chapter 2: probability concepts and applications · • the probability, p, of any event or state...
TRANSCRIPT
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Chapter 2: Probability Concepts and
Applications
Textbook: pp. 39-80
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Learning Objectives
After completing this chapter, students will be able to:
• Understand the basic foundations of probability analysis.
• Describe statistically dependent and independent events.
• Use Bayes’ theorem to establish posterior probabilities.
• Describe and provide examples of both discrete and
continuous random variables.
• Explain the difference between discrete and continuous
probability distributions.
• Calculate expected values and variances and use the
normal table.
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Learning Objectives
After completing this chapter, students will be able to:
• Understand the binomial distribution.
• Understand the normal distribution and use the normal
table.
• Understand the F distribution.
• Understand the exponential distribution and its relation to
queuing theory.
• Understand the Poisson distribution and its relation to
queuing theory.
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• Life is uncertain; we are not sure what the future will
bring
• Probability is a numerical statement about the
likelihood that an event will occur
Introduction
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• The probability, P, of any event or state of nature
occurring is greater than or equal to 0 and less than or
equal to 1. That is,
0 ≤ P(event) ≤ 1
• A probability of 0 indicates that an event is never
expected to occur. A probability of 1 means that an
event is always expected to occur.
• The sum of the simple probabilities for all possible
outcomes of an activity must equal 1. Regardless of
how probabilities are determined, they must adhere to
these two rules.
Two Basic Rules of Probability
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• Objective Approach
• Relative frequency approach
• Classical or logical method
Types of Probability (1 of 3)
P(event) =Number of occurrences of the event
Total numbers of trials or outcomes
P(head) =
P(head) =
1
2
13
52
Number of ways of getting a head
Number of possible outcomes (head or tail)
Number of chances of drawing a spade
Number of possible outcomes
2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King, Ace
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• Historical demand for white latex paint at = 0, 1, 2, 3, or
4 gallons per day
• Observed frequencies over the past 200 days
Diversey Paint Example (1 of 2)
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• Historical demand for white latex paint at = 0, 1, 2, 3, or
4 gallons per day
• Observed frequencies over the past 200 days
Diversey Paint Example (2 of 2)
Individual
probabilities are all
between 0 and 1
0 ≤ P (event) ≤ 1
Total of all event
probabilities equals 1
∑ P (event) = 1.00
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• Objective Approach
o Classical or logical method
• Perform a series of trials
Types of Probability (2 of 3)
1head =
2
13spade =
52
01
= .25=4
= 25%
Number of ways of getting a headP
Number of possible outcomes head or tail
Number of chances of drawing a spadeP
Number of possible outcomes
( )
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• Subjective Approach
o Based on the experience and judgment of the
person making the estimate
• Opinion polls
• Judgment of experts
• Delphi method
Types of Probability (3 of 3)
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• Events are said to be mutually exclusive if only one of
the events can occur on any one trial
o Tossing a coin will result in either a head or a tail
o Rolling a die will result in only one of six possible
outcomes
Mutually Exclusive and Collectively
Exhaustive Events (1 of 2)
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• Events are said to be
collectively exhaustive if
the list of outcomes
includes every possible
outcome
• Both heads and tails as
possible outcomes of coin
flips
• All six possible outcomes of
the roll of a die:
Mutually Exclusive and Collectively
Exhaustive Events (2 of 2)
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Venn Diagrams
A B
Events that are mutually exclusive
Events that are notmutually exclusive
A B
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• Draw one card from a deck of 52 playing cards
A = event that a 7 is drawn
B = event that a heart is drawn
P (a 7 is drawn) = P(A)= 4/52 = 1/13
P (a heart is drawn) = P(B) = 13/52 = 1/4
• These two events are not mutually exclusive since a 7
of hearts can be drawn
• These two events are not collectively exhaustive
