Question

A variable has a density curve whose equation is $y=1$for $0<x<1$and $y=0$and otherwisse,

(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 1equals x.

What percentage of all possible observations of the variable

(c) lie between $\frac{1}{4}$and $\frac{3}{4}$.

(d) exceed $\frac{7}{8}$.

Short answer

(a) The density curve of this variable is:

(b) Thus, the area under this density curve to the left of any number x between 0 and 1equals x.

(c) The percentage of all possible observations that lie between $\frac{1}{4},\frac{3}{4}$is $50\%$.

(d) The percentage of all possible observations that exceed$\frac{7}{8}is1.25\%$

Step-by-step solution

Part (a) Step 1. Given Information. 

A variable has a density curve whose equation is $y=1,0<x<1,y=0$.

Part (a) Step 2. Graph the density function. 

Graph the density function$y=1,0<x<1,y=0$

Part (b) Step 1. The area between 0, 1. 

From part (a),

Length,$l=x$

Breadth,$b=1$

Thus the area of the rectangle is given by,

$$\begin{gathered} A=lb \\ =x\left(1\right) \\ =x \end{gathered}$$

Part (c) Step 1. Find the percentage. 

Area between $\frac{1}{4},\frac{3}{4}=$[Area to the left of $\frac{3}{4}]-$[Area to the left of $\frac{1}{4}]$

$$\begin{gathered} =\frac{3}{4}-\frac{1}{4} \\ =\frac{2}{4} \\ =0.5 \end{gathered}$$

The percentage of all possible observations that lie between$\frac{1}{4},\frac{3}{4}is50\%$

Part (d) Step 1. Find the percentage. 

[Area to the right of $\frac{7}{8}]=1-$[Area to the left of $\frac{7}{8}]$

$$\begin{gathered} =1-\frac{7}{8} \\ =\frac{1}{8} \\ =0.125 \end{gathered}$$

The percentage of all possible observations that exceeds $\frac{7}{8}is1.25\%$.