CH 32 - Exponential Random Variables

Given Formula

Time until the first success occurs	$E(X)=\frac{1}{\lambda }={\int }_{-\infty }^{\infty }x{f}_X(x)dx$
X = time until the next even occurs, $X\ge 0$	$Var(X)=\frac{1}{{\lambda }^2}$
$\lambda$ = the average rate	$Var(X)=\frac{1}{{\lambda }^2}$
$f_X(x)=\lambda e^{-2x},x>0$, and $f_X(x)=0$ otherwise
$F_X(x)=1-e^{-2x}$, $x>0$, and $F_X(x)=0$ otherwise	SD(X)= \

1. What is this exponential problem? (other than it says it has an exponential distribution). because we are waiting until the $1^{st}$ success

2. What does X represent in this scenario?

X is the time waited until the next

3. What is the expected time for receiving the next text message?

$E(X)=5\text{min}$

4. What is the value of $\lambda$?

$$E(X)=\frac{1}{\lambda }\to 5=\frac{1}{\lambda }$$

5. What is the CDF for the wait time of the first text message? Write it in function form and graph it.

$$F_X(x)={\int }_{-\infty }^{x}f(t)dt={\int }_0^x\frac{1}{5}{e}^{-\frac{1}{5}x}dx=1-e^{-\frac{1}{5}x}$$

6. Find the probability that he will not receive a text in the next 10 minutes.

$$P(x>10)={\int }_{10}^{\infty }\frac{1}{5}{e}^{-\frac{1}{5}x}$$

which becomes:

$$1-{\int }_0^{10}\frac{1}{5}{e}^{-\frac{1}{5}x}dx=1-f(x)=1-\left(1-e^{-\frac{10}{5}}\right)=0.1353$$

7. Find the probability that the next text arrives between $9:07$ and $9:10$ PM

$$P(7<x<10)={\int }_7^{10}\frac{1}{5}{e}^{-\frac{1}{5}x}dx=F(10)-F(7)=0.113$$

8. Given that the first text has not arrived by $9:05$pm, what is the probability that the first text will arrive by $9:10$pm?

$$P(X<10\mid x>5)$$

Given the important formula we memorized ( for a given b):

$$P(A\mid B)=\frac{P(A\cap B)}{P(B)}$$

we can plug this in:

$$=\frac{P(X<10\cap X>5)}{P(X>5)}=\frac{P(5<X<10)}{P(X>5)}$$

We can then solve this with either $f_X(x)$ or $F_X(x)$

For $F_X(x)$:

$$F_X(x)=\frac{F(10)-F(5)}{1-F(5)}$$

For $f_X(x)$:

$$f_X(x)=\frac{{\int }_5^{10}\frac{1}{5}e}{}$$