Question

The time between the arrival of electronic messages at your computer is exponentially distributed with a mean of four hours.

Answer 1

Answer:

(a) The probability that you do not receive a message during a two-hour period  is $P(X>2)=\frac{1}{\sqrt{e}} \approx 0.607$

(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 $P(X>4+2|X>4)=P(X>2)=\frac{1}{\sqrt{e} }\approx 0.607$

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,

$P(a\leq X\leq b)=\int\limits^b_a {f(x)} \, dx$

X is said to have an exponential distribution with parameter λ ( λ > 0) if the pdf of X is

$f(x;\lambda)=\left \{{{\lambda e^{-\lambda x}\quad x\geq 0} \atop {0}\quad \:otherwise} \right.$

If the random variable X has an exponential distribution with parameter λ,

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

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:

$\mu=4=\frac{1}{\lambda}\\\\\lambda = \frac{1}{4}$

$P(X>2)=\int\limits^{\infty}_{2}{\frac{1}{4}e^{-\frac{1}{4} x}} \, dx$

Compute the indefinite integral

$\int \frac{1}{4}e^{-\frac{1}{4}x}dx$

$\frac{1}{4}\cdot \int \:e^{-\frac{x}{4}}dx\\\\\mathrm{Apply\:u \:substitution:} \:{u=-\frac{x}{4}}\\\\\frac{1}{4}\cdot \int \:-4e^udu\\\\\frac{1}{4}\left(-4e^u\right)\\\\\mathrm{Substitute\:back}\:u=-\frac{x}{4}\\\\\frac{1}{4}\left(-4e^{-\frac{x}{4}}\right)\\\\-e^{-\frac{x}{4}}$

Compute the boundaries

$-e^{-\frac{x}{4}}|_2^{\infty}=0-(-\frac{1}{\sqrt{e} }) =\frac{1}{\sqrt{e}}$

$P(X>2)=\frac{1}{\sqrt{e}} \approx 0.607$

(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,

$P(X\geq t_1 +t_2|X\geq t_1)=P(X\geq t_2)$

Applying this property we get

$P(X>4+2|X>4)=P(X>2)=\frac{1}{\sqrt{e} }\approx 0.607$