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Week 5 Sep 29 – Oct

3. Four Mini-Lectures . QMM 510. Fall . 2014 . 7-. 2. Continuous Probability Distributions . ML 5.1. . Chapter Contents. 7.1 Describing a Continuous Distribution. 7.2 Uniform Continuous Distribution .

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Presentation on theme: " Week 5 Sep 29 – Oct "— Presentation transcript:

Slide1

Week 5 Sep 29 – Oct 3

Four Mini-Lectures QMM 510Fall 2014

Slide2

7-2

Continuous Probability Distributions ML 5.1

Chapter Contents7.1 Describing a Continuous Distribution7.2 Uniform Continuous Distribution 7.3 Normal Distribution7.4 Standard Normal Distribution7.5 Normal Approximations7.6 Exponential Distribution7.7 Triangular Distribution (Optional)

Chapter 7

So many topics, so little time …

Slide3

7-3

Discrete Variable – each value of X has its own probability P(X).Continuous Variable – events are intervals and probabilities are areas under continuous curves. A single point has no probability.

Events as Intervals

Chapter 7

Continuous Distributions

Slide4

7-4

Continuous PDF:Denoted f(x)Must be nonnegativeTotal area under curve = 1Mean, variance, and shape depend onthe PDF parametersPDF reveals the shape of the distribution

PDF – Probability Density Function

Chapter 7

Continuous Distributions

Slide5

7-5

Continuous CDF:Denoted F(x)Shows P(X ≤ x), thecumulative proportion below x.Shows the area to the left of any given point on the PDF.There are Excel functions for either the PDF or CDF.

CDF – Cumulative Distribution Function

Chapter 7

Continuous Distributions

Slide6

7-6

Continuous probability functions:Unlike discrete distributions, the probability at any singlepoint is 0.The entire area under any PDF, by definition, is 1.Mean is the balancepoint of the distribution.

Probabilities as Areas

Chapter 7

Continuous Distributions

Slide7

7-7

Expected Value and Variance

Chapter 7

The mean and variance of a continuous random variable are analogous to

E(X)

and Var(X ) for a discrete random variable. Here the integral sign replaces the summation sign. Calculus is required to compute the integrals.

Continuous Distributions

Slide8

7-8

Characteristics of the Normal Distribution

Normal or Gaussian (or bell-shaped) distribution was named for German mathematician Karl Gauss (1777 – 1855).

Defined by two parameters, µ and .Denoted N(µ, ).Domain is –  < X < +  (continuous scale).Almost all (99.7%) of the area under the normal curve is included in the range µ – 3 < X < µ + 3.Symmetric and unimodal about the mean.

Chapter 7

Normal Distribution

Slide9

7-9

Characteristics of the Normal Distribution

Chapter 7

Normal

Distribution

Slide10

7-10

Normal PDF

f

(x) reaches a maximum at µ and has points of inflection at µ ± 

Bell-shaped curve

Note: All normal distributions have the same shape but differ in the axis scales.

Chapter 7

Characteristics of the Normal Distribution

Normal

Distribution

Excel function for PDF (height of the function at x) is =NORM.DIST(x, µ, , 0)

0 for PDF, 1 for CDF

Slide11

7-11

Normal

CDF has a “lazy-S” shape

Chapter 7

Characteristics of the Normal Distribution

Normal

Distribution

Excel function for CDF (area to left of x) is =NORM.DIST(x, µ, , 1)

0 for PDF, 1 for CDF

Slide12

7-12

Characteristics of the Standard Normal Distribution

Chapter 7

Standard

Normal Distribution

Since for every value of µ and , there is a different normal distribution, we transform a normal random variable to a standard normal distribution with µ = 0 and  = 1 using the formula z = (x - µ)/.

Slide13

7-13

Characteristics of the Standard Normal

Standard normal PDF

f(x) reaches a maximum at z = 0 and has points of inflection at +1.

Shape is unaffected by the transformation. It is still a bell-shaped curve.Standard normal tables or Excel functions can be used to find the desired probabilities.

Figure 7.11

Chapter 7

Standard

Normal Distribution

Excel function for CDF (area to left of

z

) is =NORM.DIST(

z

,

1)

Slide14

7-14

Characteristics of the Standard Normal

Standard normal CDF

Chapter 7

A common scale from

3 to +3 is used.Entire area under the curve is unity.The probability of an event P(z1 < Z < z2) is a definite integral of f(z).However, standard normal tables or Excel functions can be used to find the desired probabilities.

