Question

g (12 points) The time between incoming phone calls at a call center is a random variable with exponential density p(x) = 1 r e −x/r on [0, [infinity]), where r = 20 ln(2). a. Verify that the function p(x) is a Probability Density Function.

Answer 1

Answer:

$(1)p(x)\geq 0\\(2)\int_{0}^{\infty} p(x) dx=0$

Explanation:

A function f(x) is a Probability Density Function if it satisfies the following conditions:

$(1)f(x)\geq 0\\(2)\int_{0}^{\infty} f(x) dx=0$

Given the function:

$p(x)=\dfrac{1}{r}e^{-x/r} \: on\: [0,\infty), where\:r=\dfrac{20}{ln(2)}$

(1)p(x) is greater than zero since the range of exponents of the Euler's number will lie in $[0,\infty).$

(2)

$\int_{0}^{\infty} p(x)=\int_{0}^{\infty} \dfrac{1}{r}e^{-x/r}\\=\dfrac{1}{r} \int_{0}^{\infty} e^{-x/r}\\=-\dfrac{r}{r}\left[e^{-x/r}\right]_{0}^{\infty}\\=-\left[e^{-\infty/r}-e^{-0/r}\right]\\=-e^{-\infty}+e^{-0}\\SInce \: e^{-\infty} \rightarrow 0\\e^{-0}=1\\\int_{0}^{\infty} p(x)=1$

The function p(x) satisfies the conditions for a probability density function.