Week 5
Sep 29 – Oct 3
Four Mini-Lectures
QMM 510
Fall 2014
Chapter Contents
7.1 Describing a Continuous Distribution
7.2 Uniform Continuous Distribution
7.3 Normal Distribution
7.4 Standard Normal Distribution
7.5 Normal Approximations
7.6 Exponential Distribution
7.7 Triangular Distribution (Optional)
7-2
So many
topics, so little
time …
Chapter 7
Continuous Probability Distributions ML 5.1
Chapter 7
Continuous Distributions
Events as Intervals
•
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.
7-3
Chapter 7
Continuous Distributions
PDF – Probability Density Function
Continuous PDF:
•
Denoted f(x)
•
Must be nonnegative
•
Total area under
curve = 1
•
Mean, variance, and
shape depend on
the PDF parameters
•
PDF reveals the shape
of the distribution
7-4
Chapter 7
Continuous Distributions
CDF – Cumulative Distribution Function
Continuous CDF:
•
•
•
•
Denoted F(x)
Shows P(X ≤ x), the
cumulative 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.
7-5
Chapter 7
Continuous Distributions
Probabilities as Areas
• Continuous probability functions:
• Unlike discrete distributions,
the probability at any single
point is 0.
• The entire area under
any PDF, by definition, is 1.
• Mean is the balance
point of the distribution.
7-6
Chapter 7
Continuous Distributions
Expected Value and Variance
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.
7-7
Chapter 7
Normal Distribution
Characteristics of the Normal Distribution
7-8
•
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.
Characteristics of the Normal Distribution
7-9
Chapter 7
Normal Distribution
Characteristics of the Normal Distribution
•
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.
•
Excel function for PDF (height of the function at x) is
=NORM.DIST(x, µ, , 0)
0 for PDF, 1 for CDF
7-10
Chapter 7
Normal Distribution
Characteristics of the Normal Distribution
•
•
Normal CDF has a “lazy-S” shape
Excel function for CDF (area to left of x) is
=NORM.DIST(x, µ, , 1)
0 for PDF, 1 for CDF
7-11
Chapter 7
Normal Distribution
Characteristics of the 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)/.
•
7-12
Chapter 7
Standard Normal Distribution
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.
•
7-13
Excel function for CDF (area to left of z) is
=NORM.DIST(z, 1) Figure 7.11
Chapter 7
Standard Normal Distribution
Characteristics of the Standard Normal
•
Standard normal CDF
•
•
•
•
7-14
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.
Chapter 7
Standard Normal Distribution
Normal Areas from Appendix C-1
7-15
•
Appendix C-1 allows you to find the area under the curve
z.
•
For example, find P(0 < Z < 1.96):
Chapter 7
Standard Normal Distribution
from 0 to
Normal Areas from Appendix C-1
•
•
•
7-16
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
Basis for the Empirical Rule
•
•
•
7-17
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
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)
=NORM.S.DIST(1.96,1)
7-18
P(Z < 1.96)
P(1.96 < Z < 1.96)
=NORM.S.DIST(-1.96,1)
=NORM.S.DIST(1.96,1)NORM.S.DIST(-1.96,1)
Chapter 7
Standard Normal Distribution
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
•
•
•
7-19
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
Chapter 7
Standard Normal Distribution
Finding Areas Using Standardized Variables
•
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?
Appendix C-2: Cumulative Standard Normal Distribution (continued)
This table shows the normal area less than z .
•
John’s score is 1.57 standard
deviations above the mean.
•
P(X < 86) = P(Z < 1.57) = .9418 (from
Appendix C-2)
•
John is approximately in the 94
percentile.
