91Ó°ÊÓ

The distribution has been modeled according to the uniform distribution on the interval$0≤x≤10min$

Such that$a=0$and$b=0$

Waiting time, $2.5≤x≤5.3min$

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,

$f\left(x\right)=\frac{1}{b-a}=\frac{1}{10-0}=\frac{1}{10}=0.1$

With

$0≤x≤10$

The likelihood that the waiting time will be between $2.5$ and $5.3$ minutes will be the area beneath the density curve between $2.5$ and $5.3$ minutes.

Note that

The area underneath the density curve will be the rectangle.

With

Width, $W=5.3−2.5=2.8$

And

Height, $H=f\left(x\right)=0.1$

Then

$P(2.5<x<5.3)=$Area of rectangle

$$\begin{gathered} =W×H \\ =2.8×0.1 \\ =0.28 \\ =28\% \end{gathered}$$

Therefore,

Sally has to wait for the bus for between $2.5$and $5.3$minutes for about $28\%$of the days.

$70$percent of the data values should be smaller than the $70th$percentile, according to the attribute for the $70th$percentile.

Let

$x$be the $70th$percentile.

The likelihood that the time is between $0$and $x$will equal the area beneath the density curve between $0$and $x$

Note that

The rectangle will be the area beneath the density curve.

With

Width, $W=x−0=x$

And

Height, $H=f\left(x\right)=0.1$

Then

$$\begin{gathered} P(0andx)=Areaofrec\tan gle \\ =W×H \end{gathered}$$

We know that

$x$is the $70th$percentile.

Then

The probability has to be equal to $70\%$or $0.7$

$0.1x=0.7$

Divide the above equation by $0.1$

That becomes

$x=\frac{0.7}{0.1}=7$

Therefore,

The $70th$percentile will be the $7$minutes.