Standard

Normal Distribution

Slide15

7-15

Normal Areas from Appendix C-1

Appendix C-1 allows you to find the area under the curve from 0 to

z.

For example, find P(0 < Z < 1.96):

Chapter 7

Standard

Normal Distribution

Slide16

7-16

Normal Areas from Appendix C-1

Now find

P(1.96 < Z < 1.96).Due to symmetry, P(1.96 < Z) is the same as P(Z < 1.96).

So, P(1.96 < Z < 1.96) = .4750 + .4750 = .9500, or 95% of the area under the curve.

Chapter 7

Standard

Normal Distribution

Slide17

7-17

Basis for the Empirical Rule

Approximately 68% of the area under the curve is between

+ 1 Approximately 95% of the area under the curve is between + 2 Approximately 99.7% of the area under the curve is between + 3

Chapter 7

Standard

Normal Distribution

Slide18

7-18

Normal Areas from Appendix C-2

Appendix C-2 allows you to find the area under the curve from the left of

z (similar to Excel).

This table is the CDF (not the PDF). For example,

P(Z < 1.96)

P(Z < 1.96)

P(1.96 < Z < 1.96)

Chapter 7

Standard

Normal Distribution

=NORM.S.DIST(1.96,1)

=NORM.S.DIST(-1.96,1)

=NORM.S.DIST(1.96,1

)-

NORM.S.DIST

(-1.96,1)

Slide19

7-19

Normal Areas from Appendices C-1 and C-2

Appendices C-1 and C-2 yield identical results.

Use whichever table is easiest.

Finding z for a Given Area

Appendices C-1 and C-2 can be used to find the z-value corresponding to a given probability.For example, what z-value defines the top 1% of a normal distribution?This implies that 49% of the area lies between 0 and z, which gives z = 2.33 by looking for an area of 0.4900 in Appendix C-1.

Chapter 7

Standard Normal Distribution

Slide20

7-20

Finding Areas

Using Standardized Variables

John’s score is 1.57 standard deviations above the mean.

Chapter 7

P(X < 86) = P(Z < 1.57) = .9418 (from Appendix C-2)John is approximately in the 94th percentile.

John took an economics exam and scored 86 points. The class mean was 75 with a standard deviation of 7. What percentile is John in? That is, what is P(X < 86) where X represents the exam scores?

Standard

Normal Distribution

Slide21

7-21

Finding Areas by Using Standardized Variables

You

can use Excel, Minitab, TI83/84, etc. to compute these probabilities directly. The Excel functions are shown:

Chapter 7

Standard

Normal Distribution

Without

standardizing:=NORM.DIST(x, µ, , 1)=NORM.DIST(86, 75, 7, 1)=.9420

With standardizing:=NORM.S.DIST(z, 1)=NORM.S.DIST(1.57, 1)=.9418

Slight difference is due to rounding z to 1.57

Slide22

7-22

Inverse Normal

How can we find the various normal percentiles (5th, 10th, 25th, 75th,

90th, 95th, etc.) known as the inverse normal? That is, how can we find X for a given area?

Chapter 7

Solving for x in z = (x − μ)/ gives x = μ + zσ

Inverse Normal ML 5.2

We simply turn the standardizing transformation around:

Slide23

7-23

Inverse

Normal: Excel

Chapter 7

Inverse Normal

Distribution

Finding x:

Finding z:

Slide24

7-24

Inverse

Normal: Example

Chapter 7

John’s economics professor decides that any student who scores below the 10th percentile must retake the exam. The exam scores are normal with μ = 75 and σ = 7. What is the score that would require a student to retake the exam? We need to find the value of x that satisfies P(X < x) = .10. The z-score for with the 10th percentile is z = −1.28.

Inverse Normal Distribution

Slide25

7-25

Inverse Normal

Chapter 7

The logical steps

to solve the problem are:Use Appendix C to find z = −1.28 to satisfy P(Z < −1.28) = .10.Substitute z = −1.28 into z = (x − μ)/σ to get −1.28 = (x − 75)/7Solve for x to get x = 75 − (1.28)(7) = 66.03 (or 66 after rounding)Students who score below 66 points on the economics exam will be required to retake the exam.