7-20
th
z
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
0.00
0.5000
0.5398
0.5793
0.6179
0.6554
0.6915
0.7257
0.7580
0.7881
0.8159
0.8413
0.8643
0.8849
0.9032
0.9192
0.9332
0.9452
0.9554
0.9641
0.9713
0.9772
0.01
0.5040
0.5438
0.5832
0.6217
0.6591
0.6950
0.7291
0.7611
0.7910
0.8186
0.8438
0.8665
0.8869
0.9049
0.9207
0.9345
0.9463
0.9564
0.9649
0.9719
0.9778
0.02
0.5080
0.5478
0.5871
0.6255
0.6628
0.6985
0.7324
0.7642
0.7939
0.8212
0.8461
0.8686
0.8888
0.9066
0.9222
0.9357
0.9474
0.9573
0.9656
0.9726
0.9783
0.03
0.5120
0.5517
0.5910
0.6293
0.6664
0.7019
0.7357
0.7673
0.7967
0.8238
0.8485
0.8708
0.8907
0.9082
0.9236
0.9370
0.9484
0.9582
0.9664
0.9732
0.9788
0.04
0.5160
0.5557
0.5948
0.6331
0.6700
0.7054
0.7389
0.7704
0.7995
0.8264
0.8508
0.8729
0.8925
0.9099
0.9251
0.9382
0.9495
0.9591
0.9671
0.9738
0.9793
0.05
0.5199
0.5596
0.5987
0.6368
0.6736
0.7088
0.7422
0.7734
0.8023
0.8289
0.8531
0.8749
0.8944
0.9115
0.9265
0.9394
0.9505
0.9599
0.9678
0.9744
0.9798
0.06
0.5239
0.5636
0.6026
0.6406
0.6772
0.7123
0.7454
0.7764
0.8051
0.8315
0.8554
0.8770
0.8962
0.9131
0.9279
0.9406
0.9515
0.9608
0.9686
0.9750
0.9803
0.07
0.5279
0.5675
0.6064
0.6443
0.6808
0.7157
0.7486
0.7794
0.8078
0.8340
0.8577
0.8790
0.8980
0.9147
0.9292
0.9418
0.9525
0.9616
0.9693
0.9756
0.9808
0.08
0.5319
0.5714
0.6103
0.6480
0.6844
0.7190
0.7517
0.7823
0.8106
0.8365
0.8599
0.8810
0.8997
0.9162
0.9306
0.9429
0.9535
0.9625
0.9699
0.9761
0.9812
0.09
0.5359
0.5753
0.6141
0.6517
0.6879
0.7224
0.7549
0.7852
0.8133
0.8389
0.8621
0.8830
0.9015
0.9177
0.9319
0.9441
0.9545
0.9633
0.9706
0.9767
0.9817
•
Finding Areas by Using Standardized Variables
You can use Excel, Minitab, TI83/84, etc. to compute these probabilities
directly. The Excel functions are shown:
Without standardizing:
=NORM.DIST(x, µ, , 1)
=NORM.DIST(86, 75, 7, 1)
=.9420
7-21
With standardizing:
=NORM.S.DIST(z, 1)
=NORM.S.DIST(1.57, 1)
=.9418
Slight
difference
is due to
rounding z
to 1.57
Chapter 7
Standard Normal Distribution
ML 5.2
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?
• We simply turn the standardizing transformation around:
Solving for x in z = (x − μ)/ gives x = μ + zσ
7-22
Chapter 7
Inverse Normal
Inverse Normal: Excel
Finding x:
7-23
Finding z:
Chapter 7
Inverse Normal Distribution
Inverse Normal: Example
• 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.
7-24
Chapter 7
Inverse Normal Distribution
Inverse Normal
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)/7
• Solve 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.
or use Excel to obtain z:
=NORM.S.INV(0.1) = 1.282
7-25
or use Excel to solve in one step:
=NORM.INV(0.1,75,7) = 66.03
Chapter 7
Inverse Normal Distribution
Normal Approximation to the Binomial
Chapter 7
Normal Approximations
•
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.
7-26
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.
7-27
Chapter 7
Normal Approximations
Example: Coin Flips
•
If we flip a coin n = 32 times and  = .50, are the requirements for a
normal approximation to the binomial distribution met?
