Question

What are the mean and standard deviation of the probability density function given by `p(x)=k(1-xe^(x) )` for `x in [0,3]`, in terms of k, with k being a constant such that the cumulative density across the range of x is equal to 1?

Answer 1

a) Mean: `mu ~~ 2.46`
b) Standard Deviation: `sigma ~~ .425`

Explanation:

The recipe is:

`color(red)"Step 1"`
Determine the function pdf, p(x) such that:
`P(x) =int_(x_1)^(x_2)p(x) dx= 1 " over " [x_1, x_2]` Normalization
`1 = kint_(0)^(3)(1-xe^x)dx = k(x-(1-x)e^x); k=1/(2(1-e^3))`
so your pdf is `p(x) = 1/(2(1-e^3)) (1-xe^x); [0, 3]`

`color(red)"Step 1"`
Determine the mean `mu` and standard deviation `sigma`
a) Mean:
`mu=1/(2-2e^3)int_0^3x(1-xe^x)dx=1/(2-2e^3)[x/2-(x^2-2x+2)]`
`mu = (10e^3-13)/(4e^3-4) ~~ 2.46`
b) var: `sigma^2 = 1/(2-2e^3)int_0^pi(x-2.4607)^2(1-xe^x) = .187`
`sigma = .425`