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, $\text{W}=x-1$

And

Height, $\text{H}=f\left(x\right)=0.25$

Then

  $\begin{array}{l}P\left(X<x\right)=P\left(1<X<x\right)\text{=Area}\text{of}\text{rectangle}\\ =\text{W}\times \text{H}\\ =\left(x-1\right)\times 0.25\\ =0.25x-0.25\end{array}$

We know that

xis the 70^th percentile.

Then

The probability has to be equal to 38% or 0.38.

  $0.25x-0.25=0.38$

Add 0.25 to both sides.

  $0.25x=0.63$

Divide the above equation by 0.1.

That becomes

  $x=\frac{0.63}{0.25}=2.52$

Therefore,

The 38^thpercentile of the distribution will be 2.52 seconds, which means that for 38% of the time, the subjects have to click the button in less than 2.52 seconds.