1.
Unit 4 : Probability concepts
Probability distributions

2.
BASIC PROBABILITY &
PROBABILITY DISTRIBUTIONS

3.
3
Basic Concepts
Random Experiment is a process leading
to at least two possible outcomes with
uncertainty as to which will occur.
A coin is thrown
A consumer is asked which of two
products he or she prefers
The daily change in an index of stock
market prices is observed

4.
4
Sample Spaces
Collection of all possible outcomes
e.g.: All six faces of a die:
e.g.: All 52 cards
a deck of
bridge cards

5.
5
Events and Sample Spaces
An event is a set of basic outcomes from
the sample space, and it is said to occur
if the random experiment gives rise to
one of its constituent basic outcomes.
Simple Event
Outcome With 1 Characteristic
Joint Event
2 Events Occurring Simultaneously
Compound Event
One or Another Event Occurring

6.
6
Simple Event
A: Male
B: Over age 20
C: Has 3 credit cards
D: Red card from a deck of bridge cards
E: Ace card from a deck of bridge cards

7.
7
Joint Event
D and E, (DE):
Red, ace card from a bridge deck
A and B, (AB):
Male, over age 20
among a group of
survey respondents

8.
8
Intersection
• Let A and B be two events in the sample space S.
Their intersection, denoted AB, is the set of all
basic outcomes in S that belong to both A and B.
• Hence, the intersection AB occurs if and only if
both A and B occur.
• If the events A and B have no common basic
outcomes, their intersection AB is said to be the
empty set.

9.
9
Union
•Let A and B be two events in the
sample space S. Their union, denoted
AB, is the set of all basic outcomes in
S that belong to at least one of these
two events.
•Hence, the union AB occurs if and
only if either A or B or both occurs

10.
10
Event Properties
Mutually Exclusive
Two outcomes that cannot occur at the same time
E.g. flip a coin, resulting in head and tail
Collectively Exhaustive
One outcome in sample space must occur
E.g. Male or Female

11.
Special Events
Null Event
Club & Diamond on
1 Card Draw
Complement of Event
For Event A, All
Events Not In A:
A' or A

12.
12
What is Probability?
1. Numerical measure of likelihood that the
event
will occur
Simple Event
Joint Event
Compound
2. Lies between 0 & 1
3. Sum of events is 1

13.
13
Concept of Probability
A Priori classical probability, the probability of success
is based on prior knowledge of the process involved.
i.e. the chance of picking a black card from a deck of
bridge cards
Empirical classical probability, the outcomes are
based on observed data, not on prior knowledge of
a process.
i.e. the chance that individual selected at random from
the Kalosha employee survey if satisfied with his or her
job. (.89)

14.
14
Concept of Probability
Subjective probability, the chance of
occurrence assigned to an event by a
particular individual, based on his/her
experience, personal opinion and
analysis of a particular situation.
i.e. The chance of a newly designed style of
mobile phone will be successful in market.

15.
15
(There are 2 ways to get one 6 and the other 4)
e.g. P( ) = 2/36
Computing Probabilities
• The probability of an event E:
• Each of the outcomes in the sample space is
equally likely to occur
number of event outcomes
( )
total number of possible outcomes in the sample space
P E
X
T



16.
16
Presenting Probability &
Sample Space
1. Listing
S = {Head, Tail}
2. Venn Diagram
3. Tree Diagram
4. Contingency
Table

17.
17
S
Ā
A
Venn Diagram
Example: ABC Employee Survey
Event: A = Satisfied, Ā = Dissatisfied
P(A) = 356/400 = .89, P(Ā) = 44/400 = .11

18.
18
Kalosha
Employee
Tree Diagram
Satisfied
Not
Satisfied
Advanced
Not
Advanced
Advanced
Not
Advanced
P(A)=.89
P(Ā)=.11
.485
.405
.035
.075
Example: ABC Employee Survey Joint
Probability

19.
19
Event
Event B1 B2 Total
A1 P(A1  B1) P(A1  B2) P(A1)
A2 P(A2  B1) P(A2  B2) P(A2)
Total P(B1) P(B2) 1
Joint Probability
Using Contingency Table
Joint Probability Marginal (Simple)
Probability

20.
20
Joint Probability
Using Contingency Table
ABC Employee Survey
Satisfied
Not
Satisfied
Total
Advanced
Not
Advanced
Total
.485 .035
.405 .075
.52
.48
.89 .11 1.00
Joint Probability
Simple Probability

