Discrete Distributions

Dr. Himani Gupta

Applications of theBinomial Distribution

A manufacturing plant labels items as either defective or acceptable

A firm bidding for contracts will either get a contract or not

A marketing research firm receives survey responses of “yes I will buy” or “no I will not”

New job applicants either accept the offer or reject it Your team either wins or loses the football game at

the company picnic

• If an experiment consists of a single trial and the outcome of the trial can only be either a success* or a failure, then the trial is called a Bernoulli trial.

• The number of success X in one Bernoulli trial, which can be 1 or 0, is a Bernoulli random variable.

• Note: If p is the probability of success in a Bernoulli experiment, the E(X) = p and V(X) = p(1 – p).

* The terms success and failure are simply statistical terms, and do not have positive or negative implications. In a production setting, finding a defective product may be termed a “success,” although it is not a positive result.

Bernoulli Random Variable

Consider a Bernoulli Process in which we have a sequence of n identical trials satisfying the following conditions:

1. Each trial has two possible outcomes, called success *and failure. The two outcomes are mutually exclusive and exhaustive.

2. The probability of success, denoted by p, remains constant from trial to trial. The probability of failure is denoted by q, where q = 1-p.

3. The n trials are independent. That is, the outcome of any trial does not affect the outcomes of the other trials.

A random variable, X, that counts the number of successes in n Bernoulli trials, where p is the probability of success* in any given trial, is said to follow the binomial probability distribution with parameters n (number of trials) and p (probability of success). We call X the binomial random variable.

* The terms success and failure are simply statistical terms, and do not have positive or negative implications. In a production setting, finding a defective product may be termed a “success,” although it is not a positive result.

The Binomial Random Variable

Dr. Himani Gupta

The Binomial Distribution:Properties A fixed number of observations, n

ex. 15 tosses of a coin; ten light bulbs taken from a warehouse

Two mutually exclusive and collectively exhaustive categories ex. head or tail in each toss of a coin; defective or not

defective light bulb; having a boy or girl Generally called “success” and “failure” Probability of success is p, probability of failure is 1 – p

Constant probability for each observation ex. Probability of getting a tail is the same each time we

toss the coin

Dr. Himani Gupta

The Binomial Distribution:Properties Observations are independent

The outcome of one observation does not affect the outcome of the other

Two sampling methods Infinite population without replacement Finite population with replacement

Suppose we toss a single fair and balanced coin five times in succession, and let X represent the number of heads.

There are 25 = 32 possible sequences of H and T (S and F) in the sample space for this experiment. Of these, there are 10 in which there are exactly 2 heads (X=2):

HHTTT HTHTT HTTHT HTTTH THHTT THTHT THTTH TTHHT TTHTH TTTHH

The probability of each of these 10 outcomes is p2q3 = (1/2)2(1/2)3=(1/32), so the probability of 2 heads in 5 tosses of a fair and balanced coin is:

P(X = 2) = 10 * (1/32) = (10/32) = 0.3125

10 (1/32)

Number of outcomeswith 2 heads

Probability of eachoutcome with 2 heads

Binomial Probabilities (Introduction)

10 (1/32)

Number of outcomeswith 2 heads

Probability of eachoutcome with 2 heads

P(X=2) = 10 * (1/32) = (10/32) = .3125Notice that this probability has two parts:

In general:

1. The probability of a given sequenceof x successes out of n trials with probability of success p and probability of failure q is equal to:

pxq(n-x) nCxnx

nx n x

!

!( )!

2. The number of different sequences of n trials that result in exactly x successes is equal to the number of choices of x elements out of a total of n elements. This number is denoted:

Binomial Probabilities (continued)

N um ber of successes, x P robability P (x )

0

1

2

3

n 1 .00

nn

p qn

np q

nn

p qn

np q

nn n n

p q

n

n

n

n

n n n

!!( ) !

!!( ) !

!!( ) !

!!( ) !

!!( ) !

( )

( )

( )

( )

( )

0 0

1 1

2 2

3 3

0 0

1 1

2 2

3 3

The binomial probability distribution:

where :p is the probability of success in a single trial,q = 1-p,n is the number of trials, andx is the number of successes.

P xnx

p q nx n x p qx n x x n x( ) !!( )!

