Question

A variable has a density curve whose equation is$y=2x$for $0<x<1$and $y=0$and otherwisse,

(a) Graph the denisty curve of this variable.

(b) Show that the area under this density curve to the left of any number x between 0 and 1equals $x^2$.

What percentage of all possible observations of the variable

(c) lie between $\frac{1}{2}$and $\frac{3}{4}$.

(d) are atleast$\frac{1}{4}$.

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 $x^2$.

(c) The percentagee of all possible observations that lie between $\frac{1}{2},\frac{3}{4}$is $31.25\%$.

(d) The percentagee of all possible observations that are atleast $\frac{1}{4}$is$93.75\%$

Step-by-step solution

Part (a) Step 1. Given Information.

A variable has a density curve whose equation is$y=2x$.

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

Graph the density function $y=2x$ for 0<x<1andy=0.

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

From part (a),

Base, $b=x$

Height, $h=2x$

Thus the area of the triangle is given by,

$$\begin{gathered} A=\frac{1}{2}\times x\times 2x \\ =x^2 \end{gathered}$$

Part (c) Step 1. Find the percentage.

Area between $\frac{1}{2},\frac{3}{4}=$[Area to the left of $\frac{3}{4}]-$[Area to the left of $\frac{1}{2}]$

$$\begin{gathered} ={\left[\frac{3}{4}\right]}^2-{\left[\frac{1}{2}\right]}^2=\frac{9}{16}-\frac{1}{4} \\ =\frac{5}{16} \\ =0.3125 \end{gathered}$$

The percentagee of all possible observations that lie between $\frac{1}{2},\frac{3}{4}$is$31.25\%$.

Part (d) Step 1. Find the percentage.

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

localid="1652597970533" $$\begin{gathered} =1-{\left(\frac{1}{4}\right)}^2 \\ =1-\frac{1}{16} \\ =\frac{15}{16} \\ =0.9375 \end{gathered}$$

The percentagee of all possible observations that lie atleast$\frac{1}{4}$ is$93.75\%$