Respuesta :
Answer:
(a) The probability that you do not receive a message during a two-hour period is [tex]P(X>2)=\frac{1}{\sqrt{e}} \approx 0.607[/tex]
(b) If you have not had a message in the last four hours, the probability that you do not receive a message in the next two hours is [tex]P(X>4+2|X>4)=P(X>2)=\frac{1}{\sqrt{e} }\approx 0.607[/tex]
Step-by-step explanation:
Let X be a continuous random variable. Then a probability distribution or probability density function (pdf) of X is a function f(x) such that for any two numbers a and b with a ≤ b,
[tex]P(a\leq X\leq b)=\int\limits^b_a {f(x)} \, dx[/tex]
X is said to have an exponential distribution with parameter λ ( λ > 0) if the pdf of X is
[tex]f(x;\lambda)=\left \{{{\lambda e^{-\lambda x}\quad x\geq 0} \atop {0}\quad \:otherwise} \right.[/tex]
If the random variable X has an exponential distribution with parameter λ,
[tex]\mu=E(X)=\frac{1}{\lambda}[/tex]
From the information given:
- The mean is four hours.
- X is the time between the arrival of electronic messages.
(a) To find the probability that you do not receive a message during a two-hour period you must:
[tex]\mu=4=\frac{1}{\lambda}\\\\\lambda = \frac{1}{4}[/tex]
[tex]P(X>2)=\int\limits^{\infty}_{2}{\frac{1}{4}e^{-\frac{1}{4} x}} \, dx[/tex]
Compute the indefinite integral
[tex]\int \frac{1}{4}e^{-\frac{1}{4}x}dx[/tex]
[tex]\frac{1}{4}\cdot \int \:e^{-\frac{x}{4}}dx\\\\\mathrm{Apply\:u \:substitution:} \:{u=-\frac{x}{4}}\\\\\frac{1}{4}\cdot \int \:-4e^udu\\\\\frac{1}{4}\left(-4e^u\right)\\\\\mathrm{Substitute\:back}\:u=-\frac{x}{4}\\\\\frac{1}{4}\left(-4e^{-\frac{x}{4}}\right)\\\\-e^{-\frac{x}{4}}[/tex]
Compute the boundaries
[tex]-e^{-\frac{x}{4}}|_2^{\infty}=0-(-\frac{1}{\sqrt{e} }) =\frac{1}{\sqrt{e}}[/tex]
[tex]P(X>2)=\frac{1}{\sqrt{e}} \approx 0.607[/tex]
(b) To find the probability that you do not receive a message in the next two hours if you have not had a message in the last four hours you must:
We can use the lack of memory property.
For an exponential random variable X,
[tex]P(X\geq t_1 +t_2|X\geq t_1)=P(X\geq t_2)[/tex]
Applying this property we get
[tex]P(X>4+2|X>4)=P(X>2)=\frac{1}{\sqrt{e} }\approx 0.607[/tex]