21.
21
Use of Venn Diagram
Fig. 3.1: AB, Intersection of events A & B, mutually
exclusive
Fig. 3.2: AB, Union of events A & B
Fig. 3.3: Ā, Complement of event A
Fig. 3.4 and 3.5:
The events AB and ĀB are mutually exclusive,
and their union is B.
(A  B)  (Ā  B) = B

22.
22
Use of Venn Diagram
Let E1, E2,…, Ek be K mutually exclusive and collective
exhaustive events, and let A be some other event.
Then the K events E1  A, E2  A, …, Ek  A are
mutually exclusive, and their union is A.
(E1  A)  (E2  A)  …  (Ek  A) = A

23.
23
Compound Probability
Addition Rule
1. Used to Get Compound Probabilities for Union of
Events
2. P(A or B) = P(A  B)
= P(A) + P(B)  P(A  B)
3. For Mutually Exclusive Events:
P(A or B) = P(A  B) = P(A) + P(B)
4. Probability of Complement
P(A) + P(Ā) = 1. So, P(Ā) = 1  P(A)

24.
24
Addition Rule: Example
A hamburger chain found that 75% of all customers
use mustard, 80% use ketchup, 65% use both.
What is the probability that a particular customer
will use at least one of these?
A = Customers use mustard
B = Customers use ketchup
AB = a particular customer will use at least one
of these
Given P(A) = .75, P(B) = .80, and P(AB) = .65,
P(AB) = P(A) + P(B)  P(AB)
= .75 + .80  .65= .90

25.
25
Conditional Probability
1. Event Probability Given that Another
Event Occurred
2. Revise Original Sample Space to
Account for New Information
Eliminates Certain Outcomes
3. P(A | B) = P(A and B) , P(B)>0
P(B)

26.
26
Example
Recall the previous hamburger chain
example, what is the probability that a
ketchup user uses mustard?
P(A|B) = P(AB)/P(B)
= .65/.80 = .8125
Please pay attention to the difference
from the joint event in wording of the
question.

27.
27
S
Black
Ace
Conditional Probability
Black ‘Happens’: Eliminates All Other Outcomes and
Thus Increase the Conditional Probability
Event (Ace and Black)
(S)
Black
Draw a card, what is the probability of black ace?
What is the probability of black ace when black happens?

28.
28
Color
Type Red Black Total
Ace 2 2 4
Non-Ace 24 24 48
Total 26 26 52
Conditional Probability Using Contingency Table
Conditional Event: Draw 1 Card. Note Black Ace
Revised
Sample
Space
P(Ace|Black) =
P(Ace and Black)
P(Black)
= = 2/26
2/52
26/52

29.
29
Statistical Independence
1. Event Occurrence Does Not Affect Probability of
Another Event
e.g. Toss 1 Coin Twice, Throw 3 Dice
2. Causality Not Implied
3. Tests For Independence
P(A | B) = P(A), or P(B | A) = P(B),
or P(A and B) = P(A)P(B)

30.
30
Statistical Independence
ABC Employee Survey
Satisfied
Not
Satisfied
Total
Advanced
Not
Advanced
Total
.485 .035
.405 .075
.52
.48
.89 .11 1.00
P(A1)
Note: (.52)(.89) = .4628  .485
P(B1)
P(A1and B1)

31.
31
Multiplication Rule
1. Used to Get Joint Probabilities for
Intersection of Events (Joint Events)
2. P(A and B) = P(A  B)
P(A  B) = P(A)P(B|A)
= P(B)P(A|B)
3. For Independent Events:
P(A and B) = P(AB) = P(A)P(B)

32.
32
Practice of Randomized Response
Question
a) Is the second last digit of your office phone
number odd?
b) Recall your undergraduate course work, have you
ever cheated in midterm or final exams?
Respondents, please do as follows:
1. Flip a coin.
2. If the result is the national emblem answer
question a); otherwise answer question b).
Please circle “Yes” or “No” below as your answer.
Yes No

33.
33
Bayes’ Theorem
1. Permits Revising Old
Probabilities Based on
New Information
2. Application of
Conditional Probability
3. Mutually Exclusive
Events
New
Information
Revised
Probability
Apply Bayes'
Theorem
Prior
Probability

34.
34
Bayes’s Theorem Example
Fifty percent of borrowers repaid their loans. Out of
those who repaid, 40% had a college degree. Ten
percent of those who defaulted had a college degree.
What is the probability that a randomly selected
borrower who has a college degree will repay the loan?
B1= repay, B2= default, A=college degree
P(B1) = .5, P(A|B1) = .4, P(A|B2) = .1, P(B1|A) =?
)
(
)
|
(
)
(
)
|
(
)
(
)
|
(
)
|
(
2
2
1
1
1
1
1
B
P
B
A
P
B
P
B
A
P
B
P
B
A
P
A
B
P


8
.
25
.
2
.
)
5
)(.
1
(.
)
5
)(.
4
(.
)
5
)(.
4
(.





