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The time, in minutes, it takes to reboot a certain system is a continuous variable with the density:

$$f(x) = \begin{cases} C(10-x)^2 & \text{if }0 < x < 10 \\ 0 & \text{otherwise}\end{cases}$$

  1. Compute $C$
  2. Compute the probability that it takes between 1 and 2 minutes to reboot.

For part 2 I believe I should take the integral of $f(x)$ from $1$ to $2$ but for part 1 I am not sure.

Though answers do help, I'm looking for more of a how to then the solutions so I can do this on my own for the test.

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up vote 2 down vote accepted

The first step is to remember that for any continuous probability density function, $f(x)$, we have: $$\int_{-\infty}^{\infty}{f(x)\:dx}=1$$

That is, the integral over the entire domain of the P.D.F is equivalent to $1$. In this case $f(x)$ is only non-zero in $(0,10)$, so you have:

$$C\int_{0}^{10}{(10-x)^{2}\:dx}=1 \implies C\left[100x-10x^{2}+\frac{x^{3}}{3}\right]_{0}^{10}=1 \\ \therefore C=\frac{1}{\frac{1000}{3}}=\frac{3}{1000}$$

Then to calculate the probability that $x$ lies in a particular interval $[a,b]$ you can compute: $$\int_{a}^{b}{f(x)\:dx}$$

Which as you correctly identified in your case means: $$\frac{3}{1000}\int_{1}^{2}{(10-x)^{2}\:dx}$$

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Thanks I didn't realize calculating C was essential to the second part which is what must have been throwing me off! – jrquick Nov 6 '12 at 21:30
@JeremyQuick No problem, glad to have helped! – Shaktal Nov 6 '12 at 21:30

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