Answer:
[tex]f(v) = \left \{ {{\frac{1}{3}v^{-\frac{2}{3}}\ 9^3 \le v \le 10^3} \atop {0, elsewhere}} \right.[/tex]
Step-by-step explanation:
Given
[tex]9 < x < 10[/tex] --- interval
Required
The probability density of the volume of the cube
The volume of a cube is:
[tex]v = x^3[/tex]
For a uniform distribution, we have:
[tex]x \to U(a,b)[/tex]
and
[tex]f(x) = \left \{ {{\frac{1}{b-a}\ a \le x \le b} \atop {0\ elsewhere}} \right.[/tex]
[tex]9 < x < 10[/tex] implies that:
[tex](a,b) = (9,10)[/tex]
So, we have:
[tex]f(x) = \left \{ {{\frac{1}{10-9}\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.[/tex]
Solve
[tex]f(x) = \left \{ {{\frac{1}{1}\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.[/tex]
[tex]f(x) = \left \{ {{1\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.[/tex]
Recall that:
[tex]v = x^3[/tex]
Make x the subject
[tex]x = v^\frac{1}{3}[/tex]
So, the cumulative density is:
[tex]F(x) = P(x < v^\frac{1}{3})[/tex]
[tex]f(x) = \left \{ {{1\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.[/tex] becomes
[tex]f(x) = \left \{ {{1\ 9 \le x \le v^\frac{1}{3} - 9} \atop {0\ elsewhere}} \right.[/tex]
The CDF is:
[tex]F(x) = \int\limits^{v^\frac{1}{3}}_9 1\ dx[/tex]
Integrate
[tex]F(x) = [v]\limits^{v^\frac{1}{3}}_9[/tex]
Expand
[tex]F(x) = v^\frac{1}{3} - 9[/tex]
The density function of the volume F(v) is:
[tex]F(v) = F'(x)[/tex]
Differentiate F(x) to give:
[tex]F(x) = v^\frac{1}{3} - 9[/tex]
[tex]F'(x) = \frac{1}{3}v^{\frac{1}{3}-1}[/tex]
[tex]F'(x) = \frac{1}{3}v^{-\frac{2}{3}}[/tex]
[tex]F(v) = \frac{1}{3}v^{-\frac{2}{3}}[/tex]
So:
[tex]f(v) = \left \{ {{\frac{1}{3}v^{-\frac{2}{3}}\ 9^3 \le v \le 10^3} \atop {0, elsewhere}} \right.[/tex]
