Answer:
![(1)p(x)\geq 0\\(2)\int_(0)^(\infty) p(x) dx=0](https://img.qammunity.org/2021/formulas/physics/college/cr32xujcvn3kz2mx6sy1szoncr4mcmkxih.png)
Step-by-step explanation:
A function f(x) is a Probability Density Function if it satisfies the following conditions:
![(1)f(x)\geq 0\\(2)\int_(0)^(\infty) f(x) dx=0](https://img.qammunity.org/2021/formulas/physics/college/rw7cttm3abjrxebrut9bxng2eog893mh31.png)
Given the function:
![p(x)=(1)/(r)e^(-x/r) \: on\: [0,\infty), where\:r=(20)/(ln(2))](https://img.qammunity.org/2021/formulas/physics/college/e4xj03nzi273tan9qj0uief0mhg4yqt5z0.png)
(1)p(x) is greater than zero since the range of exponents of the Euler's number will lie in
![[0,\infty).](https://img.qammunity.org/2021/formulas/physics/college/qcr2xpmwd6qzfqsd4pynv5dqipt1p1ylqw.png)
(2)
![\int_(0)^(\infty) p(x)=\int_(0)^(\infty) (1)/(r)e^(-x/r)\\=(1)/(r) \int_(0)^(\infty) e^(-x/r)\\=-(r)/(r)\left[e^(-x/r)\right]_(0)^(\infty)\\=-\left[e^(-\infty/r)-e^(-0/r)\right]\\=-e^(-\infty)+e^(-0)\\SInce \: e^(-\infty) \rightarrow 0\\e^(-0)=1\\\int_(0)^(\infty) p(x)=1](https://img.qammunity.org/2021/formulas/physics/college/kl8cueia10z2k2forsqnh0oxft0o6byvmo.png)
The function p(x) satisfies the conditions for a probability density function.