The diameter of a grain of sand in a sand roof, measured in mm, can be considered as a random variable X with probability density
$f(x) = k(x-x^2)$ for $0\leq x \leq 1$
$f(x) = 0$ otherwise
Decide the constant $k$ such that this becomes a legitimate probability density.
For $f(x)$ to be legitimate, it must satisfy the following:
- $f(x)\geq 0$ $\forall$ $X$
- $\int_{-\infty}^{\infty} f(x)dx$ = [the area under the graph of $f$]=1
Now, we can see that $f(x) \geq 0$ for all $k \geq 1$ because the numbers in the given interval $[0,1]$ yields a positive number for $(x-x^2)$ for all $x$. But how do I prove that $f(x)$satisfies the second demand?
We can see that $\int_{-\infty}^{\infty}f(x)dx$ = $\int_{0}^1 k(x-x^2)dx $ (we are restricting the integral to the given domain)
We then get:
$=k\big[\frac{x^2}{2}-\frac{x^3}{3}\big]_0^1=k\big[\frac{1}{2}-\frac{1}{3}\big]=1\Rightarrow k=6.$