Question

What are the mean and standard deviation of the probability density function given by `p(x)=x^4-x^3+1` for `x in [0,1]`, 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

1) `P(x) = Kint_0^1 x^4-x^3+1 dx`
`= K[x^5/5 - x^4/4 + x]_0^1 = K [1/5-1/4+1]= 19/20`
Thus `K= 20/19`

2) `mu = 20/19 int_0^1 x(x^4-x^3 +1) dx`
`= 29/19int (x^5-x^4+x) dx = 28/57 ~~ 0.491`

3) `var = 20/19 int_0^1 (x-28/57)^2(x^4-x^3 +1) dx`

`var = sqrt[(1,922)/(22,743)] ~~ .291`

Explanation:

Given : Probability distribution, `p(x)=x^4-x^3+1` for `x in [0,1]`

Required: The mean and standard deviation

Solution Strategy:
1) First ensure you have a probability distribution function in the interval [0,1] by normalizing p(x), `P(x) = int_0^1 p(x) dx = 1`
2) Use the mean integral equation, `mu= int_0^1 x p(x) dx`
3) Use the variance integral, `var = int_0^1 (x-mu)^2 p(x) dx`

Solution:
1) `P(x) = Kint_0^1 x^4-x^3+1 dx`
`= K[x^5/5 - x^4/4 + x]_0^1 = K [1/5-1/4+1]= 19/20`
Thus `K= 20/19`

2) `mu = 20/19 int_0^1 x(x^4-x^3 +1) dx`
`= 29/19int (x^5-x^4+x) dx = 28/57 ~~ 0.491`

3) `var = 20/19 int_0^1 (x-28/57)^2(x^4-x^3 +1) dx`

`var = sqrt[(1,922)/(22,743)] ~~ .291`