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The Exponential Probability Distribution

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The Exponential Probability Distribution

Exponential Probability Density Function

for x > 0, > 0

where = mean e = 2.71828

Cumulative Exponential Distribution Function

where x0 = some specific value of x

f x e x( ) / 1

f x e x( ) / 1

P x x e x( ) / 0 1 o P x x e x( ) / 0 1 o

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Mean and Variance

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Exponential Distribution

.0

1.0

2.0

0 1 2 3 4 5 t

l =

l =

l = 1/2

f(x)=l e-l x, (x > 0).

μλ 1

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5

Exponential Random Variables: Pdf

A simple integration shows that the area under f is 1:

00

0

lim lim

lim 0 1 1

t tx x x

t t

t

t

e dx e dx e

e e

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Random Times

Examples:

• Time until decay of a radioactive atom.

• Lifetime of a circuit.

• Time until a phone call arrives.

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The time between arrivals of cars at Al’s Carwash

follows an exponential probability distribution with a

mean time between arrivals of 3 minutes. Al would like

to know the probability that the time between two

successive arrivals will be 2 minutes or less.

P(x < 2) = 1 - 2.71828-2/3 = 1 - .5134 = .4866

Example: Al’s Carwash

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Example: Al’s Carwash

Graph of the Probability Density Function

xx

F (x )F (x )

.1.1

.3.3

.4.4

.2.2

1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10

P(x < 2) = area = .4866P(x < 2) = area = .4866

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11/5/2003 9

Suppose the wait time X for service at the post office has an exponential distribution with mean 3 minutes. If you enter the post office immediately behind another customer, what is the probability you wait over 5 minutes? Since E(X)=1/=3 minutes, then =1/3, so X~exponential(1/3). We want .

1 55

3 3

( 5) 1 ( 5) 1 (5)

1 1 0.189

P X P X F

e e

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11/5/2003 10

Under the same conditions, what is the probability of waiting between 2 and 4 minutes? Here we calculate .

4 2

3 3

2 4

3 3

(2 4) (4) (2) 1 1

0.250

P X F F e e

e e

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Thank you