35.
35
Event Prior
Prob
Cond.
Prob
Joint
Prob
Post.
Prob
Bi P(Bi) P(A|Bi) P(Bi  A) P(Bi |A)
B1 .5 .4 .20 .20/.25 = .8
B2 .5 .1 .05 .05/.25 = .2
1.0 P(A) = 0.25 1.0
Bayes’ Theorem Example
Table Solution
Default
Repay P(College)
X =

36.
36
Permutation and Combination
Counting Rule 1
Example：In TV Series: ABC tossed 50 coins, the
number of outcomes is 2·2 ·… ·2 ＝ 250. What is the
probability of all coins with heads up?
If any one of n different mutually exclusive and
collectively exhaustive events can occur on each of
r trials, the number of possible outcomes is equal to：
n·n ·… ·n ＝ nr

37.
37
Permutation and Combination
Applicatin: If a license plate consists of 3 letters
followed by 3 digits, the total number of
outcomes would be? (most states in the US)
Application: China License Plates
How many licenses can be issued?
Style 1992: one letter or digit plus 4 digits.
Style 2002: 1) three letters + three digits
2) three digits + three letters
3) three digits + three digits

38.
38
Permutation and Combination
Counting Rule 2
Example: The number of ways that 5 books could
be arranged on a shelf is: (5)(4)(3)(2)(1) = 120
The number of ways that all n objects can be
arranged in order is：
= n(n -1)(n -2)(2)(1) = n!
Where n! is called factorial and 0! is
defined as 1.
n
n
P

39.
39
Permutation and Combination
Counting Rule 3: Permutation
Example：What is the number of ways of arranging 3
books selected from 5 books in order? (5)(4)(3)
= 60
The number of ways of arranging r objects selected
from n objects in order is：
)!
(
!
r
n
n
P
n
r



40.
40
Permutation and Combination
Counting Rule 4: Combination
Example：The number of combinations of 3 books
selected from 5 books is
(5)(4)(3)/[(3)(2)(1)] = 10
Note: 3! possible arrangements in order are irrelevant
The number of ways that arranging r objects selected
from n objects, irrespective of the order, is equal to
)!
(
!
!
r
n
r
n
r
n
Cn
r












41.
PROBABILITY DISTRIBUTIONS
BINOMIAL
POISSON
41

42.
Chap 5-42
Discrete Random Variables
• Can only assume a countable number of values
Examples:
• Roll a die twice
Let X be the number of times 4 comes up
(then X could be 0, 1, or 2 times)
• Toss a coin 5 times.
Let X be the number of heads
(then X = 0, 1, 2, 3, 4, or 5)

43.
Chap 5-43
• Variance of a discrete random variable
• Standard Deviation of a discrete random variable
where:
E(X) = Expected value of the discrete random variable X
Xi = the ith outcome of X
P(Xi) = Probability of the ith occurrence of X
Discrete Random Variable
Summary Measures




N
1
i
i
2
i
2
)
P(X
E(X)]
[X
σ
(continued)





N
1
i
i
2
i
2
)
P(X
E(X)]
[X
σ
σ

44.
Chap 5-44
• Example: Toss 2 coins, X = # heads,
compute standard deviation (recall E(X) = 1)
Discrete Random Variable
Summary Measures
)
P(X
E(X)]
[X
σ i
2
i

 
0.707
0.50
(0.25)
1)
(2
(0.50)
1)
(1
(0.25)
1)
(0
σ 2
2
2








(continued)
Possible number of heads = 0,
1, or 2

45.
Chap 5-45
The Covariance
• The covariance measures the strength of the linear
relationship between two variables
• The covariance:
)
Y
X
(
P
)]
Y
(
E
Y
)][(
X
(
E
X
[
σ
N
1
i
i
i
i
i
XY 




where: X = discrete variable X
Xi = the ith outcome of X
Y = discrete variable Y
Yi = the ith outcome of Y
P(XiYi) = probability of occurrence of the
ith outcome of X and the ith outcome of Y