( ) ( )

The Binomial Probability Distribution

n=5p

x 0.01 0.05 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 0.95 0.99

0 .951 .774 .590 .328 .168 .078 .031 .010 .002 .000 .000 .000 .000

1 .999 .977 .919 .737 .528 .337 .187 .087 .031 .007 .000 .000 .000

2 1.000 .999 .991 .942 .837 .683 .500 .317 .163 .058 .009 .001 .000

3 1.000 1.000 1.000 .993 .969 .913 .813 .663 .472 .263 .081 .023 .001

4 1.000 1.000 1.000 1.000 .998 .990 .969 .922 .832 .672 .410 .226 .049

h F(h) P(h)

0 0.031 0.031

1 0.187 0.156

2 0.500 0.313

3 0.813 0.3134 0.969 0.1565 1.000 0.031

1.000

Cumulative Binomial Probability Distribution and

Binomial Probability Distribution of H,the

Number of Heads Appearing in Five Tosses of

a Fair Coin

F x P X x P i

P F F

all i x( ) ( ) ( )

( ) ( ) ( ). ..

P(X) = F(x) - F(x - 1)

For example:

3 3 2813 500313

Deriving Individual Probabilities from Cumulative Probabilities

The Cumulative Binomial Probability Table

002.0)3()3(

)()()(

XPF

iPxXPxFxiall

n=15p

.50 .60 .700 .000 .000 .0001 .000 .000 .0002 .004 .000 .0003 .018 .002 .0004 .059 .009 .001

... ... ... ...

60% of Brooke shares are owned by LeBow. A random sample of 15 shares is chosen. What is the probability that at most three of them will be found to be owned by LeBow? Check by excel if table values are not available.

60% of Brooke shares are owned by LeBow. A random sample of 15 shares is chosen. What is the probability that at most three of them will be found to be owned by LeBow? Check by excel if table values are not available.

Calculating Binomial Probabilities -Example

Mean of a binomial distribution:

Variance of a binomial distribution:

Standard deviation of a binomial distribution:

= SD(X) = npq

2

E X np

V X npq

( )

( )

118.125.1)(

25.1)5)(.5)(.5()(

5.2)5)(.5()(

2

:coinfair a of tossesfivein heads

ofnumber thecountsH ifexample,For

HSD

HV

HE

H

H

H

Mean, Variance, and Standard Deviation of the Binomial Distribution

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

i

43210

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0.0

x

P(x)

Binomial Probability: n=4 p=0.5

43210

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0.0

x

P(x)

Binomial Probability: n=4 p=0.1

43210

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0.0

x

P(x)

Binomial Probability: n=4 p=0.3

109876543210

05

04

03

02

01

00

x

P(x)

Binomial Probability: n=10 p=0.1

109876543210

05

04

03

02

01

00

x

P(x)

Binomial Probability: n=10 p=0.3

109876543210

05

04

03

02

01

00

x

P(x

)

Binomial Probabil ty: n=10 p=0.5

20191817161514131211109876543210

0.2

0.1

0.0

x

P(x)

Binomial Probability: n=20 p=0.1

20191817161514131211109876543210

0.2

0.1

0.0

x

P(x)

Binomial Probability: n=20 p=0.3

20191817161514131211109876543210

0.2

0.1

0.0

x

P(x)

Binomial Probability: n=20 p=0.5

Binomial distributions become more symmetric as n increases and as p 0.5.

p = 0.1 p = 0.3 p = 0.5

n = 4

n = 10

n = 20

Shape of the Binomial Distribution

Dr. Himani Gupta

Example

.32805)(5)(.1)(.9

.1)(1(.1)1)!(51!

5!

)(1X)!(nX!

n!1)P(X

4

151

XnX

pp

What is the probability of one success in five observations if the probability of success is .1?

X = 1, n = 5, and p = .1

Dr. Himani Gupta

Example

Suppose the probability of purchasing a defective computer is 0.02. What is the probability of purchasing 2 defective computers is a lot of 10?

X = 2, n = 10, and p = .02

.01531)(.8508)(45)(.0004

.02)(1(.02)2)!(102!

10!

)(1X)!(nX!

n!2)P(X

2102

XnX

pp

Dr. Himani Gupta

ProblemA manufacturing company of south Maharashtrafound that after launching a golden handshake scheme for voluntary retirement, 10% of workers are unemployed. What is the probability of obtaining three or fewer unemployed workers in a random sample of 30 in a survey conducted by the company?

