# 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?

Apr 13, 2016

1) $P \left(x\right) = K {\int}_{0}^{1} {x}^{4} - {x}^{3} + 1 \mathrm{dx}$
$= K {\left[{x}^{5} / 5 - {x}^{4} / 4 + x\right]}_{0}^{1} = K \left[\frac{1}{5} - \frac{1}{4} + 1\right] = \frac{19}{20}$
Thus $K = \frac{20}{19}$

2) $\mu = \frac{20}{19} {\int}_{0}^{1} x \left({x}^{4} - {x}^{3} + 1\right) \mathrm{dx}$
$= \frac{29}{19} \int \left({x}^{5} - {x}^{4} + x\right) \mathrm{dx} = \frac{28}{57} \approx 0.491$

3) $v a r = \frac{20}{19} {\int}_{0}^{1} {\left(x - \frac{28}{57}\right)}^{2} \left({x}^{4} - {x}^{3} + 1\right) \mathrm{dx}$

$v a r = \sqrt{\frac{1 , 922}{22 , 743}} \approx .291$

#### Explanation:

Given : Probability distribution, $p \left(x\right) = {x}^{4} - {x}^{3} + 1$ for $x \in \left[0 , 1\right]$

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), $P \left(x\right) = {\int}_{0}^{1} p \left(x\right) \mathrm{dx} = 1$
2) Use the mean integral equation, $\mu = {\int}_{0}^{1} x p \left(x\right) \mathrm{dx}$
3) Use the variance integral, $v a r = {\int}_{0}^{1} {\left(x - \mu\right)}^{2} p \left(x\right) \mathrm{dx}$

Solution:
1) $P \left(x\right) = K {\int}_{0}^{1} {x}^{4} - {x}^{3} + 1 \mathrm{dx}$
$= K {\left[{x}^{5} / 5 - {x}^{4} / 4 + x\right]}_{0}^{1} = K \left[\frac{1}{5} - \frac{1}{4} + 1\right] = \frac{19}{20}$
Thus $K = \frac{20}{19}$

2) $\mu = \frac{20}{19} {\int}_{0}^{1} x \left({x}^{4} - {x}^{3} + 1\right) \mathrm{dx}$
$= \frac{29}{19} \int \left({x}^{5} - {x}^{4} + x\right) \mathrm{dx} = \frac{28}{57} \approx 0.491$

3) $v a r = \frac{20}{19} {\int}_{0}^{1} {\left(x - \frac{28}{57}\right)}^{2} \left({x}^{4} - {x}^{3} + 1\right) \mathrm{dx}$

$v a r = \sqrt{\frac{1 , 922}{22 , 743}} \approx .291$