Question

What are the mean and standard deviation of the probability density function given by `p(x)=kabs(x-9)` for `x in [0,18]`, with k being a constant such that the cumulative density is equal to 1?

Answer 1

`"Mean, mu = 9`
`sigma^2 = 81/2; sigma = sqrt(81/2) = 9sqrt2/2`

Explanation:

First Normalize the distribution to 1, by integrating (compute the area) over [0, 18]
`F(x) = int_0^18k|x-9| dx = 1`
Now you can easily see the integral the area of the two triangles on [0, 18]. Thus the integral is equal to:
`F(0leq x leq 18) = 1 = k/2(9xx9 + 9xx9) = 81k; k = 1/81`
so the Probability distribution function:
`f(x) = 1/81|x-9|`, with this in mind we can proceed to calculate the mean and standard deviation - `mu, sigma`.

`"Mean, " mu= int _(xo)^(x1) xf(x) dx`
`mu = 1/81 int_0^18|x-9| dx = int_0^9x(-x+9) dx + int_9^18x (x-9) dx` Now you can integrate this by brute force or note that the mean is half way through the range [0, 18] `=> mu = 9` Why? Symmetry

`"Sigma^2, " sigma^= int _(xo)^(x1) (x-mu)^2f(x) dx`
`sigma^2= 1/81 int_0^18(x-mu)^2|x-9| dx`
`sigma^2 = int_0^9(x-9)^2(-x+9) dx + int_9^18(x-9)^2 (x-9) dx`
`sigma^2 = 81/2; sigma = sqrt(81/2) = 9sqrt2/2`