Dr. Himani Gupta

We have to find out the probability of getting (a) zero unemployed, x = 0; (b) one unemployed, x = 1; (c) two unemployed x = 2; and (d) three unemployed, x = 3 workers.

Dr. Himani Gupta

According to the U.S. Census Bureau, approximately 6% of all workers in jackson, Mississippi, are unemployed. In conducting a random telephone survey in Jackson, what is the probability of getting two or fewer unemployed workers in a sample of 20?

Problem

Dr. Himani Gupta

Answernpq

P X P X P X P X

2006942 0 1 22901 3703 2246 8850

.

.( ) ( ) ( ) ( )

. . . .

P X( ))!

( )( )(. ) .. .

020!

0!(20 01 1 2901 2901

0 20 006 94

P X( )!( )!

( )(. )(. ) .. .

120!

1 20 120 06 3086 3703

1 20 106 94

P X( )!( )!

( )(. )(. ) .. .

220!

2 20 2190 0036 3283 2246

2 20 206 94

Dr. Himani Gupta

The Binomial DistributionUsing Binomial Tables

n = 10

x … p=.20 p=.25 p=.30 p=.35 p=.40 p=.45 p=.50

012345678910

……………………………

0.10740.26840.30200.20130.08810.02640.00550.00080.00010.00000.0000

0.05630.18770.28160.25030.14600.05840.01620.00310.00040.00000.0000

0.02820.12110.23350.26680.20010.10290.03680.00900.00140.00010.0000

0.01350.07250.17570.25220.23770.15360.06890.02120.00430.00050.0000

0.00600.04030.12090.21500.25080.20070.11150.04250.01060.00160.0001

0.00250.02070.07630.16650.23840.23400.15960.07460.02290.00420.0003

0.00100.00980.04390.11720.20510.24610.20510.11720.04390.00980.0010

109876543210

… p=.80 p=.75 p=.70 p=.65 p=.60 p=.55 p=.50 xExamples: n = 10, p = .35, x = 3: P(x = 3|n =10, p = .35) = .2522

n = 10, p = .75, x = 2: P(x = 2|n =10, p = .75) = .0004

Dr. Himani Gupta

Poisson Distribution: Applications Arrivals at queuing systems

airports -- people, airplanes, automobiles, baggage banks -- people, automobiles, loan applications computer file servers -- read and write operations

The number of scratches in a car’s paint The number of mosquito bites on a person The number of computer crashes in a day

Defects in manufactured goods number of defects per 1,000 feet of extruded copper wire number of blemishes per square foot of painted surface number of errors per typed page

The Poisson probability distribution is useful for determining the probability of a number of occurrences over a given period of time or within a given area or volume. That is, the Poisson random variable counts occurrences over a continuous interval of time or space. It can also be used to calculate approximate binomial probabilities when the probability of success is small (p 0.05) and the number of trials is large (n 20).

Poisson Distribution:

P x ex

x

( ) !

for x = 1,2,3,...

where is the mean of the distribution (which also happens to be the variance) ande is the base of natural logarithms (e=2.71828...).

The Poisson Distribution

Dr. Himani Gupta

The Poisson DistributionProperties

Apply the Poisson Distribution when: You wish to count the number of times an event occurs in a

given area of opportunity The probability that an event occurs in one area of opportunity is

the same for all areas of opportunity The number of events that occur in one area of opportunity is

independent of the number of events that occur in the other areas of opportunity

The probability that two or more events occur in an area of opportunity approaches zero as the area of opportunity becomes smaller

The average number of events per unit is (lambda)

Dr. Himani Gupta

Example

Suppose that, on average, 5 cars enter a parking lot per minute. What is the probability that in a given minute, 7 cars will enter?

So, X = 7 and λ = 5

0.1047!5e

X!λeP(7)

75xλ

So, there is a 10.4% chance 7 cars will enter the parking in a given minute.

Dr. Himani Gupta

Bank customer arrive randomly on weekday afternoons at an average of 3.2 customer every 4 minutes. A. What is the probability of having 10 customers every 8 minutes.B. What is the probability of having 6 customers every 8 minutes.C. What is the probability of having more than 7 customers in a 4-minute interval on a weekday afternoon.

3 2

6 4

1010

0 05286 4

.

!

!.

.

customers / 4 minutesX = 10 customers / 8 minutesAdjusted

= . customers / 8 minutes

P(X) =

( = ) =

X

106.4

e

eX

P X

3 2

6 4

66

0 15866 4

.