since there are other cards in the deck besides 7s and
hearts
Drawing a Card
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Differences (mutually exclusive / collectively
exhaustive)
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Differences (mutually exclusive / collectively
exhaustive)
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• Intersection – the set of all outcomes that are common to
both events
• Intersection of event A and event B = A and B
= A ∩ B
= AB
• Probability notation
P(Intersection of event A and event B) = P(A and B)
= P(A ∩ B)
= P(AB)
Sometimes called joint probability
Unions and Intersections of Events (1 of 3)
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• Union – the set of all outcomes that are contained in
either of two events
Union of event A and event B = A or B
• Probability notation
P(Union of event A and event B) = P(A or B)
= P(A ∪ B)
Unions and Intersections of Events (2 of 3)
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• In the previous example
• Intersection of event A and event B
(A and B) = the 7 of hearts is drawn
P(A and B) = P(7 of hearts is drawn) = 1/52
• Union of event A and event B
(A or B) = either a 7 or a heart is drawn
P(A or B) = P(any 7 or any heart is drawn) = 16
/52
Unions and Intersections of Events (3 of 3)
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• General rule for union of two events,
additive rule
P(A or B) = P(A) + P(B) – P(A and B)
o Union of two events, a 7 or a heart
P(A or B) = P(A) + P(B) – P(A and B)
= 4/52 + 13/52 – 1/52
= 16/52
Probability Rules (1 of 5)
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• Conditional probability – probability that an event
occurs given another event has already happened
o Probability of a 7 given a heart has been drawn
Probability Rules (2 of 5)
P(A | B) = = = 1/13
P(AB)
P(B)
1/52
13/52
P(A | B) =P(AB)
P(B)
P(AB) = P(A | B) P(B)
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Which sets are independent?1. (a) Your education
(b) Your income level
2. (a) Draw a jack of hearts from a full 52-card deck
(b) Draw a jack of clubs from a full 52-card deck
3. (a) Snow in Zhengzhou
(b) Rain in Bangor / Wales
Probability Rules (3 of 5)
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• Independent one event has no effect on the other event
P(A | B) = P(A)
P(A and B) = P(A)P(B)
• For a fair coin tossed twice
A = event that a head is the result of the first toss
B = event that a head is the result of the second toss
P(A) = 0.5 and P(B) = 0.5
P(AB) = P(A)P(B) = 0.5(0.5) = 0.25
Probability Rules (5 of 5)
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• A bucket contains 3 black balls and 7 green balls
Draw a ball from the bucket, replace it, and draw a
second ball!
1. The probability of a black ball drawn on first draw is:
P(B) = 0.30
2. The probability of two green balls drawn is:P(GG) = P(G) x P(G) = 0.7 x 0.7 = 0.49
Independent Events (1 of 2)
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• A bucket contains 3 black balls and 7 green balls
Draw a ball from the bucket, replace it, and draw a
second ball!
3. The probability of a black ball drawn on the second
draw if the first draw is green is:
P(B | G) = P(B) = 0.30
4. The probability of a green ball drawn on the second
draw if the first draw is green is:
P(G | G) = P(G) = 0.70
Independent Events (2 of 2)
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• An urn contains the following 10 balls:
o 4 are white (W) and lettered (L)
o 2 are white (W) and numbered (N)
o 3 are yellow (Y) and lettered (L)
o 1 is yellow (Y) and numbered (N)
P(WL) = 4/10 = 0.4 P(YL) = 3/10 = 0.3
P(WN) = 2/10 = 0.2 P(YN) = 1/10 = 0.1
P(W) = 6/10 = 0.6 P(L) = 7/10 = 0.7
P(Y) = 4/10 = 0.4 P(N) = 3/10 = 0.3
Dependent Events (1 of 3)
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Dependent Events (2 of 3)
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• The conditional probability that the ball drawn is
lettered, given that it is yellow
We can verify P(YL) using the joint probability formula:
Dependent Events (3 of 3)
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• Bayes’ theorem is used to incorporate additional
information and help create posterior probabilities from
original or prior probabilities
Revising Probabilities with Bayes’ Theorem (1 of 7)
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• A cup contains two dice identical in appearance but one
is fair (unbiased) and the other is loaded (biased)
o The probability of rolling a 3 on the fair die is 1/6 or 0.166
o The probability of tossing the same number on the loaded
die is 0.60
o We select one by chance, toss it, and get a 3
o What is the probability that the die rolled was fair?
o What is the probability that the loaded die was rolled?