Inverse Normal Distribution

or use Excel to solve in one step:

=

NORM.INV(0.1,75,7) = 66.03

or use Excel to obtain

z

:

=NORM.S.INV(0.1) = 1.282

Slide26

7-26

Normal Approximation to the Binomial

Binomial probabilities are difficult to calculate when

n is large.Use a normal approximation to the binomial distribution.As n becomes large, the binomial bars become smaller and the PDF approaches a continuous distribution.

Chapter 7

Normal Approximations

Slide27

7-27

Normal Approximation to the Binomial

Rule of thumb: when

n ≥ 10 and n(1  ) ≥ 10, then it is appropriate to use the normal approximation to the binomial distribution.Set the mean and standard deviation for the binomial distribution equal to the normal µ and , respectively.

Chapter 7

Normal

Approximations

Slide28

7-28

Example: Coin Flips

Yes, because:

n = 32 x .50 = 16 (at least 10 “successes”) n(1  ) = 32 x (1 .50) = 16 (at least 10 “failures”)

Chapter 7

When translating a discrete scale into a continuous scale, care must be taken about individual points.For example, find the probability of more than 17 heads in 32 flips of a fair coin. This can be written as P(X  18). However, “more than 17” actually falls between 17 and 18 on a discrete scale.

Normal Approximations

If we

flip

a coin

n

= 32 times and

= .50, are the requirements for a normal approximation to the binomial

distribution

met?

Slide29

7-29

Example: Coin Flips

Since the cutoff point for “more than 17” is halfway between 17 and 18, we add 0.5 to the lower limit and find

P(X > 17.5).This addition to X is called the Continuity Correction.At this point, the problem can be completed as any normal distribution problem.

Chapter 7

Normal

Approximations

Slide30

7-30

Chapter 7

Example: Coin Flips

P

(

X

> 17) =

P(X ≥ 18)  P(X ≥ 17.5) = P(Z > 0.53) = 0.2981

Normal

Approximations

Slide31

7-31

Normal Approximation to the Poisson

The normal approximation to the Poisson distribution works best when

 is large (e.g., when  exceeds the values in Appendix B).Set the normal µ and  equal to the mean and standard deviation for the Poisson distribution.

Chapter 7

Example: Utility Bills

On Wednesday between 10 a.m. and noon customer billing inquiries arrive at a mean rate of 42 inquiries per hour at Consumers Energy. What is the probability of receiving more than 50 calls in an hour?  = 42, which is too big to use the Poisson table.Use the normal approximation with  = 42 and  = 6.48074.

Normal

Approximations

Slide32

7-32

Example: Utility Bills

To find

P(X > 50) calls, use the continuity-corrected cutoff point halfway between 50 and 51 (i.e., X = 50.5).At this point, the problem can be completed as any normal distribution problem.

Chapter 7

Normal

Approximations

Slide33

7-33

Bottom Line:

Chapter 7

Normal

Approximations

With Excel, we do not need these approximations for calculations.They are still useful when Excel is not available.They are taught to show the logical connection between discrete and continuous distributions.

Slide34

7-34

Characteristics of the Exponential Distribution

If events per unit of time follow a Poisson

distribution (e.g., customer arrivals), the waiting time until the next event (e.g., customer arrival) follows the exponential distribution.The time until the next event is a continuous variable.

Chapter 7

Exponential Distribution

ML 5.3

Note: We seek tail probabilities such as P(X  x) or P(X ≤ x).

Slide35

7-35

Characteristics of the Exponential Distribution

Probability of waiting

more than x

Probability of waiting less than or equal to x

Chapter 7

Exponential Distribution

Note:

A point has no area so

P

(

X

x

) is the same as

P

(

X

<

x

)

and similarly

P

(

X

>

x

) is the same as

P

(

X

x).

Slide36

7-36

Example: Customer Waiting Time

Between 2 p.m. and 4 p.m. on Wednesday, patient insurance inquiries arrive at Blue Choice insurance at a mean rate of 2.2 calls per minute.

What is the probability of waiting more than 30 seconds (i.e., 0.50 minutes) for the next call?Set  = 2.2 events/min and x = 0.50 min P(X > 0.50) = e–x = e–(2.2)(0.5) = .3329 or a 33.29% chance of waiting more than 30 seconds for the next call.

Chapter 7

Exponential Distribution

Slide37

7-37

Example: Customer Waiting Time

P

(X > 0.50) = e–x = e–(2.2)(0.5) = .3329

P(X ≤ 0.50) = 1-.3329 = .6671

Chapter 7

Exponential Distribution

Given

λ

= 2.2 inquiries per minute, what

is the probability of waiting more than 30 seconds

(

i.e., 0.50 minutes) for the next call

?