!

!.

.

customers / 4 minutesX = 6 customers / 8 minutesAdjusted

= . customers / 8 minutes

P(X) =

( = ) =

X

66.4

e

eX

P X

Dr. Himani Gupta

Poisson Distribution: Probability Table

X 0.5 1.5 1.6 3.0 3.2 6.4 6.5 7.0 8.00 0.6065 0.2231 0.2019 0.0498 0.0408 0.0017 0.0015 0.0009 0.00031 0.3033 0.3347 0.3230 0.1494 0.1304 0.0106 0.0098 0.0064 0.00272 0.0758 0.2510 0.2584 0.2240 0.2087 0.0340 0.0318 0.0223 0.01073 0.0126 0.1255 0.1378 0.2240 0.2226 0.0726 0.0688 0.0521 0.02864 0.0016 0.0471 0.0551 0.1680 0.1781 0.1162 0.1118 0.0912 0.05735 0.0002 0.0141 0.0176 0.1008 0.1140 0.1487 0.1454 0.1277 0.09166 0.0000 0.0035 0.0047 0.0504 0.0608 0.1586 0.1575 0.1490 0.12217 0.0000 0.0008 0.0011 0.0216 0.0278 0.1450 0.1462 0.1490 0.13968 0.0000 0.0001 0.0002 0.0081 0.0111 0.1160 0.1188 0.1304 0.13969 0.0000 0.0000 0.0000 0.0027 0.0040 0.0825 0.0858 0.1014 0.1241

10 0.0000 0.0000 0.0000 0.0008 0.0013 0.0528 0.0558 0.0710 0.099311 0.0000 0.0000 0.0000 0.0002 0.0004 0.0307 0.0330 0.0452 0.072212 0.0000 0.0000 0.0000 0.0001 0.0001 0.0164 0.0179 0.0263 0.048113 0.0000 0.0000 0.0000 0.0000 0.0000 0.0081 0.0089 0.0142 0.029614 0.0000 0.0000 0.0000 0.0000 0.0000 0.0037 0.0041 0.0071 0.016915 0.0000 0.0000 0.0000 0.0000 0.0000 0.0016 0.0018 0.0033 0.009016 0.0000 0.0000 0.0000 0.0000 0.0000 0.0006 0.0007 0.0014 0.004517 0.0000 0.0000 0.0000 0.0000 0.0000 0.0002 0.0003 0.0006 0.002118 0.0000 0.0000 0.0000 0.0000 0.0000 0.0001 0.0001 0.0002 0.0009

Dr. Himani Gupta

Poisson Distribution: Using the Poisson Tables

X 0.5 1.5 1.6 3.00 0.6065 0.2231 0.2019 0.04981 0.3033 0.3347 0.3230 0.14942 0.0758 0.2510 0.2584 0.22403 0.0126 0.1255 0.1378 0.22404 0.0016 0.0471 0.0551 0.16805 0.0002 0.0141 0.0176 0.10086 0.0000 0.0035 0.0047 0.05047 0.0000 0.0008 0.0011 0.02168 0.0000 0.0001 0.0002 0.00819 0.0000 0.0000 0.0000 0.002710 0.0000 0.0000 0.0000 0.000811 0.0000 0.0000 0.0000 0.000212 0.0000 0.0000 0.0000 0.0001

1 64 0 0551

.( ) .P X

Dr. Himani Gupta

Poisson Distribution: Using the Poisson Tables

X 0.5 1.5 1.6 3.00 0.6065 0.2231 0.2019 0.04981 0.3033 0.3347 0.3230 0.14942 0.0758 0.2510 0.2584 0.22403 0.0126 0.1255 0.1378 0.22404 0.0016 0.0471 0.0551 0.16805 0.0002 0.0141 0.0176 0.10086 0.0000 0.0035 0.0047 0.05047 0.0000 0.0008 0.0011 0.02168 0.0000 0.0001 0.0002 0.00819 0.0000 0.0000 0.0000 0.002710 0.0000 0.0000 0.0000 0.000811 0.0000 0.0000 0.0000 0.000212 0.0000 0.0000 0.0000 0.0001

1 65 6 7 8 90047 0011 0002 0000 0060

.( ) ( ) ( ) ( ) ( )