Revising Probabilities with Bayes’ Theorem (2 of 7)
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• The probability of the die being fair or loaded is
P(fair) = 0.50 P(loaded) = 0.50
and that
P(3 | fair) = 0.166 P(3 | loaded) = 0.60
• The probabilities of P(3 and fair) and P(3 and loaded)
are
P(3 and fair) = P(3 | fair) x P(fair) = (0.166)(0.50) = 0.083
P(3 and loaded) = P(3 | loaded) x P(loaded) = (0.60)(0.50)
= 0.300
Revising Probabilities with Bayes’ Theorem (3 of 7)
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• The probability of the die being fair or loaded is
P(fair) = 0.50 P(loaded) = 0.50
and that
P(3 | fair) = 0.166 P(3 | loaded) = 0.60
• The probabilities of P(3 and fair) and P(3 and loaded)
are
P(3 and fair) = P(3 | fair) x P(fair) = (0.166)(0.50) = 0.083
P(3 and loaded) = P(3 | loaded) x P(loaded) = (0.60)(0.50)
= 0.300
The sum of these probabilities gives us the unconditional
probability of tossing a 3
P(3) = 0.083 + 0.300 = 0.383
Revising Probabilities with Bayes’ Theorem (4 of 7)
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• If a 3 does occur, the probability that the die rolled was
the fair one is
• The probability that the die was loaded is
Revising Probabilities with Bayes’ Theorem (5 of 7)
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• If a 3 does occur, the probability that the die rolled was
the fair one is
• The probability that the die was loaded is
• These are the revised or posterior probabilities for the
next roll of the die
• We use these to revise our prior probability estimates
Revising Probabilities with Bayes’ Theorem (6 of 7)
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• Event B has occurred
• A 3 is rolled
Revising Probabilities with Bayes’ Theorem (7 of 7)
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• We can compute revised probabilities more directly by
using
where
A’ = the complement of the event A;
for example, if A is the event “fair die”, then A’ is “loaded
die”
General Form of Bayes’ Theorem (1 of 2)
P(A | B) =P(B | A)P(A)
P(B | A)P(A)+P(B | ¢A )P( ¢A )
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• Conditional probability
From the previous example
Replace A with “fair die”, A’ with “loaded die”, B with “3
rolled”
General Form of Bayes’ Theorem (2 of 2)
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• Additional information from a second experiment
o If you can afford it, perform experiments several
times
o We roll the die again and again get a 3
Further Probability Revisions (1 of 4)
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• Additional information from a second experiment
• If you can afford it, perform experiments several times
• We roll the die again and again get a 3
Further Probability Revisions (2 of 4)
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• Additional information from a second experiment
• If you can afford it, perform experiments several times
• We roll the die again and again get a 3
Further Probability Revisions (3 of 4)
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• After the first roll of the die
probability the die is fair = 0.22
probability the die is loaded = 0.78
• After the second roll of the die
probability the die is fair = 0.067
probability the die is loaded = 0.933
Further Probability Revisions (4 of 4)
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• A random variable assigns a real number to every
possible outcome or event in an experiment
X = number of refrigerators sold during the day
• Discrete random variables can assume only a finite or
limited set of values
• Continuous random variables can assume any one of
an infinite set of values
Random Variables (1 of 3)
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Random Variables --- that are Numbers (2 of 3)
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Random Variables --- that are not Numbers (3 of 3)
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• For discrete random variables, probability value
assigned to each event
o Statistics class of 100 students
o Quiz with five problems with 1 point for each correct
answer
o Lowest score = 1, highest score = 5
• Examples follows the three rules
o Events are mutually exclusive and collectively exhaustive
o Individual probability values between 0 and 1
o Total probability sums to 1
Probability Distributions (1 of 4)
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• Developed using relative frequency approach
Probability Distributions --- Quiz Scores (2 of 4)
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Probability Distributions --- Class Distribution(3 of 4)
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Probability Distributions --- Class Distribution (4 of 4)
• Central tendency of the
distribution is the mean
or expected value
• Amount of variability is
the variance
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• Expected value is a measure of the central tendency of
the distribution
where
Expected Value of a Discrete Probability
Distribution (1 of 2)
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• For the quiz scores
Expected Value of a Discrete Probability
Distribution (2 of 2)
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where
Xi = random variable’s possible values
E(Xi) = expected value of the random variable
[Xi – E(X)] = difference between each value of the
random variable and the expected
value
E(X) = probability of each possible value of
the random variable
Variance of a Discrete Probability Distribution(1 of 4)
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• For quiz scores
Variance = (5 – 2.9)2(0.1) + (4 – 2.9)2(0.2) + (3 – 2.9)2(0.3)
+ (2 – 2.9)2(0.3) + (1 – 2.9)2(0.1)
= (2.1)2(0.1) + (1.1)2(0.2) + (0.1)2(0.3)
+ (–0.9)2(0.3) + (–1.9)2(0.1)
= 0.441 + 0.242 + 0.003 + 0.243 + 0.361
= 1.29
Variance of a Discrete Probability Distribution(2 of 4)
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• Standard deviation is the square root of the variance
where
Variance of a Discrete Probability Distribution(3 of 4)
s = Variance = s 2
= square root
s = standard deviation
For this example:
s = Variance
= 1.29 =1.14
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• The fundamental rules for continuous random variables
must be modified
o The sum of the probability values must still equal 1
o The probability of each individual value of the random
variable occurring must equal 0 or the sum would be
infinitely large
• The probability distribution is defined by a continuous
mathematical function called the probability density
function or just the probability function represented
by f (X)
Probability Distribution of a Continuous
Random Variable (1 of 3)
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Graph of Sample Density Function:
Probability Distribution of a Continuous
Random Variable (2 of 3)
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• For any continuous distribution, the probability does not
change if a single point is added to the range of values
that is being considered.
