Question

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.

Answer 1

Answer:

(a) 0.20

(b) 31%

(c) 2.52 seconds

Step-by-step explanation:

The random variable Y models the amount of time the subject has to wait for the light to flash.

The density curve represents that of an Uniform distribution with parameters a = 1 and b = 5.

So, $Y\sim Unif(1,5)$

(a)

The area under the density curve is always 1.

The length is 5 units.

Compute the height as follows:

$\text{Area under the density curve}=\text{length}\times \text{height}$

                                          $1=5\times\text{height}\\\\\text{height}=\frac{1}{5}\\\\\text{height}=0.20$

Thus, the height of the density curve is 0.20.

(b)

Compute the value of P (Y > 3.75) as follows:

$P(Y>3.75)=\int\limits^{5}_{3.75} {\frac{1}{b-a}} \, dy \\\\=\int\limits^{5}_{3.75} {\frac{1}{5-1}} \, dy\\\\=\frac{1}{4}\times [y]^{5}_{3.75}\\\\=\frac{5-3.75}{4}\\\\=0.3125\\\\\approx 0.31$

Thus, the light will flash more than 3.75 seconds after the subject clicks "Start" 31% of the times.

(c)

Compute the 38th percentile as follows:

$P(Y$

Thus, the 38th percentile is 2.52 seconds.