46.
Chap 5-46
The Sum of
Two Random Variables
• Expected Value of the sum of two random variables:
• Variance of the sum of two random variables:
• Standard deviation of the sum of two random variables:
XY
2
Y
2
X
2
Y
X σ
2
σ
σ
σ
Y)
Var(X 



 
)
Y
(
E
)
X
(
E
Y)
E(X 


2
Y
X
Y
X σ
σ 
 

47.
Probability Distributions
Continuous
Probability
Distributions
Binomial
Hypergeometric
Poisson
Probability
Distributions
Discrete
Probability
Distributions
Normal
Uniform
Exponential

48.
Binomial Probability Distribution
 A fixed number of observations, n
 e.g., 15 tosses of a coin; ten light bulbs taken from a warehouse
 Two mutually exclusive and collectively exhaustive categories
 e.g., head or tail in each toss of a coin; defective or not defective light
bulb
 Generally called “success” and “failure”
 Probability of success is p, probability of failure is 1 – p
 Constant probability for each observation
 e.g., Probability of getting a tail is the same each time we toss the coin

49.
Binomial Probability Distribution
(continued)
 Observations are independent
 The outcome of one observation does not affect the
outcome of the other
 Two sampling methods
 Infinite population without replacement
 Finite population with replacement

50.
Binomial Distribution Settings
• A manufacturing plant labels items as either
defective or acceptable
• A firm bidding for contracts will either get a
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

51.
Rule of Combinations
• The number of combinations of selecting X objects out of n objects is
X)!
(n
X!
n!
Cx
n


where:
n! =(n)(n - 1)(n - 2) . . . (2)(1)
X! = (X)(X - 1)(X - 2) . . . (2)(1)
0! = 1 (by definition)

52.
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)
n = sample size (number of trials
or observations)
p = probability of “success”
P(X)
n
X ! n X
p (1-p)
X n X
!
( )!



Example: Flip a coin four
times, let x = # heads:
n = 4
p = 0.5
1 - p = (1 - 0.5) = 0.5
X = 0, 1, 2, 3, 4
Binomial Distribution Formula

53.
Calculating a Binomial Probability
What is the probability of one success in five
observations if the probability of success is .1?
X = 1, n = 5, and p = 0.1
0.32805
.9)
(5)(0.1)(0
0.1)
(1
(0.1)
1)!
(5
1!
5!
p)
(1
p
X)!
(n
X!
n!
1)
P(X
4
1
5
1
X
n
X












54.
n = 5 p = 0.1
n = 5 p = 0.5
Mean
0
.2
.4
.6
0 1 2 3 4 5
X
P(X)
.2
.4
.6
0 1 2 3 4 5
X
P(X)
0
Binomial Distribution
• The shape of the binomial distribution depends on the
values of p and n
 Here, n = 5 and p = 0.1
 Here, n = 5 and p = 0.5

55.
Binomial Distribution Characteristics
• Mean
 Variance and Standard Deviation
np
E(x)
μ 

p)
-
np(1
σ2

p)
-
np(1
σ 
Where n = sample size
p = probability of success
(1 – p) = probability of failure

56.
Using Binomial Tables
n = 10
x … p=.20 p=.25 p=.30 p=.35 p=.40 p=.45 p=.50
0
1
2
3
4
5
6
7
8
9
10
…
…
…
…
…
…
…
…
…
…
…
0.1074
0.2684
0.3020
0.2013
0.0881
0.0264
0.0055
0.0008
0.0001
0.0000
0.0000
0.0563
0.1877
0.2816
0.2503
0.1460
0.0584
0.0162
0.0031
0.0004
0.0000
0.0000
0.0282
0.1211
0.2335
0.2668
0.2001
0.1029
0.0368
0.0090
0.0014
0.0001
0.0000
0.0135
0.0725
0.1757
0.2522
0.2377
0.1536
0.0689
0.0212
0.0043
0.0005
0.0000
0.0060
0.0403
0.1209
0.2150
0.2508
0.2007
0.1115
0.0425
0.0106
0.0016
0.0001
0.0025
0.0207
0.0763
0.1665
0.2384
0.2340
0.1596
0.0746
0.0229
0.0042
0.0003
0.0010
0.0098
0.0439
0.1172
0.2051
0.2461
0.2051
0.1172
0.0439
0.0098
0.0010
10
9
8
7
6
5
4
3
2
1
0
… p=.80 p=.75 p=.70 p=.65 p=.60 p=.55 p=.50 x
Examples:
n = 10, p = 0.35, x = 3: P(x = 3|n =10, p = 0.35) = 0.2522
n = 10, p = 0.75, x = 2: P(x = 2|n =10, p = 0.75) = 0.0004

