Mean and Standard Deviation of two uniform distribution functions?

T1's Question from Y!Answers:

I have two uniform distribution functions, one between X=2 and X=8, and another between X=9 and X=10.

The probability of any number is 1/7. Before X=2, between X=8 and X=9 and after X=10, the probability is 0.

I need to find the mean and standard deviation of X.

Answer:

The pdf of your distribution is:

\[f(x) = \left\{ \begin{array}{ll} 0,& x\lt 2\\
1/7,& 2 \le x \le 8 \\
0,& 8 \lt x \lt 9\\
1/7,& 9 \le x \le 10\\
0,& 10 \lt x \\ \end{array}\right. \]

The mean is:
\[ \mu = E(x) = \int_{-infty}^{infty} x f(x) dx\]
which we can decompose into a sum of integrals over intervals where \(f(x)\) is non-zero:
\[E(x) = \int_2^8 \frac{x}{7} dx  + \int_9^{10} \frac{x}{7} dx \]
so:
\[ \mu = E(x) = (1/7) (8^2/2 - 2^2/2) + (1/7) (10^2/2 - 9^2/2) = 5.5\]

The variance is \(E( (x-\mu)^2 ) = E(x^2) - \mu^2\).

Now use the same method as before to evaluate \(E(x^2)\) by decomposing the integral into the sum of integrals over the intervals where \(f(x)\) is non-zero.

The standard deviation is then the square root of the variance.