Question

Quick, click! An Internet reaction time test asks subjects to click their mouse button as soon as a light flashes on the screen. The light is programmed to go on at a randomly selected time after the subject clicks “Start.” The density curve models the amount of time the subject has to wait for the light to flash.

a. What height must the density curve have? Justify your answer.

b. About what percent of the time will the light flash more than $3.75$ seconds after the subject clicks “Start”?

c. Calculate and interpret the 38th percentile of this distribution.

Short answer

Part (a) Height of the density curve is $0.25$

Part (b) $31.75\%$ of the time that the light flashes more than $3.75$ seconds.

Part (c) $38th$ percentile of the distribution is $2.52$ seconds.

Step-by-step solution

Given information

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

Such that

$a=1$

And

$b=5$

Part (a) Step 2: Concept

A density curve is always on or above the horizontal axis.

Part (a) Step 3: Calculation

The density curve is reciprocal to the difference in the borders in a uniform distribution.

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 $0.25$

Part (b) Step 1: Calculation

The likelihood that the duration until light flashes is between $3.75<X<5$ equals the area beneath the density curve between $3.75<X<5$ (maximum time is $5$ seconds).

Note that

The rectangle will be the area beneath the density curve.

With

Width, $W=5-3.75=1.25$

And

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

Then

$$\begin{gathered} P(X>3.75)=P(3.75<X<5)=Areaofrec\tan gle \\ =W\times H \\ =1.25\times 0.25 \\ =03125 \\ =31.25\% \end{gathered}$$

Therefore,

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

Part (c) Step 1: Calculation

$38$ percent of the data values should be smaller than the $38th$ percentile, according to the attribute for the $38th$ percentile.

Let

$x$be the $38th$percentile.

The likelihood that the time is between the lower boundary and $x$ will be the area beneath the density curve between $1$ and $x$

Note that

The area underneath the density curve will be the rectangle.

With

Width, $W=x-1$

And

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

Then

$$\begin{gathered} P(X<x)=P(1<X<x)=Areaofrec\tan gle \\ =W\times H \\ =(x-1)\times 0.25 \\ =0.25x-0.25 \end{gathered}$$

We know that

$x$ is the $70th$ 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 $38th$ percentile of the distribution will be $2.52$ seconds, implying that the participants must click the button in fewer than $2.52$ seconds 38 percent of the time.