. . . . .P X P X P X P X P X

Dr. Himani Gupta

Poisson Distribution: Using the Poisson Tables

1 62 1 2 1 0 11 2019 3230 4751

.( ) ( ) ( ) ( )

. . .P X P X P X P X

X 0.5 1.5 1.6 3.00 0.6065 0.2231 0.2019 0.04981 0.3033 0.3347 0.3230 0.14942 0.0758 0.2510 0.2584 0.22403 0.0126 0.1255 0.1378 0.22404 0.0016 0.0471 0.0551 0.16805 0.0002 0.0141 0.0176 0.10086 0.0000 0.0035 0.0047 0.05047 0.0000 0.0008 0.0011 0.02168 0.0000 0.0001 0.0002 0.00819 0.0000 0.0000 0.0000 0.002710 0.0000 0.0000 0.0000 0.000811 0.0000 0.0000 0.0000 0.000212 0.0000 0.0000 0.0000 0.0001

Dr. Himani Gupta

Poisson Approximation of the Binomial Distribution Binomial probabilities are difficult to

calculate when n is large. Under certain conditions binomial

probabilities may be approximated by Poisson probabilities.

Poisson approximation

If and the approximation is acceptable .n n p 20 7,

Use n p.

Dr. Himani Gupta

ExampleTelephone manufacturers now offer 1000 different choices for a telephone (as combinations of color, type, options, portability, etc.). A company is opening a large regional office, and each of its 200 managers is allowed to order his or her own choice of a telephone. Assuming independence of choices and that each of the 1000 choices is equally likely, what is the probability that a particular choice will be made by none, one, two, or three of the managers?

n = 200 = np = (200)(0.001) = 0.2p = 1/1000 = 0.001

ExampleTelephone manufacturers now offer 1000 different choices for a telephone (as combinations of color, type, options, portability, etc.). A company is opening a large regional office, and each of its 200 managers is allowed to order his or her own choice of a telephone. Assuming independence of choices and that each of the 1000 choices is equally likely, what is the probability that a particular choice will be made by none, one, two, or three of the managers?

n = 200 = np = (200)(0.001) = 0.2p = 1/1000 = 0.001

P e

P e

P e

P e

( ) .!

( ) .!

( ) .!

( ) .!

.

.

.

.

0 20

1 21

2 22

3 23

0 2

1 2

2 2

3 2

= 0.8187

= 0.1637

= 0.0164

= 0.0011

The Poisson Distribution -Example

20191817161514131211109876543210

0.15

0.10

0.05

0.00

X

P(x

)

=10

109876543210

0.2

0.1

0.0

X

P( x

)

=4

76543210

0.4

0.3

0.2

0.1

0.0

X

P(x)

=1.5

43210

0.4

0.3

0.2

0.1

0.0

X

P(x)

=1.0

The Poisson Distribution (continued)

Dr. Himani Gupta

The Poisson DistributionUsing Poisson Tables

X

0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90

01234567

0.90480.09050.00450.00020.00000.00000.00000.0000

0.81870.16370.01640.00110.00010.00000.00000.0000

0.74080.22220.03330.00330.00030.00000.00000.0000

0.67030.26810.05360.00720.00070.00010.00000.0000

0.60650.30330.07580.01260.00160.00020.00000.0000

0.54880.32930.09880.01980.00300.00040.00000.0000

0.49660.34760.12170.02840.00500.00070.00010.0000

0.44930.35950.14380.03830.00770.00120.00020.0000

0.40660.36590.16470.04940.01110.00200.00030.0000

Example: Find P(X = 2) if = .50

.07582!(0.50)e

X!λe2)P(X

20.50Xλ

Dr. Himani Gupta

The Poisson DistributionShape

0.00

0.10

0.20

0.30

0.40

0.50

0.60

0.70

0 1 2 3 4 5 6 7

x

P(x)

P(X = 2) = .0758

X P(X)

01234567

0.60650.30330.07580.01260.00160.00020.00000.0000

= .50

Dr. Himani Gupta

The Poisson DistributionShape

0.00

0.05

0.10

0.15

0.20

0.25

1 2 3 4 5 6 7 8 9 10 11 12

x

P(x)

0.00

0.10

0.20

0.30

0.40

0.50

0.60

0.70

0 1 2 3 4 5 6 7

x

P(x)

= 0.50 = 3.00

The shape of the Poisson Distribution depends on the parameter :

Dr. Himani Gupta

Poisson Approximation of the Binomial Distribution

X Error0 0.2231 0.2181 -0.00511 0.3347 0.3372 0.00252 0.2510 0.2555 0.00453 0.1255 0.1264 0.00094 0.0471 0.0459 -0.00115 0.0141 0.0131 -0.00106 0.0035 0.0030 -0.00057 0.0008 0.0006 -0.00028 0.0001 0.0001 0.00009 0.0000 0.0000 0.0000

Poisson 1 5.