57.
The Poisson Distribution
• Apply the Poisson Distribution when:
• You wish to count the number of times an event occurs in a
given area of opportunity
• The probability that an event occurs in one area of
opportunity is the same for all areas of opportunity
• The number of events that occur in one area of
opportunity is independent of the number of events that
occur in the other areas of opportunity
• The probability that two or more events occur in an area of
opportunity approaches zero as the area of opportunity
becomes smaller
• The average number of events per unit is  (lambda)

58.
Poisson Distribution Formula
where:
X = number of events in an area of opportunity
 = expected number of events
e = base of the natural logarithm system (2.71828...)
!
X
e
)
X
(
P
x





59.
Poisson Distribution Characteristics
• Mean
 Variance and Standard Deviation
λ
μ 
λ
σ2

λ
σ 
where  = expected number of events

60.
Using Poisson Tables
X

0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90
0
1
2
3
4
5
6
7
0.9048
0.0905
0.0045
0.0002
0.0000
0.0000
0.0000
0.0000
0.8187
0.1637
0.0164
0.0011
0.0001
0.0000
0.0000
0.0000
0.7408
0.2222
0.0333
0.0033
0.0003
0.0000
0.0000
0.0000
0.6703
0.2681
0.0536
0.0072
0.0007
0.0001
0.0000
0.0000
0.6065
0.3033
0.0758
0.0126
0.0016
0.0002
0.0000
0.0000
0.5488
0.3293
0.0988
0.0198
0.0030
0.0004
0.0000
0.0000
0.4966
0.3476
0.1217
0.0284
0.0050
0.0007
0.0001
0.0000
0.4493
0.3595
0.1438
0.0383
0.0077
0.0012
0.0002
0.0000
0.4066
0.3659
0.1647
0.0494
0.0111
0.0020
0.0003
0.0000
Example: Find P(X = 2) if  = 0.50
0.0758
2!
(0.50)
e
X!
e
2)
P(X
2
0.50
X
λ






λ

61.
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
 =
0.50
0
1
2
3
4
5
6
7
0.6065
0.3033
0.0758
0.0126
0.0016
0.0002
0.0000
0.0000
P(X = 2) = 0.0758
Graphically:
 = 0.50

62.
Poisson Distribution Shape
• The shape of the Poisson Distribution depends
on the parameter  :
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)
 = 0.50  = 3.00

63.
63
Continuous Probability Distributions
•Continuous random variable
• Values from an interval of numbers
• Absence of gaps
•Continuous probability distribution
• Probability distribution of continuous
random variable (probability of an
interval)
• The most important continuous probability
distribution: The normal distribution

64.
64
Normal Distribution
Probability Density Function
• ‘Bell-Shaped’ &
Symmetrical
• Mean & Median Are
Equal
• Random Variable Has
Infinite Range Mean
Median
X
f(X)
Application: From discrete to continuous
Working Years of BiMBA Classes

65.
65
Normal Distribution
Probability Density Function
f(X) : probability density function of X
 : mean
 : standard deviation
 = 3.14159; e = 2.71828
X = value of random variable (- < X < )
2
2
1
2
1
)
( 




















 






X
e
X
f

66.
66
Z=0
z=1
Z
Standardize the
Normal Distribution
One table!
Normal
Distribution
Standardized
Normal Distribution
X


Z =
X – 

f(X)
f (Z)

67.
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Standardized Normal Distribution
6.2 5
0.12
10
X
Z


 
  
Normal Distribution Standardized
Normal Distribution
Shaded Area Exaggerated
10
  1
Z
 
5
 
6.2 X Z
0
Z
 
0.12
f(X)

68.
68
Standardized Normal Distribution
Z .00 .01
0.0 .5000 .5040 .5080
.5398 .5438
0.2 .5793 .5832 .5871
0.3 .6179 .6217 .6255
.5478
.02
0.1 .5478
Cumulative Standardized Normal
Distribution Table (Portion)
Probabilities
Only One Table is Needed
0 1
Z Z
 
 
Z = 0.12
0

69.
69
Normal Distribution Standardized
Normal Distribution
Shaded Area Exaggerated
10
 
