91Ó°ÊÓ

To graph the density curve of this variable.

A density curve exists for the amount of coffee dispensed (in $floz$) into a paper cup by a coffee machine:
$y=\left\{\begin{array}{lr}2,& 5.75<x<6.25\\ 0,& \text{otherwise}\end{array}\right.$
The $y$density curve is shown below.

To show that the area under this density curve to the left of any number $x$between $5.75$and $6.25$ equals $2x-11.5$.

The area of a rectangle with width $2$and length $x-5.75$ is the area under the curve for $5.75x6.25$.
As a result, the area beneath the curve for $5.75x6.25$ is:
$2(x-5.75)=2x-11.5$

As a result, the area under this density curve to the left of any number $x$between $5.75$ and $6.25$ equals $2x-11.5$.

To find the percentage of cups dispensed by this machine contain less than $6floz$.

The area of a rectangle with width $2$and length $6-5.75$is equal to the proportion of observations smaller than $6floz$.
$2(6-5.75)=0.5$
$=50\%$

As a result, the percentage of cups dispensed by this machine contain less than $6floz$ is $50\%$.

To the percentage of cups dispensed by this machine contain between $5.9$ and$6.1floz$.

The area of a rectangle with width $2$and length $6.1-5.9$is equal to the proportion of observations between $5.9$and $6.1floz$.
$2(6.1-5.9)=0.4$
$=40\%$
As a result, the percentage of cups dispensed by this machine contain between $5.9$and $6.1floz$ is $40\%$.

To find the percentage of cups dispensed by this machine contain at least $5.8fl\mathrm{oz}$.

The area of a rectangle with width 2 and length $6.25-5.8$ is equal to the proportion of observations at least $5.8floz$.
$2(6.25-5.8)=0.9$

$=90\%$

As a result, the percentage of cups dispensed by this machine contain at least $5.8fl\mathrm{oz}$is $90\%$.