• The following probabilities are all exactly the same:
P(5.22 < X < 5.26) = P(5.22 < X ≤ 5.26) = P(5.22 ≤ X < 5.26)
= P(5.22 ≤ X ≤ 5.26)
Probability Distribution of a Continuous
Random Variable (3 of 3)
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• Many business experiments can be characterised by
the Bernoulli process
• The Bernoulli process is described by the binomial
probability distribution
o Each trial has only two possible outcomes
o The probability of each outcome stays the same
from one trial to the next
o The trials are statistically independent
o The number of trials is a positive integer
The Binomial Distribution (1 of 3)
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• The binomial distribution is used to find the probability of
a specific number of successes in n trials
We need to know
n = number of trials
p = the probability of success on any single trial
We let
r = number of successes
q = 1 – p = the probability of a failure
The Binomial Distribution (2 of 3)
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• The binomial formula is
Probability of r success in n trials
The symbol ! means factorial, and n! = n(n − 1)(n − 2)…(1)
4! = (4)(3)(2)(1) = 24
Also, 1! = 1 and 0! = 0 by definition
The Binomial Distribution (3 of 3)
!
!( )!
r n rnp q
r n r
n = number of trials p = the probability of success on any single trial
r = number of successes q = 1 – p = the probability of a failure
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Find the probability of getting 4 heads in 5 tosses of a coin
n = 5, r = 4, p = 0.5, and q = 1 − 0.5 = 0.5
P(4 successes in 5 trials)
Solving Problems with the Binomial Formula (1 of 3)
4 5-45!= 0.5 0.5
4!(5-4)!
5(4)(3)(2)(1)= (0.0625)(0.5) = 0.15625
4(3)(2)(1)1!
n = number of trials p = the probability of success on any single trial
r = number of successes q = 1 – p = the probability of a failure
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Binomial Distribution for n = 5, p = 0.50
Solving Problems with the Binomial Formula (2 of 3)
n = number of trials p = the probability of success on any single trial
r = number of successes q = 1 – p = the probability of a failure
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Binomial Distribution for n = 5, p = 0.50
Solving Problems with the Binomial Formula (3 of 3)
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MSA Electronics is experimenting with the manufacture of
a new transistor:
o Every hour a random sample of 5 transistors is taken
o The probability of one transistor being defective is
0.15
o What is the probability of finding 3, 4, or 5 defective?
