chap 5-1 a course in business statistics, 4th © 2006 prentice-hall, inc. a course in business...
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Chap 5-3 A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc. Probability Distributions Continuous Probability Distributions Binomial Hypergeometric Poisson Probability Distributions Discrete Probability Distributions Normal Uniform ExponentialTRANSCRIPT
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Chap 5-1A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
A Course In Business Statistics
4th Edition
Chapter 5Discrete and Continuous Probability Distributions
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Chap 5-2A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
Chapter GoalsAfter completing this chapter, you should be
able to: Apply the binomial distribution to applied problems Compute probabilities for the Poisson and
hypergeometric distributions Find probabilities using a normal distribution table
and apply the normal distribution to business problems
Recognize when to apply the uniform and exponential distributions
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Chap 5-3A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
Probability Distributions
Continuous Probability
Distributions
Binomial
Hypergeometric
Poisson
Probability Distributions
Discrete Probability
Distributions
Normal
Uniform
Exponential
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Chap 5-4A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
A discrete random variable is a variable that can assume only a countable number of valuesMany possible outcomes: number of complaints per day number of TV’s in a household number of rings before the phone is answered
Only two possible outcomes: gender: male or female defective: yes or no spreads peanut butter first vs. spreads jelly first
Discrete Probability Distributions
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Chap 5-5A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
Continuous Probability Distributions
A continuous random variable is a variable that can assume any value on a continuum (can assume an uncountable number of values) thickness of an item time required to complete a task temperature of a solution height, in inches
These can potentially take on any value, depending only on the ability to measure accurately.
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Chap 5-6A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
The Binomial Distribution
Binomial
Hypergeometric
Poisson
Probability Distributions
Discrete Probability
Distributions
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Chap 5-7A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
The Binomial Distribution Characteristics of the Binomial Distribution:
A trial has only two possible outcomes – “success” or “failure”
There is a fixed number, n, of identical trials The trials of the experiment are independent of each
other The probability of a success, p, remains constant from
trial to trial If p represents the probability of a success, then
(1-p) = q is the probability of a failure
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Chap 5-8A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
Binomial Distribution Settings
A manufacturing plant labels items as either defective or acceptable
A firm bidding for a contract will either get the 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
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Chap 5-9A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
Counting Rule for Combinations
A combination is an outcome of an experiment where x objects are selected from a group of n objects
)!xn(!x!nCn
x
where:n! =n(n - 1)(n - 2) . . . (2)(1)
x! = x(x - 1)(x - 2) . . . (2)(1) 0! = 1 (by definition)
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Chap 5-10A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
P(x) = probability of x successes in n trials, with probability of success p on each trial
x = number of ‘successes’ in sample, (x = 0, 1, 2, ..., n) p = probability of “success” per trial q = probability of “failure” = (1 – p) n = number of trials (sample size)
P(x)n
x ! n xp qx n x!
( )!
Example: Flip a coin four times, let x = # heads:
n = 4
p = 0.5
q = (1 - .5) = .5
x = 0, 1, 2, 3, 4
Binomial Distribution Formula
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Chap 5-11A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
n = 5 p = 0.1
n = 5 p = 0.5
Mean
0.2.4.6
0 1 2 3 4 5X
P(X)
.2
.4
.6
0 1 2 3 4 5X
P(X)
0
Binomial Distribution The shape of the binomial distribution depends on the
values of p and n
Here, n = 5 and p = .1
Here, n = 5 and p = .5
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Chap 5-12A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
Binomial Distribution Characteristics
Mean
Variance and Standard Deviation
npE(x)μ
npqσ2
npqσ
Where n = sample sizep = probability of successq = (1 – p) = probability of failure
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Chap 5-13A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
n = 5 p = 0.1
n = 5 p = 0.5
Mean
0.2.4.6
0 1 2 3 4 5X
P(X)
.2
.4
.6
0 1 2 3 4 5X
P(X)
0
0.5(5)(.1)npμ
0.6708
.1)(5)(.1)(1npqσ
2.5(5)(.5)npμ
1.118
.5)(5)(.5)(1npqσ
Binomial CharacteristicsExamples
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Chap 5-14A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
Using Binomial Tablesn = 10
x p=.15 p=.20 p=.25 p=.30 p=.35 p=.40 p=.45 p=.50
012345678910
0.19690.34740.27590.12980.04010.00850.00120.00010.00000.00000.0000
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=.85 p=.80 p=.75 p=.70 p=.65 p=.60 p=.55 p=.50 x
Examples: 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
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Chap 5-15A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
