Class is over! Mr. Shrager does not always let his statistics class out on time. In fact, he seems to end class according to his own “internal clock.” The density curve models the distribution of the amount of time after class ends (in minutes) when Mr. Shrager dismisses the class. (A negative value indicates he ended class early.)

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

b. About what proportion of the time does Mr. Shrager dismiss class within 1 minute of its scheduled end time?

c. Calculate and interpret the $20th$percentile of the distribution.

The uniform distribution on interval $-1\le x\le 4$ was used to simulate the distribution.

Such that

$a=-1$

And

$b=4$

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

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

In the space between the two limits,

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

With

−1 ≤ x ≤ 4

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

Thus, the height of the density curve is $0.2$

$b=4$

The likelihood that the dismissal time occurs between $-1<X<1$will be the area beneath the density curve between $-1<X<1$

Note that

The rectangle will be the area beneath the density curve.

With

Width, $W=1-(-1)=1+1=2$

And

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

Then

$$\begin{gathered} P(-1<X<1)=Areaofrec\tan gle \\ =W\times H \\ =2\times 0.2 \\ =0.4 \end{gathered}$$

Therefore,

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

$20$ percent of the data values should be smaller than the $20th$ percentile, according to the attribute for the $20th$ percentile.

Let

$x$ be the $20th$ percentile.

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

Note that

The rectangle will be the area beneath the density curve.

With

Width, $W=x-(-1)=x+1$

And

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

Then

We know that

$x$ is the $20th$ 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

$x=0/0.2=0$

Therefore,

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