n = 5, p = 0.15, and r = 3, 4, or 5
Solving Problems with Binomial Tables (1 of 4)
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A Sample Table for the Binomial Distribution:
We find the three probabilities in the table for
n = 5, p = 0.15, and r = 3, 4, and 5 and add them together
Solving Problems with Binomial Tables (2 of 4)
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Solving Problems with Binomial Tables (3 of 4)
A Sample Table for the Binomial Distribution:
We find the three probabilities in the table for
n = 5, p = 0.15, and r = 3, 4, and 5 and add them together
P(3 or more defects) = P(3) + P(4) + P(5)
= 0.0244 + 0.0022 + 0.0001
= 0.0267
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Expected value is:
Expected value (mean) = np
Variance = np(1 − p)
For the MSA example:
Expected value = np 5(0.15) = 0.75
Variance = np(1 − p) = 5(0.15)(0.85) = 0.6375
Solving Problems with Binomial Tables (4 of 4)
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• One of the most popular and useful continuous
probability distributions
o The probability density function
o Completely specified by the mean, μ, and the
standard deviation, σ
The Normal Distribution (1 of 4)
2
2
( )
21
( ) =2
x
f X e
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Normal Distribution with Different Values for μ
The Normal Distribution (2 of 4)
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Normal Distribution with Different Values for σ
The Normal Distribution (3 of 4)
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• Symmetrical with the midpoint representing the mean
• Shifting the mean does not change the shape
• Values on the X axis measured in the number of
standard deviations away from the mean
• As standard deviation becomes larger, curve flattens
• As standard deviation becomes smaller, curve becomes
steeper
The Normal Distribution (4 of 4)
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Step 1:
• Convert the normal distribution into a standard normal
distribution:
o Mean of 0 and a standard deviation of 1
o The new standard random variable is Z
where
X = value of the random variable we want to measure
μ = mean of the distribution
σ = standard deviation of the distribution
Z = number of standard deviations from X to the mean, μ
Using the Standard Normal Table (1 of 4)
=X
Z
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For μ = 100, σ = 15, find the probability that X is less than
130
Using the Standard Normal Table (2 of 4)
130 100= =
15
30= = 2 std dev
15
XZ
Normal Distribution Showing the
Relationship Between Z Values and X
Values:
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Step 2:
• Look up the probability from a table of normal curve
areas
• Use Appendix A or Table 2.10
• Column on the left is Z value
• Row at the top has second decimal places for Z values
Using the Standard Normal Table (3 of 4)
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Standardised Normal Distribution Function:
Using the Standard Normal Table (4 of 4)
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Builds three- and four-unit
apartment buildings:
o Total construction time follows
a normal distribution
o For triplexes, μ = 100 days and
σ = 20 days
o Contract calls for completion in
125 days
o Late completion will incur a
severe penalty fee
o Probability of completing in 125
days?
Haynes Construction Company (1 of 6)
Normal Distribution for
Haynes Construction
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• Compute Z
Haynes Construction Company (2 of 6)
125 – 100= =
20
25= =1.25
20
XZ
From Appendix A,
for Z = 1.25
area = 0.89435
Normal Distribution for
Haynes Construction
The probability is about 0.89
that Haynes will not violate the contract!
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If finished in 75 days or less, bonus = $5,000
o Probability of bonus?
o Because the distribution is symmetrical, equivalent
to Z = 1.25 … so area = 0.89435
Haynes Construction Company (3 of 6)
75 – 100
20
–25–1.25
20
XZ
Probability That Haynes Will
Receive the Bonus by Finishing in
75 Days or Less
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If finished in 75 days or less, bonus = $5,000
o Probability of bonus?
o Because the distribution is symmetrical, equivalent
to Z = 1.25 … so area = 0.89435
Haynes Construction Company (4 of 6)
75 – 100
20
–25–1.25
20
XZ
Probability That Haynes Will
Receive the Bonus by Finishing in
75 Days or Less
P(X > 125) = 1.0 − P(X ≤ 125)
= 1.0 − 0.89435 = 0.10565
The probability of completing the contract in 75 days or
less is about 11%
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Probability of completing between 110 and 125 days?
P(110 < X < 125) = P(X ≤ 125) − P(X < 110)
P(X ≤ 125) = 0.89435
Haynes Construction Company (5 of 6)
110 – 100
20
100.5
20
XZ
Probability That Haynes Will
Complete in 110 to 125 Days
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Probability of completing between 110 and 125 days?