Using PHStat Select PHStat / Probability & Prob. Distributions / Binomial…
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Chap 5-16A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
Using PHStat Enter desired values in dialog box
Here: n = 10p = .35
Output for x = 0 to x = 10 will be generated by PHStat
Optional check boxesfor additional output
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Chap 5-17A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
P(x = 3 | n = 10, p = .35) = .2522
PHStat Output
P(x > 5 | n = 10, p = .35) = .0949
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Chap 5-18A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
The Poisson Distribution
Binomial
Hypergeometric
Poisson
Probability Distributions
Discrete Probability
Distributions
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Chap 5-19A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
The Poisson Distribution
Characteristics of the Poisson Distribution: The outcomes of interest are rare relative to the
possible outcomes The average number of outcomes of interest per time
or space interval is The number of outcomes of interest are random, and
the occurrence of one outcome does not influence the chances of another outcome of interest
The probability of that an outcome of interest occurs in a given segment is the same for all segments
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Chap 5-20A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
Poisson Distribution Formula
where:t = size of the segment of interest x = number of successes in segment of interest = expected number of successes in a segment of unit sizee = base of the natural logarithm system (2.71828...)
!xe)t()x(P
tx
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Chap 5-21A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
Poisson Distribution Characteristics
Mean
Variance and Standard Deviation
λtμ
λtσ2
λtσ where = number of successes in a segment of unit size
t = the size of the segment of interest
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Chap 5-22A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
Using Poisson Tables
X
t
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 = .05 and t = 100
.07582!
e(0.50)!xe)t()2x(P
0.502tx
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Chap 5-23A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
Graph of Poisson Probabilities
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)X
t =0.50
01234567
0.60650.30330.07580.01260.00160.00020.00000.0000
P(x = 2) = .0758
Graphically: = .05 and t = 100
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Chap 5-24A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
Poisson Distribution Shape The shape of the Poisson Distribution
depends on the parameters and t:
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)
t = 0.50 t = 3.0
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Chap 5-25A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
The Hypergeometric Distribution
Binomial
Poisson
Probability Distributions
Discrete Probability
Distributions
Hypergeometric
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Chap 5-26A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
The Hypergeometric Distribution
“n” trials in a sample taken from a finite population of size N
Sample taken without replacement Trials are dependent Concerned with finding the probability of “x”
successes in the sample where there are “X” successes in the population
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Chap 5-27A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
Hypergeometric Distribution Formula
Nn
Xx
XNxn
CCC)x(P
.
WhereN = Population sizeX = number of successes in the populationn = sample sizex = number of successes in the sample
n – x = number of failures in the sample
(Two possible outcomes per trial)
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Chap 5-28A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
Hypergeometric Distribution Formula
0.3120
(6)(6)C
CCC
CC2)P(x 103
42
61
Nn
Xx
XNxn
■ Example: 3 Light bulbs were selected from 10. Of the 10 there were 4 defective. What is the probability that 2 of the 3 selected are defective?
N = 10 n = 3 X = 4 x = 2
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Chap 5-29A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
Hypergeometric Distribution in PHStat
Select:PHStat / Probability & Prob. Distributions / Hypergeometric …
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Chap 5-30A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
Hypergeometric Distribution in PHStat
Complete dialog box entries and get output …
N = 10 n = 3X = 4 x = 2
P(x = 2) = 0.3
(continued)
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Chap 5-31A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
The Normal Distribution
Continuous Probability
Distributions
Probability Distributions
Normal
Uniform
Exponential
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Chap 5-32A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
The Normal Distribution ‘Bell Shaped’ Symmetrical Mean, Median and Mode
are EqualLocation is determined by the mean, μSpread is determined by the standard deviation, σ
The random variable has an infinite theoretical range: + to
Mean = Median = Mode
x
f(x)
μ
σ
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Chap 5-33A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
By varying the parameters μ and σ, we obtain different normal distributions
Many Normal Distributions
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Chap 5-34A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
The Normal Distribution Shape
x
f(x)
μ
σ
Changing μ shifts the distribution left or right.
