# What are the mean and standard deviation of the probability density function given by (p(x))/k=x^8-x^16 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?

Dec 15, 2016

The mean is $\mu = 0.7875$
The standard deviation is $\sigma = 0.33$

#### Explanation:

The probability density function is $p \left(x\right) = k \left({x}^{8} - {x}^{16}\right)$

We use $\int {x}^{n} \mathrm{dx} = {x}^{n + 1} / \left(n + 1\right) + C \left(n \ne - 1\right)$

Therefore,

${\int}_{0}^{1} k \left({x}^{8} - {x}^{16}\right) \mathrm{dx} = 1$

$k {\left[{x}^{9} / 9 - {x}^{17} / 17\right]}_{0}^{1} = 1$

$k \left(\frac{1}{9} - \frac{1}{17}\right) = 1$

$k = 17 \cdot \frac{9}{8} = 19.125$

The mean is $E \left(x\right) = \mu = {\int}_{0}^{1} x p \left(x\right) \mathrm{dx}$

$= 19.125 \cdot {\int}_{0}^{1} \left({x}^{9} - {x}^{17}\right)$

$= 19.125 \cdot {\left[{x}^{10} / 10 - {x}^{17} / 17\right]}_{0}^{1}$

$= 19.125 \cdot \left(\frac{1}{10} - \frac{1}{17}\right)$

$= 19.125 \cdot \frac{7}{170} = 0.7875$

The standard deviation is

$\sigma = \sqrt{{\int}_{0}^{1} {x}^{2} p \left(x\right) \mathrm{dx} - {\mu}^{2}}$

${\int}_{0}^{1} {x}^{2} p \left(x\right) \mathrm{dx} = k {\int}_{0}^{1} \left({x}^{10} - {x}^{18}\right) \mathrm{dx}$

$= k {\left[{x}^{11} / 11 - {x}^{19} / 19\right]}_{0}^{1}$

$= 19.125 \cdot \left[\frac{1}{11} - \frac{1}{19}\right]$

$= 19.125 \cdot \frac{8}{19 \cdot 11} = 0.73$

Therefore,

$\sigma = \sqrt{0.73 - {0.7875}^{2}} = 0.33$