Probability Distributions

Continuous Random Variables

Continuous Random Variables A random variable was a numerical value

associated with the outcome of an experiment. Finite discrete random variables were ones in which

the values were countable whole numbered values A continuous random variable is a random

variable that can assume any value in some interval of numbers, and are thus NOT countable. Examples:

The time that a train arrives at a specified stop The lifetime of a transistor A randomly selected number between 0 and 1 Let R be a future value of a weekly ratio of closing prices for

IBM stock Let W be the exact weight of a randomly selected student

Continuous Random Variables A random variable is said to be continuous if there is a function fX(x)

with the following properties: Domain: all real numbers Range: fX(x)≥0 The area under the entire curve is 1

Such a function fX(x) is called the probability density function (abbreviated p.d.f.)

The fact that the total area under the curve fX(x) is 1 for all X values of the random variable tells us that all probabilities are expressed in terms of the area under the curve of this function. Example: If X are values on the interval from [a,b], then the P(a≤X≤b) =

area under the graph of fX(x) over the interval [a,b]

A

a b

fX

Continuous Random Variables

Because all probabilities for a continuous random variable are described in terms of the area under the p.d.f. function, the P(X=x) = 0. Why: the area of the p.d.f. for a single value is zero

because the width of the interval is zero! That is, for any continuous random variable, X,

P(X = a) = 0 for every number a. This DOES NOT imply that X cannot take on the value a, it simply means that the probability of that event is 0.

Continuous Random Variables Rather than considering the probability of X taking on a

given single value, we look for the probability that X assumes a value in an interval.

Suppose that a and b are real numbers with a < b. Recall that X a is the event that X assumes a value in the interval(, a]. Likewise, a < X b and b < X are the events that X assumes values in (a, b] and (b, ), respectively. These three events are mutually exclusive and at least one of them must happen. Thus, P(X a) + P(a < X b) + P(b < X) = 1.

Since we are interested in the probability that X takes a value in an interval, we will solve for P(a < X b).

Continuous Random Variables

Because X is a continuous random variable, P(X = a) = 0 and P(X = b) = 0. Thus, it makes no difference whether or not we include the end points in an interval.

)()(

11

11

1

1

aFbFbXaP

aXPbXPbXaP

bXPaXPbXaP

bXPaXPbXaP

XbPaXPbXaP

XbPbXaPaXP

XX

( ) ( ) ( )

( )

( )

( ).

X XA F b F a P a X b

P a X b

P a X b

P a X b

The cumulative distribution function

The same probability information is often given in a different form, called the cumulative distribution function, (c.d.f), FX(x)

FX(x)=P(Xx)

0 FX(x) 1, for all x Domain is all real numbers

Example

The p.d.f. of T, the weekly CPU time (in hours) used by an accounting firm, is given below.

4 if1

40 if)4(

0 if0

)( 264

3

t

ttt

t

tfT

Example (cont)

The graph of the p.d.f. is given below:

-0.1

0

0.1

0.2

0.3

0.4

0.5

-4 -2 0 2 4 6

t

f T (t )

Example (cont)

is equal to the area between the graph of and the t-axis over the interval.

)21( TP

-0.1

0

0.1

0.2

0.3

0.4

0.5

-4 -2 0 2 4 6

t

f T (t )

Another Example

The c.d.f. of T (for the previous example) is given below.

Find

4 1

40 )316(

0 0

)( 3256

1

tif

tiftt

tif

tFT

)21( TP

The graph of the c.d.f.

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

1.2

-4 -2 0 2 4 6

t

F T (t )

Solution

2617.0

)1316)(1(2561

)2316)(2(2561

)1()2(

)1()2(

)1()2()21(

33

TT FF

TPTP

TPTPTP

Expected Value—Continuous Random Variable For discrete finite random variables, the

expected value was determined by taking each value of the random variable and multiplying it by the corresponding probability as stated by:

x

X xfxXEall

)()(

Expected Value (cont)

For a continuous random variable, the process is more tricky.

For example, suppose we wanted to know the expected value of X which is defined on the interval [0,2].

Unfortunately, the P(X=a) for any number is 0

So when we try to compute the expected value we’re adding up a whole bunch of zeros

X=x fX(x) x fX(x)

0 fX(0) 0 fX(0)

1 fX(1) 1 fX(1)

2 fX(2) 2 fX(2)

Expected Value (cont)

We need some other way of looking for the expected value for a continuous random variable.

