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The uniform distribution on interval $鈭�1鈮�x鈮�4$ was used to simulate the distribution.

Such that

$a=鈭�1$

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

鈭�1 鈮� x 鈮� 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 $鈭�1<X<1$will be the area beneath the density curve between $鈭�1<X<1$

Note that

The rectangle will be the area beneath the density curve.

With

Width, $W=1鈭�(鈭�1)=1+1=2$

And

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

Then

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

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, $W=x鈭�(鈭�1)=x+1$

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.