Q. 5 - Elementary Statistics

Question

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?

Step-by-Step Solution

Answer

B ) : Area is $x -\frac{x^2}{4}$

C) : The percentage of all observations that lie between 0.5 and is 1 is 31.25% 

D) : The percentage of all observations that are at least 1.5 is 6.25% 

Step 1. Given

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

Step 2. Part ( a )

Graph the density of the given function.

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

Step 3. Part ( b )

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 = $=1-\left(1-x+\frac{x^2}{4}\right) = x - \frac{x^2}{4}$

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.

Step 4. Part ( c )

Find the percentage of all observations lie between 0.5 and 1

$$\begin{gathered} \left(Areab between \frac{1}{2} and 1\right) = \left(Area to the left of 1\right) - \left(\left(Area to the left of \frac{1}{2}\right)\right) \\ =\left(1-\frac{1}{4}\right)-\left(\frac{1}{2}-\frac{{\left({\displaystyle \frac{1}{2}}\right)}^2}{4}\right)[ from part (b ) ] \\ = \frac{3}{4}-\left(\frac{1}{2}-\frac{1}{16}\right)= \frac{12-8+1}{16}= 0.3125 \end{gathered}$$

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

Step 5. Part ( d )

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

$$\begin{gathered} \left(Area to the right of 1.5\right) = 1 - \left(\left(Area to the left of 1.5\right)\right) \\ =1-\left(1.5-\frac{{(1.5)}^2}{4}\right)[ from part (b ) ] \\ = 1-\left(1.5-\frac{2.25}{4}\right)= 0.0625 \end{gathered}$$

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