A variable has the density curve with equation y=1-x/2 for 0x< 2. and y=0 otherwise.

a. Graph the density curve of this variable.

b. Show that the area under this density curve to the left of any number x between 0 and 2 equals x-x²/4. Hint. First find the area to the right of x and then apply Property 2 of Key Fact 6.1. on Page 252. What percentage of all possible observations of the variable

c. lie between 1/2 and 12

d. are at least 1.5?

A variable has the density curve with equation y=1-x/2 for 0x< 2. and y=0

Graph the density of the given function.

It is given that the variable (x) has density function $y=1-\frac{x}{2},0<x<2andy=0$otherwise This graph of the density function is given below

Show that the area under the density curve to the left of any number x is equal to $x-\frac{x^2}{4}$

between 0 and 1

For the density function in part (a), the area to the night of any number x has the base of the triangle as (2-x) and the height as

$1-\frac{x}{2}$Thus, the area of the triangle is

Area to the right =

$$\begin{gathered} \frac{1}{2}(Base)(Height)=\frac{1}{2}(2-x)(1-\frac{x}{2}) \\ =\frac{1}{2}\left(2-x-x+\frac{x^2}{2}\right) \\ =\frac{1}{2}\left(2-2x+\frac{x^2}{2}\right) \end{gathered}$$

Thus

Area to the left = 1-Area to the right = role="math" localid="1652689033473" $=1-\left(1-x+\frac{x^2}{4}\right)=x-\frac{x^2}{4}$

Hence, the area under the density curve to the left of any number x is equal to$x-\frac{x^2}{4}$

between 0 and 1.

Find the percentage of all observations lie between 0.5 and 1

$$\begin{gathered} \left(Areabbetween\frac{1}{2}and1\right)=\left(Areatotheleftof1\right)-\left(\left(Areatotheleftof\frac{1}{2}\right)\right) \\ =\left(1-\frac{1}{4}\right)-\left(\frac{1}{2}-\frac{{\left({\displaystyle \frac{1}{2}}\right)}^2}{4}\right)[frompart(b)] \\ =\frac{3}{4}-\left(\frac{1}{2}-\frac{1}{16}\right)=\frac{12-8+1}{16}=0.3125 \end{gathered}$$

Thus, the percentage of all observations that lie between 0.5 and is 1 is 31.25%

Find the percentage of all observations for the variables is at at least 1.5

$$\begin{gathered} \left(Areatotherightof1.5\right)=1-\left(\left(Areatotheleftof1.5\right)\right) \\ =1-\left(1.5-\frac{{(1.5)}^2}{4}\right)[frompart(b)] \\ =1-\left(1.5-\frac{2.25}{4}\right)=0.0625 \end{gathered}$$

Thus, the percentage of all observations that are at least 1.5 is 6.25%