3 Four MiniLectures QMM 510 Fall 2014 7 2 Continuous Probability Distributions ML 51 Chapter Contents 71 Describing a Continuous Distribution 72 Uniform Continuous Distribution ID: 776450
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Week 5 Sep 29 – Oct 3
Four Mini-Lectures QMM 510Fall 2014
Slide27-2
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 …
Slide37-3
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
Slide47-4
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
Slide57-5
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
Slide67-6
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
Slide77-7
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
Slide87-8
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
Slide97-9
Characteristics of the Normal Distribution
Chapter 7
Normal
Distribution
Slide107-10
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
Slide117-11
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
Slide127-12
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)/.
Slide137-13
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)
Slide147-14
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
Slide157-15
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
Slide167-16
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
Slide177-17
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
Slide187-18
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)
Slide197-19
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
Slide207-20
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
Slide217-21
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
Slide227-22
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:
Slide237-23
Inverse
Normal: Excel
Chapter 7
Inverse Normal
Distribution
Finding x:
Finding z:
Slide247-24
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
Slide257-25
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
Slide267-26
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
Slide277-27
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
Slide287-28
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?
Slide297-29
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
Slide307-30
Chapter 7
Example: Coin Flips
P
(
X
> 17) =
P(X ≥ 18) P(X ≥ 17.5) = P(Z > 0.53) = 0.2981
Normal
Approximations
Slide317-31
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
Slide327-32
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
Slide337-33
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.
Slide347-34
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).
Slide357-35
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).
Slide367-36
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
Slide377-37
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
?
Slide387-38
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
Slide397-39
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%.
Slide407-40
Chapter 7
Mean Time Between Events
Exponential Distribution
Slide417-41
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.
Slide427-42
Characteristics of the Triangular Distribution
Chapter 7
Other Continuous Distributions
ML 5.4
Slide437-43
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
Slide447-44
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
Slide457-45
Triangular Distribution: Example T(15, 20, 30)
Chapter 7
Other Continuous Distributions
Slide467-46
Triangular Distribution: Example T(15, 20, 30)
Chapter 7
Other Continuous Distributions
Slide477-47
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
Slide487-48
Characteristics of the Uniform Distribution
Chapter 7
Uniform
Continuous Distribution
Slide497-49
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
Slide507-50
P
(20 <
X < 25) = (25 – 20)/(30 – 15) = 5/15 = 0.3333 = 33.33%
Chapter 7
Example: Anesthesia Effectiveness
Uniform
Continuous Distribution
Slide517-51
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).
Slide527-52
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