Changing σ increases or decreases the spread.
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Chap 5-35A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
Finding Normal Probabilities
Probability is the area under thecurve!
a b x
f(x) P a x b( )
Probability is measured by the area under the curve
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Chap 5-36A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
f(x)
xμ
Probability as Area Under the Curve
0.50.5
The total area under the curve is 1.0, and the curve is symmetric, so half is above the mean, half is below
1.0)xP(
0.5)xP(μ 0.5μ)xP(
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Chap 5-37A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
Empirical Rules
μ ± 1σencloses about 68% of x’s
f(x)
xμ μσμσ
What can we say about the distribution of values around the mean? There are some general rules:
σσ
68.26%
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Chap 5-38A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
The Empirical Rule μ ± 2σ covers about 95% of x’s μ ± 3σ covers about 99.7% of x’s
xμ2σ 2σ
xμ3σ 3σ
95.44% 99.72%
(continued)
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Chap 5-39A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
Importance of the Rule If a value is about 2 or more standard deviations away from the mean in a normal distribution, then it is far from the mean
The chance that a value that far or farther away from the mean is highly unlikely, given that particular mean and standard deviation
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Chap 5-40A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
The Standard Normal Distribution
Also known as the “z” distribution Mean is defined to be 0 Standard Deviation is 1
z
f(z)
0
1
Values above the mean have positive z-values, values below the mean have negative z-values
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Chap 5-41A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
The Standard Normal
Any normal distribution (with any mean and standard deviation combination) can be transformed into the standard normal distribution (z)
Need to transform x units into z units
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Chap 5-42A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
Translation to the Standard Normal Distribution
Translate from x to the standard normal (the “z” distribution) by subtracting the mean of x and dividing by its standard deviation:
σμxz
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Chap 5-43A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
Example If x is distributed normally with mean of 100
and standard deviation of 50, the z value for x = 250 is
This says that x = 250 is three standard deviations (3 increments of 50 units) above the mean of 100.
3.050
100250σ
μxz
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Chap 5-44A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
Comparing x and z units
z100
3.00250 x
Note that the distribution is the same, only the scale has changed. We can express the problem in original units (x) or in standardized units (z)
μ = 100
σ = 50
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Chap 5-45A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
The Standard Normal Table
The Standard Normal table in the textbook (Appendix D)
gives the probability from the mean (zero) up to a desired value for z
z0 2.00
.4772Example: P(0 < z < 2.00) = .4772
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Chap 5-46A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
The Standard Normal Table
The value within the table gives the probability from z = 0 up to the desired z value
z 0.00 0.01 0.02 …
0.1
0.2
.4772
2.0P(0 < z < 2.00) = .4772
The row shows the value of z to the first decimal point
The column gives the value of z to the second decimal point
2.0
.
.
.
(continued)
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Chap 5-47A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
General Procedure for Finding Probabilities
Draw the normal curve for the problem in terms of x
Translate x-values to z-values
Use the Standard Normal Table
To find P(a < x < b) when x is distributed normally:
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Chap 5-48A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
Z Table example Suppose x is normal with mean 8.0 and
standard deviation 5.0. Find P(8 < x < 8.6)
P(8 < x < 8.6)
= P(0 < z < 0.12)
Z0.12 0x8.6 8
05
88σ
μxz
0.125
88.6σ
μxz
Calculate z-values:
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Chap 5-49A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
Z Table example Suppose x is normal with mean 8.0 and
standard deviation 5.0. Find P(8 < x < 8.6)
P(0 < z < 0.12)
z0.12 0x8.6 8
P(8 < x < 8.6)
= 8 = 5
= 0 = 1
(continued)
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Chap 5-50A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
Z
0.12
z .00 .01
0.0 .0000 .0040 .0080
.0398 .0438
0.2 .0793 .0832 .0871
0.3 .1179 .1217 .1255
Solution: Finding P(0 < z < 0.12)
.0478.02
0.1 .0478
Standard Normal Probability Table (Portion)
0.00
= P(0 < z < 0.12)P(8 < x < 8.6)
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Chap 5-51A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
Finding Normal Probabilities Suppose x is normal with mean 8.0
and standard deviation 5.0. Now Find P(x < 8.6)
Z
8.68.0