Unfortunately, this requires calculus which you will learn more about in 115b

Basically we need to find the area under the entire curve for the function x fX(x)

We can’t do this without knowing some calculus so we’ll give a more geometric interpretation of the E(X) for a continuous random variable.

Expected Value—Geometric Interpretation We can use the probability density function to give a geometric

interpretation for the mean of a continuous random variable, X. Suppose that we draw the p.d.f. on a thin sheet of metal and cut out the region between the graph of f(X) and the x-axis. If we place a knife edge under a line through on the x-axis and perpendicular to that axis, then the metal sheet will balance on that edge.

X

fX

A more concrete example Consider the p.d.f. shown below for a

continuous random variable X

p.d.f. of X

0

0.1

0.2

0.3

0.4

0 1 2 3 4 5 6 7 8

x

f X(x

)

Expected Value—Geometric Interpretation Note: The mean does not occur at the highest

point of the graph! x-coordinate of highest point is the mode. In the previous example, this was at x = 1

Note: The mean does not divide the area in half! x-coordinate that does this is the median. In previous example, the median is approximately

0.3466.

Special Distribution A continuous uniform random variable

is a random variable defined on an interval such that every subinterval of having the same length has the same probability.

If X is a continuous uniform random variable on the interval , then

. 0

1

0

)(

bxif

bxaifab

axif

xfX

. 1

0

)(

bxif

bxaifabax

axif

xFX

Example—Uniform Distribution

A bus arrives at a bus stop every 10 minutes. Let W be the waiting time (in minutes) until the next bus. The p.d.f. and c.d.f. of W are given below.

10 0

100 101

0 0

)(

wif

wif

wif

wfW 0

1 1

10 10

0 0

)(

wif

wifw

wif

wFW

Uniform Distribution: p.d.f. & c.d.f.

-0.020

0.000

0.020

0.040

0.060

0.080

0.100

0.120

-5 0 5 10 15

w

f W (w )

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

1.2

-5 0 5 10 15

w

F W (w )

Questions:

Find

Find E(X)Notice that The expected value of W is given

by:

)64( WP

52100

)( WWE

.

Special Distribution

An exponential random variable may be used to model the length of time between consecutive occurrences of some event in a fixed unit of space or time.

If X is an exponential random variable with parameter, then

0.

10 0

)( / xife

xifxf xX

0. 1

0 0)( / xife

xifxF xX

Example

On average, three customers per hour use the ATM in a local grocery store. Let T be the time (in minutes) between consecutive customers. The p.d.f. and c.d.f. of T are given below.

0

201

0 0)( 20/ tife

tiftf tT

0 1

0 0)( 20/ tife

tiftF tT

Exponential Distribution: p.d.f. & c.d.f.

-0.01

0.00

0.01

0.02

0.03

0.04

0.05

0.06

-20 0 20 40 60 80 100 120

t

f T (t )

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

1.2

-20 0 20 40 60 80 100 120

t

F T (t )

Questions:

Find

The expected value cannot be determined from the p.d.f. function without using calculus so we’ll simply tell you that:

The expected value of T is given by:

)15( TP

XXE )(

20)( TTE

Summary of DistributionsRANDOM VARIABLE, X

Type Finite Continuous

ValuesA finite set of numbers

x1, x2, x3, , xnAll numbers in an interval

Probability

Probability Mass Function, fXp.m.f.

P(X = x) = fX(x)

Probability Density Function, fXp.d.f.

],[overofgraph

theunderarea

)(

baf

bXaPX

a b

fX(x)

xx

fX(x)

Summary of Distributions

RANDOM VARIABLE, X

Type Finite Continuous

ValuesA finite set of numbers

x1, x2, x3, , xnAll numbers in an interval

CumulativeProbability

Cumulative DistributionFunction, FX

c.d.f.P(X x) = F (x)

Cumulative DistributionFunction, FX

c.d.f.P(X x) = FX (x)

FX(x)

x

FX(x)

x

X

Expected Values

Finite Discrete Random Variables:

Continuous Random Variables:Uniform Distribution:Exponential Distribution:

x

X xfxXEall

)()(

2)(

baXE X

XXE )(

Additional Information

See Distribution Study Guide under “Worksheets” link on my webpage

Make sure you are able to distinguish between the various types of distributions, their expected values, the c.d.f. and p.d.f.