•
Yes, because:
n = 32 x .50 = 16
n(1  ) = 32 x (1 .50) = 16
7-28
(at least 10 “successes”)
(at least 10 “failures”)
•
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.
Chapter 7
Normal Approximations
Example: Coin Flips
•
•
•
7-29
Chapter 7
Normal Approximations
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.
Example: Coin Flips
P(X > 17) = P(X ≥ 18)  P(X ≥ 17.5) = P(Z
> 0.53) = 0.2981
7-30
Chapter 7
Normal Approximations
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.
Example: Utility Bills
•
•
•
7-31
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.
Chapter 7
Normal Approximations
Example: Utility Bills
•
•
7-32
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
Bottom Line:
•
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.
7-33
Chapter 7
Normal Approximations
ML 5.3
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.
Note: We seek
tail probabilities
such as P(X  x)
or P(X ≤ x).
7-34
Chapter 7
Exponential Distribution
Characteristics of the Exponential Distribution
Probability of waiting
less than or equal to x
Probability of waiting
more than x
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).
7-35
Chapter 7
Exponential Distribution
Example: Customer Waiting Time
7-36
•
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
Example: Customer Waiting Time
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?
P(X > 0.50) = e–x = e–(2.2)(0.5) = .3329
7-37
P(X ≤ 0.50) = 1-.3329 = .6671
Chapter 7
Exponential Distribution
Inverse Exponential
•
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%.
7-38
Chapter 7
Inverse Exponential Distribution
Inverse Exponential
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%.
7-39
Chapter 7
Inverse Exponential Distribution
Mean Time Between Events
7-40
Chapter 7
Exponential Distribution
Bottom Line:
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.
7-41
Chapter 7
Exponential Distribution
Characteristics of the Triangular Distribution
7-42
ML 5.4
Chapter 7
Other Continuous Distributions
Characteristics of the Triangular Distribution
• 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.
7-43
Chapter 7
Other Continuous Distributions
Characteristics of the Triangular Distribution
• 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.
7-44
Chapter 7
Other Continuous Distributions
Triangular Distribution: Example T(15, 20, 30)
7-45
Chapter 7
Other Continuous Distributions
Triangular Distribution: Example T(15, 20, 30)
7-46
Chapter 7
Other Continuous Distributions
Characteristics of the Uniform Distribution
If X is a random variable that is
uniformly distributed between
a and b, its PDF has
constant height.
•
•
7-47
Denoted U(a, b)
Area =
base x height =
(b  a) x 1/(b  a) = 1
Chapter 7
Uniform Continuous Distribution
Characteristics of the Uniform Distribution
7-48
Chapter 7
Uniform Continuous Distribution
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.
• Find the probability that the effectiveness of the anaesthetic
takes between 20 and 25 minutes.
7-49
Chapter 7
Uniform Continuous Distribution
Example: Anesthesia Effectiveness
P(20 < X < 25) = (25 – 20)/(30 – 15) = 5/15 = 0.3333 = 33.33%
7-50
Chapter 7
Uniform Continuous Distribution
Chapter 7
Uniform Continuous Distribution
Uses of Uniform Distribution
• 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
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).
7-51
0.84328
0.33170
0.45351
0.53490
0.46443
0.43802
0.00549
Mean = 0.494637
0.68397 0.69134
0.56953 0.04807
0.70129 0.15553
0.96473 0.62752
0.98558 0.25002
0.37406 0.08978
0.32222 0.63328
0.09071
0.65731
0.36416
0.78566
0.05013
0.29142
0.28581
St Dev = 0.271894
0.72185 0.73706
0.34992 0.79984
0.33627 0.71570
0.82808 0.34901
0.61517 0.09537
0.47772 0.25935
0.27208 0.81790
0.78645
0.97143
0.80646
0.14220
0.50000
0.36504
0.59686
Comparison of Models
• 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).
7-52
Chapter 7
Continuous Distributions