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Chap 5-52A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
Finding Normal Probabilities Suppose x is normal with mean 8.0
and standard deviation 5.0. Now Find P(x < 8.6)
(continued)
Z
0.12
.0478
0.00
.5000 P(x < 8.6)
= P(z < 0.12)
= P(z < 0) + P(0 < z < 0.12)
= .5 + .0478 = .5478
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Chap 5-53A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
Upper Tail Probabilities Suppose x is normal with mean 8.0
and standard deviation 5.0. Now Find P(x > 8.6)
Z
8.68.0
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Chap 5-54A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
Now Find P(x > 8.6)…(continued)
Z
0.12 0
Z
0.12
.0478
0
.5000 .50 - .0478 = .4522
P(x > 8.6) = P(z > 0.12) = P(z > 0) - P(0 < z < 0.12)
= .5 - .0478 = .4522
Upper Tail Probabilities
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Chap 5-55A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
Lower Tail Probabilities Suppose x is normal with mean 8.0
and standard deviation 5.0. Now Find P(7.4 < x < 8)
Z
7.48.0
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Chap 5-56A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
Lower Tail Probabilities Now Find P(7.4 < x < 8)…
Z
7.48.0
The Normal distribution is symmetric, so we use the same table even if z-values are negative:
P(7.4 < x < 8)
= P(-0.12 < z < 0)
= .0478
(continued)
.0478
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Chap 5-57A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
Normal Probabilities in PHStat
We can use Excel and PHStat to quickly generate probabilities for any normal distribution
We will find P(8 < x < 8.6) when x is normally distributed with mean 8 and standard deviation 5
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Chap 5-58A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
PHStat Dialogue Box
Select desired options and enter values
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Chap 5-59A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
PHStat Output
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Chap 5-60A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
The Uniform Distribution
Continuous Probability
Distributions
Probability Distributions
Normal
Uniform
Exponential
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Chap 5-61A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
The Uniform Distribution
The uniform distribution is a probability distribution that has equal probabilities for all possible outcomes of the random variable
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Chap 5-62A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
The Continuous Uniform Distribution:
otherwise 0
bxaifab
1
wheref(x) = value of the density function at any x valuea = lower limit of the intervalb = upper limit of the interval
The Uniform Distribution(continued)
f(x) =
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Chap 5-63A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
Uniform DistributionExample: Uniform Probability Distribution
Over the range 2 ≤ x ≤ 6:
2 6
.25
f(x) = = .25 for 2 ≤ x ≤ 66 - 21
x
f(x)
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Chap 5-64A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
The Exponential Distribution
Continuous Probability
Distributions
Probability Distributions
Normal
Uniform
Exponential
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Chap 5-65A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
The Exponential Distribution
Used to measure the time that elapses between two occurrences of an event (the time between arrivals)
Examples: Time between trucks arriving at an unloading
dock Time between transactions at an ATM Machine Time between phone calls to the main operator
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Chap 5-66A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
The Exponential Distribution
aλe1a)xP(0
The probability that an arrival time is equal to or less than some specified time a is
where 1/ is the mean time between events
Note that if the number of occurrences per time period is Poisson with mean , then the time between occurrences is exponential with mean time 1/
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Chap 5-67A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
Exponential Distribution Shape of the exponential distribution
(continued)
f(x)
x
= 1.0(mean = 1.0)
= 0.5 (mean = 2.0)
= 3.0(mean = .333)
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Chap 5-68A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
Example
Example: Customers arrive at the claims counter at the rate of 15 per hour (Poisson distributed). What is the probability that the arrival time between consecutive customers is less than five minutes?
Time between arrivals is exponentially distributed with mean time between arrivals of 4 minutes (15 per 60 minutes, on average)
1/ = 4.0, so = .25
P(x < 5) = 1 - e-a = 1 – e-(.25)(5) = .7135
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Chap 5-69A Course In Business Statistics, 4th © 2006 Prentice-Hall, Inc.
Chapter Summary
Reviewed key discrete distributions binomial, poisson, hypergeometric
Reviewed key continuous distributions normal, uniform, exponential
Found probabilities using formulas and tables
Recognized when to apply different distributions
Applied distributions to decision problems