# What are the mean and standard deviation of the probability density function given by (p(x))/k=3x+1-e^(3x+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?

Jan 23, 2016

The problem asks for the solutions for both the mean and standard deviation in terms of k. Since we are given the interval for the p.d.f. ...

#### Explanation:

mean $= k {\int}_{0}^{1} x \left(3 x + 1 - {e}^{3 x + 1}\right) \mathrm{dx}$

mean $= k {\left[{x}^{3} + {x}^{2} / 2 + \frac{{e}^{3 x + 1} \left(1 - 3 x\right)}{9}\right]}_{0}^{1}$

mean $= k \left(1.5 - \frac{2 {e}^{4}}{9} - \frac{e}{9}\right) \approx - 10.9350 k$

Next, calculate the variance ...

variance $= E \left({X}^{2}\right] - {\left[E \left(X\right)\right]}^{2}$

variance $= k {\int}_{0}^{1} {x}^{2} \left(3 x + 1 - {e}^{3 x + 1}\right) \mathrm{dx} - {\left[k \left(1.5 - \frac{2 {e}^{4}}{9} - \frac{e}{9}\right)\right]}^{2}$

variance $= \frac{k}{108} \left(117 + 8e-20 {e}^{4}\right) - {k}^{2} \left[\frac{4 {e}^{8}}{81} + \frac{4 {e}^{5}}{81} - \frac{2}{3} {e}^{4} + {e}^{2} / 81 - \frac{e}{3} + 2.25\right]$

variance $\approx k \left(- 8.8261\right) - {k}^{2} \left(119.5732\right)$

standard deviation$= \sqrt{\text{variance}}$

standard deviation ~~sqrt(k(-8.8261)-k^2(119.5732)

This is what the asker wanted ... mean and standard deviation in terms of k.

However, John Garbriel solved for k above. $k \approx - 0.0676$. So, we can go one step further and derive the approximate numerical value for the mean and s.d.:

mean $\approx \left(- 0.0676\right) \left(1.5 - \frac{2 {e}^{4}}{9} - \frac{e}{9}\right) \approx 0.7392$

s.d. $\approx \sqrt{- 0.0676 \left(- 8.8261\right) - \left(- {0.0676}^{2}\right) \left(119.5732\right)} \approx 0.2241$

Hope that helped