Binomialnp

5003.

X Error

0 0.0498 0.0498 0.00001 0.1494 0.1493 0.00002 0.2240 0.2241 0.00003 0.2240 0.2241 0.00004 0.1680 0.1681 0.00005 0.1008 0.1008 0.00006 0.0504 0.0504 0.00007 0.0216 0.0216 0.00008 0.0081 0.0081 0.00009 0.0027 0.0027 0.0000

10 0.0008 0.0008 0.000011 0.0002 0.0002 0.000012 0.0001 0.0001 0.000013 0.0000 0.0000 0.0000

Poisson 3 0.

Binomialnp

10 0000003

,.

• A discrete random variable: counts occurrences has a countable number of possible values has discrete jumps between successive values has measurable probability associated with

individual values probability is height

• A continuous random variable: measures (e.g.: height, weight, speed,

value, duration, length) has an uncountably infinite number of

possible values moves continuously from value to value has no measurable probability

associated with individual values probability is area

For example: Binomial n=3 p=.5

x P(x)0 0.1251 0.3752 0.3753 0.125

1.0003210

0.4

0.3

0.2

0.1

0.0

C1

P(x)

Binomial: n=3 p=.5For example:In this case, the shaded area epresentsthe probability that the task takes between 2 and 3 minutes.

654321

0.3

0.2

0.1

0.0

Minutes

P(x)

Minutes to Complete Task

Discrete and Continuous Random Variables - Revisited

6.56.05.55.04.54.03.53.02.52.01.51.0

0.15

0.10

0.05

0.00

Minutes

P(x)

Minutes to Complete Task: By Half-Minutes

0.0. 0 1 2 3 4 5 6 7Minutes

P(x)

Minutes to Complete Task: Fourths of aMinute

Minutes

P(x)

Minutes toCompleteTask:Eighths of aMinute

0 1 2 3 4 5 6 7

The time it takes to complete a task can be subdivided into:Half-Minute Intervals Quarter-Minute Intervals Eighth-Minute Intervals

Or even infinitesimally small intervals:When a continuous random variable has been subdivided into infinitesimally small intervals, a measurable probability can only be associated with an interval of values, and the probability is given by the area beneath the probability densityfunction corresponding to that interval. In this example, the shaded area represents P(2 X ).

When a continuous random variable has been subdivided into infinitesimally small intervals, a measurable probability can only be associated with an interval of values, and the probability is given by the area beneath the probability densityfunction corresponding to that interval. In this example, the shaded area represents P(2 X ).

Minutes to Complete Task: Probability Density Function

76543210

Minutes

f( z)

From a Discrete to a Continuous Distribution

A continuous random variable is a random variable that can take on any value in an interval of numbers.

The probabilities associated with a continuous random variable X are determined by the probability density function of the random variable. The function, denoted f(x), has the following properties.

1. f(x) 0 for all x. 2. The probability that X will be between two numbers a and b is equal to the area

under f(x) between a and b.3. The total area under the curve of f(x) is equal to 1.00.

The cumulative distribution function of a continuous random variable:

F(x) = P(X x) =Area under f(x) between the smallest possible value of X (often -) and the point x.

A continuous random variable is a random variable that can take on any value in an interval of numbers.

The probabilities associated with a continuous random variable X are determined by the probability density function of the random variable. The function, denoted f(x), has the following properties.

1. f(x) 0 for all x. 2. The probability that X will be between two numbers a and b is equal to the area

under f(x) between a and b.3. The total area under the curve of f(x) is equal to 1.00.

The cumulative distribution function of a continuous random variable:

F(x) = P(X x) =Area under f(x) between the smallest possible value of X (often -) and the point x.

Continuous Random Variables

F(x)

f(x)x

x0

0

ba

F(b)

F(a)

1

ba

}P(a X b) = Area under f(x) between a and b= F(b) - F(a)

P(a X b)=F(b) - F(a)

Probability Density Function and Cumulative Distribution Function