1
Z
 
5
 
X Z
0
Z
 
2.9 5 8.1 5
.21 .31
10 10
X X
Z Z
 
 
   
      
0.21
 .31
8.1
2.9
P(2.9 < X < 8.1) = .2049

70.
70
Z .00 .01
0.0 .5000 .5040 .5080
.5398 .5438
0.2 .5793 .5832 .5871
0.3 .6179 .6217 .6255
.6217
.02
0.1 .5478
Cumulative Standardized Normal
Distribution Table (Portion)
Shaded Area Exaggerated
0 1
Z Z
 
 
Z = 0.31
(continued)
0
P(2.9 < X < 8.1) = .2049

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71
Z .00 .01
-03 .3821 .3783 .3745
.4207 .4168
-0.1.4602 .4562 .4522
0.0 .5000 .4960 .4920
.4168
.02
-02 .4129
Cumulative Standardized Normal
Distribution Table (Portion) 0 1
Z Z
 
 
Z = -0.21
(continued)
0
.6217.4168 = .2049
P(2.9 < X < 8.1) = .2049

72.
72
P(X > 8) = .3821
Normal Distribution Standardized
Normal Distribution
Shaded Area Exaggerated
10
 
1
Z
 
5
 
X Z
0
Z
 
8 5
.30
10
X
Z


 
  
?
.30
8

73.
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Normal Distribution and 
• Refer to Fig. 6.11, 6.12, and 6.13 (p. 204-205)
68.26% of normal values fall between 1
95.44% of normal values fall between 2
99.73% of normal values fall between 3
 Check Standardized Normal Distribution Table and
find the probability of normal values fall between 4,
5, and 6

74.
74
Normal Distribution
Thinking Challenge
You work in Quality Control
for GE. Light bulb life has a
normal distribution with
= 2000 hours &  = 200
hours. What’s the probability
that a bulb will last?
a. between 2000 & 2400
hours?
b. less than 1470 hours

75.
75
Z
Z= 0
Z = 1
2.0
You work in Quality Control for GE. Light bulb life has a normal distribution with
= 2000 hours &  = 200 hours. What’s the probability that a bulb will last? a.
between 2000 & 2400
hours?
b. less than 1470 hours Solution: P(2000  X  2400)
Normal
Distribution
.4772
Standardized
Normal Distribution
Z
X







2400 2000
200
2.0
X
 = 2000
 = 200
2400

76.
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.6217
Finding Z Values for Known Probabilities
Z .00 0.2
0.0 .5000 .5040 .5080
0.1 .5398 .5438 .5478
0.2 .5793 .5832 .5871
.6179 .6255
.01
0.3
Cumulative Standardized
Normal Distribution Table
What is Z Given
Probability = 0.6217 ?
Shaded Area
Exaggerated
.6217
0 1
Z Z
 
 
.31
Z 
0 Z

77.
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Uniform Distribution
• The probability
density function:
Mean:
Variance:
1
( )
0
for a x b
b a
f x
othrewise

 
 

 



2
a b



2
2 ( )
12
b a



2 1
1 2 1 2
( ) ,
x x
P x X x a x x b
b a

     


78.
78
Exponential Distribution
Density function: (X > 0)
Mean and
Standard Deviation:
f(X)
X
 = 1.0
 = 0.5
  X
f X e 
 

1 1
 
 
 

79.
79
Exponential Distribution
f (X) = eX
P(a < X < b) = e -a – e -b
a b X

0
f (X)

80.
80
Exponential Distribution
P( X < b ) =1 - e
 : expected (average) number of “arrivals”
X : time between successive “arrivals”
a, b : value of the continuous random variable
e = 2.71828
- b
( 0 < b <∞)
The cumulative distribution function is :
P(a < X < b) = e-a – e-b
( 0 < a < b)
Or

81.
81
Example
Customers arrive at the check out line of a
supermarket at the rate of 30 per hour. What
is the probability that the arrival time between
consecutive customers to be greater than 5
minutes?
   
 
 
 
.5 5
30/60 .5/minute, 5minutes
> 1
1 1
.0821
b
b
P X b P X b
e
e




  
  
  
 

82.
82
Application
If the number of times that a particular event occurs
over an interval of time or space follows Poisson
distribution. Then the time or space between
successive occurrences of the event follows
exponential distribution.
The record shows the loss of a bike every 3 years.
a) What is the probability that the time between
successive losses of bikes is more than a year?
b) What is the probability of less than one bike loss in a
year?

83.
• Thank you