Slide38

7-38

If the mean arrival rate is 2.2 calls per minute,

what is

the 90th percentile for waiting time (the top 10% of waiting time)?Find the x-value that defines the upper 10%.

Chapter 7

Inverse Exponential Distribution

Inverse Exponential

Slide39

7-39

Inverse Exponential

Chapter 7

Inverse Exponential Distribution

If the mean arrival rate is 2.2 calls per minute,

what is

the 90th percentile for waiting time (the top 10% of waiting time)? Find the x-value that defines the upper 10%.

Slide40

7-40

Chapter 7

Mean Time Between Events

Exponential Distribution

Slide41

7-41

Chapter 7

Bottom Line:

Exponential Distribution

You may encounter the exponential model in any situation that involves customer arrivals, waiting lines, and

queueing (e.g., retail business, call center, concert, theme park, bank, grocery store, airline check-in, traffic planning).Such applications are not rare in our crowded world.Study simulation (Chapter 18) to learn more about how such situations can be modeled to plan facility capacity, predict waiting times, and study system throughput.

Slide42

7-42

Characteristics of the Triangular Distribution

Chapter 7

Other Continuous Distributions

ML 5.4

Slide43

7-43

Characteristics of the Triangular Distribution

Chapter 7

The

triangular distribution is a way of thinking about variation

that corresponds rather well to what-if analysis in business.It is not surprising that business analysts are attracted to the triangular model. Its finite range and simple form are more understandable than a normal distribution.

Other Continuous Distributions

Slide44

7-44

Characteristics of the Triangular Distribution

Chapter 7

It

is more versatile than a normal because it can be skewed in either direction. Yet it has some of the nice properties of a normal, such as a distinct mode.The triangular model is especially handy for what-if analysis when the business case depends on predicting a stochastic variable (e.g., the price of a raw material, an interest rate, a sales volume). If the analyst can anticipate the range (a to c) and most likely value (b), it will be possible to calculate probabilities of various outcomes.Many times, distributions will be skewed, so a normal wouldn’t be much help.

Other Continuous Distributions

Slide45

7-45

Triangular Distribution: Example T(15, 20, 30)

Chapter 7

Other Continuous Distributions

Slide46

7-46

Triangular Distribution: Example T(15, 20, 30)

Chapter 7

Other Continuous Distributions

Slide47

7-47

Characteristics of the Uniform Distribution

If

X is a random variable that is uniformly distributed between a and b, its PDF has constant height.

Denoted U(a, b)Area = base x height =(b  a) x 1/(b  a) = 1

Chapter 7

Uniform Continuous Distribution

Slide48

7-48

Characteristics of the Uniform Distribution

Chapter 7

Uniform

Continuous Distribution

Slide49

7-49

Example: Anesthesia Effectiveness

An oral surgeon injects a painkiller prior to extracting a tooth. Given the varying characteristics of patients, the dentist views the time for anesthesia effectiveness as a uniform random variable that takes between 15 minutes and 30 minutes.

X is U(15, 30)a = 15, b = 30, find the mean and standard deviation.

Chapter 7

Find the probability that the effectiveness of the anaesthetic takes between 20 and 25 minutes.

Uniform

Continuous Distribution

Slide50

7-50

P

(20 <

X < 25) = (25 – 20)/(30 – 15) = 5/15 = 0.3333 = 33.33%

Chapter 7

Example: Anesthesia Effectiveness

Uniform

Continuous Distribution

Slide51

7-51

Can be a conservative “what-if” baseline model.

Excel’s =RAND() function follows this model:

μ = (a + b)/2 = (0 + 1)/2 = .5000 σ = [(b - a)2/12]1/2 = [(1 - 0)2/12]1/2 = [1/12]1/2 = .2887

Chapter 7

Uses of Uniform Distribution

Uniform Continuous Distribution

Try it yourself! Calculate a bunch of =RAND() values in Excel, and look at the mean and standard deviation. They should be close to the above predictions (if sample is large).

Slide52

7-52

Comparison of Models

Chapter 7

The normal distribution is the used most often.

The exponential is useful in modeling waiting lines (queues).The triangular distribution is a way of thinking about variation that corresponds well to what-if analysis in business.The uniform distribution is a useful baseline model or for random sampling (randomizing a list).

Continuous Distributions