Fire Loss. The loss, in millions of dollars, due to a fire in a commercial building is a variable with density curve$y=1-x/2$for $0<x<2$and$y=0$otherwise. Using the fact that the area of a triangle equals one-half its base times its height, we find that the area under this density curve to the left of any number $x$between $0and2$equals $x-x^2/4$

a. Graph the density curve of this variable.

b. What percentage of losses exceed $1.5$million?

Given in the question that,

Equation of density curve$=\left\{\begin{array}{l}y=1-\frac{x}{2}\text{for}0<x<2\\ y=0\text{otherwise}\end{array}\right.$

we have to sketch the dencity curve of the given variable.

The percentage of all possible observations of the variable falling inside any particular range is equal to the corresponding area under the density curve for the variable with the density curve (expressed in percentage).

Equation of density curve$=\left\{\begin{array}{l}y=1-\frac{x}{2}\text{for}0<x<2\\ y=0\text{otherwise}\end{array}\right.$

The density curve can be drawn as:

Given in the question that,

Equation of density curve $=\left\{\begin{array}{l}y=1-\frac{x}{2}\text{for}0<x<2\\ y=0\text{otherwise}\end{array}\right.$

area under this density curve to the left of any number $x$between $0and2$equals $x-x^2/4$

we have to determine the percentage of losses exceeding$1.5$million.

The percentage of all possible observations of the variable falling inside any particular range is equal to the corresponding area under the density curve for the variable with the density curve (expressed in percentage).

area under this density curve to the left of any number $x$is$x-x^2/4$

the percentage of losses exceeding $1.5$million can be calculated as:

$\left(Areatotherightof1.5\right)=1-\left(Areatotheleftof1.5\right)$

$$\begin{gathered} =1-\left(1.5-\frac{1.5^2}{4}\right) \\ =1-0.9375 \\ =0.0625 \\ =6.25\% \end{gathered}$$