P(110 < X < 125) = P(X ≤ 125) − P(X < 110)
P(X ≤ 125) = 0.89435
Haynes Construction Company (6 of 6)
110 – 100
20
100.5
20
XZ
Probability That Haynes Will
Complete in 110 to 125 Days
P(110 ≤ X < 125) = 0.89435 − 0.69146
= 0.20289
The probability of completing between 110 and 125 days
is about 20%
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Standardised Normal Distribution Function
Standard Normal Distribution
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For a normally distributed random variable with mean μ
and standard deviation σ
o Approximately 68% of values will be within ±1σ of the
mean
o Approximately 95% of values will be within ±2σ of the
mean
o Almost all (99.7%) of values will be within ±3σ of the
mean
The Empirical Rule (1 of 2)
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Approximate Probabilities from
the Empirical Rule´:
The Empirical Rule (2 of 2)
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It is a continuous probability distribution:
o The F statistic is the ratio of two sample
variances
o F distributions have two sets of degrees of freedom
o Degrees of freedom are based on sample size and
used to calculate the numerator and denominator
df1 = degrees of freedom for the numerator
df2 = degrees of freedom for the denominator
o The probabilities of large values of F are very small
The F Distribution (1 of 4)
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The F Distribution (2 of 4)
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Consider the example:
df1 = 5
df2 = 6
= 0.05
From Appendix D, we get
F, df1, df2= F0.05, 5, 6 = 4.39
This means
P(F > 4.39) = 0.05
The probability is only 0.05 F will exceed 4.39
The F Distribution (3 of 4)
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F Value for 0.05 Probability with 5 and 6 Degrees of
Freedom:
The F Distribution (4 of 4)
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• Also called the negative exponential distribution
o A continuous distribution often used in queuing
models
o Probability function given by
where
X = random variable (service times)
μ = average number of units the service facility can
handle in a specific period of time
e = 2.718 (the base of natural logarithms)
The Exponential Distribution (1 of 2)
( ) xf X e
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The Exponential Distribution (2 of 2)
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• Installs new mufflers on automobiles and small trucks
o Can install 3 new mufflers per hour
o Service time is exponentially distributed
• What is the probability that the time to install a new
muffler would be ½ hour or less?
X = Exponentially distributed service time
μ = average number of units the served per time period = 3
per hour
t = ½ hour = 0.5 hour
P(X ≤ 0.5) = 1 − e−3(0.5) = 1 − e −1.5 = 1 = 0.2231 = 0.7769
Arnold’s Muffler Shop (1 of 3)
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Probability That the Mechanic Will Install a Muffler in 0.5
Hour:
Arnold’s Muffler Shop (2 of 3)
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Similarly:
And:
Arnold’s Muffler Shop (3 of 3)
13
13
23
23
11 1 1 0.3679 0.6321
3
21 1 1 0.1353 0.8647
3
P X e e
P X e e
P(X > 0.5) = 1 − P(X ≤ 0.5) = 1 − 0.7769 = 0.2231
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• A discrete probability distribution
o Often used in queuing models to describe arrival
rates over time
o Probability function given by:
where
P(X) = probability of exactly X arrivals or occurrences
= average number of arrivals per unit of time (the mean arrival rate)
e = 2.718, the base of natural logarithms
X = number of occurrences (0, 1, 2, 3, …)
The Poisson Distribution (1 of 3)
( )!
xeP X
X
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From Appendix C for λ = 2
The Poisson Distribution (2 of 3)
2 0
2 1 2
2 2 2
( )!
2 (0.1353)1(0) 0.1353 14%
0! 1
2 2 0.1353(2)(1) 0.2706 27%
1! 1 1
2 4 0.1353(4)(2) 0.2706 27%
2! 2(1) 2
xeP X
X
eP
e eP
e eP
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Sample Poisson Distributions with λ = 2 and λ = 4
The Poisson Distribution (3 of 3)
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• Please read Chapter 3!
Homework --- Chapter 2
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If the sale of ice cream and pizza are independent, then as
ice cream sales decrease by 60 percent during the winter
months, pizza sales will
A. increase by 60 percent.
B. increase by 40 percent.
C. decrease by 40 percent.
D. be unrelated.
Multiple Choice Question
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If P(A) = 0.3, P(B) = 0.2, P(A and B) = 0.0, what can be
said about events A and B?
A. They are independent.
B. They are mutually exclusive.
C. They are posterior probabilities.
D. They are collectively exhaustive.
Multiple Choice Question
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Disco Fever is randomly found in one half of one percent
of the general population. Testing a swatch of clothing for
the presence of polyester is 99% effective in detecting the
presence of this disease. The test also yields a false-
positive in 4% of the cases where the disease is not
present. What is the probability that the test result is
negative if the disease is absent?
A. 0.95
B. 0.96
C. 0.99
D. 0.995
Multiple Choice Question