2-1
•For exams (MD1, MD2, and Final):
You may bring one 8.5” by 11” sheet of
paper with formulas and notes written or
typed on both sides to each exam.
2-2
Types of Data
Quantitative data are measurements that are recorded
on a naturally occurring numerical scale.
Qualitative data are measurements that cannot be
measured on a natural numerical scale; they can only be
classified into one of a group of categories.
2-3
Data Presentation
Data
Presentation
Qualitative
Data
Quantitative
Data
Dot
Plot
Summary
Table
Bar
Graph
Pie
Chart
Pareto
Diagram
Stem-&-Leaf
Display
Frequency
Distribution
Histogram
2-4
Example
1.72
2.5
2.16
2.13
1.06
2.24
2.31
2.03
1.09
1.40
2.57
2.64
1.26
2.05
1.19
2.13
1.27
1.51
2.41
1.95
stem
Leaf unit=0.01
1.0
69
1.1
9
1.2
67
stem
Leaf unit=0.1
1.4
0
1
001
1.5
1
1
22
1.7
2
1
45
1.9
5
1
7
2.0
35
1
9
2.1
336
2
00111
2.2
4
2
23
2.3
1
2
455
2.4
1
2
6
2.5
07
2.6
4
2-5
Example
1.72
2.5
2.16
2.13
1.06
2.24
2.31
2.03
1.09
1.40
2.57
2.64
1.26
2.05
1.19
2.13
1.27
1.51
2.41
1.95
2-6
Example
1.72
2.5
2.16
2.13
1.06
2.24
2.31
2.03
1.09
1.40
2.57
2.64
1.26
2.05
1.19
2.13
1.27
1.51
2.41
1.95
2-7
Two Characteristics
The central tendency of the set of
measurements–that is, the tendency of the data to
cluster, or center, about certain numerical values.
Central Tendency
(Location)
Center
2-8
Two Characteristics
The variability of the set of measurements–that
is, the spread of the data, spread around the
mean.
Sample B
Sample A
Variation
(Dispersion)
Variation of Sample B
Variation of Sample A
2-9
Mean
1.
2.
3.
4.
Most common measure of central tendency
Acts as ‘balance point’
Affected by extreme values (‘outliers’)
Denoted x where
Sample
mean x 
n
x i
i 1
n

x 1  x 2 … x
n
n
2-10
Median
1. Measure of central tendency
2. Middle value in ordered sequence
•
•
If n is odd, middle value of sequence
If n is even, average of 2 middle values
3. Position of median in sequence
n 1
Positioning Point 
2
4. Not affected by extreme values
2-11
Median Example
Even-Sized Sample
• Raw Data: 10.3 4.9 8.9 11.7 6.3 7.7
• Ordered: 4.9 6.3 7.7 8.9 10.3 11.7
• Position:
1
2
3
4
5
6
n 1 6 1
Positioning Point 

 3.5
2
2
7.7  8.9
Median 
 8.30
2
2-12
Mode Example
• No Mode
Raw Data: 10.3 4.9 8.9 11.7 6.3 7.7
• One Mode
Raw Data: 6.3 4.9 8.9
6.3 4.9 4.9
• More Than 1 Mode
Raw Data: 21 28
41
28
43
43
2-13
Shape
• Describes how data are distributed
• A data set is said to be skewed if one tail of the distribution
has more extreme observations than the other tail.
Left-Skewed Mode
Mean Median
Symmetric
Mean = Median = Mode
Right-Skewed
Mode
Median Mean
2-14
Example
•
•
•
•
Mean=45
Median=68
Mode=94
Is this data-set skewed? If it is, which direction
is the skewness?
2-15
Example
•
•
•
•
Mean=45
Median=68
Mode=94
Skewed to the left.
2-16
Sample Variance Formula
n
s 
2
 x
i 1
i
 x
2
n 1
x1  x    x2  x 


2
2

  xn  x 
2
n 1
n – 1 in denominator!
2-17
A shortcut formula for variance
2-18
Sample Standard Deviation
Formula
s  s2
n

 x
i 1
i
 x
n 1
 x1  x    x2  x 
2

2
2

  xn  x 
2
n 1
2-19
Thinking Challenge 1
• Why do we need to take square root of
variance to have a meaningful measure?
• Otherwise we would have a squared unit.
2-20
Interpreting Standard Deviation:
Chebyshev’s Theorem
x  3s
x  2s
xs
x
xs
x  2s
x  3s
No useful information
At least 3/4 of the data
At least 8/9 of the data
2-21
Interpreting Standard Deviation:
Empirical Rule
 – 3  – 2
–

–
 – 2
 – 3
Approximately 68% of the measurements
Approximately 95% of the measurements
Approximately 99.7% of the measurements
2-22
Empirical Rule Example
• According to the Empirical Rule, approximately 68%
of the data will lie in the interval (x – s, x + s),
(15.5 – 3.34, 15.5 + 3.34) = (12.16, 18.84)
• Approximately 95% of the data will lie in the interval
(x – 2s, x + 2s),
(15.5 – 2∙3.34, 15.5 + 2∙3.34) = (8.82, 22.18)
• Approximately 99.7% of the data will lie in the interval
(x – 3s, x + 3s),
(15.5 – 3∙3.34, 15.5 + 3∙3.34) = (5.48, 25.52)
2-23
Numerical Measures of
Relative Standing: Percentiles
• Describes the relative location of a
measurement compared to the rest of the data
• Descriptive measures of the relationship of a
measurement to the rest of the data are called
measures of relative standing.
• The pth percentile is a number such that p% of
the data falls below it and (100 – p)% falls
above it
• Median = 50th percentile
2-24
Percentile Example
• You scored 560 on the GMAT exam. This
score puts you in the 58th percentile.
• What percentage of test takers scored lower
than you did?
• What percentage of test takers scored higher
than you did?
2-25
Percentile Example
• What percentage of test takers scored lower
than you did?
58% of test takers scored lower than 560.
• What percentage of test takers scored higher
than you did?
(100 – 58)% = 42% of test takers scored
higher than 560.
2-26
Quartiles
• Percentiles that partition a data set into four
categories, each category contains exactly 25
percent of the measurements, are called
quartiles.
2-27
Example
1.06
1.09
1.19
1.26
1.27
1.4
1.51
1.72
1.95
2.03
2.05
2.13
2.13
2.16
2.24
2.31
2.41
2.5
2.57
2.64
Position for median=21/2=10.5
Median=(2.03+2.05)/2=2.04
Q1=median of the first half with
position=5.5→Q1=(1.27+1.4)/21.3
Q3=median of the second half with
position=5.5→Q3=(2.24+2.31)/22.3
2-28
Numerical Measures of
Relative Standing: z–Scores
• Describes the relative location of a
measurement (x) compared to the rest of the
data
• Sample z–score
Population z–score
xx
z
s
z
x µ

• Measures the number of standard deviations
away from the mean a data value is located
2-29
• The value of z-score reflects the relative standing of
the measurement.
• A large positive z-score implies that the measurement
is larger than almost all other measurements.
• A large value in negative magnitude indicates that the
measurements is smaller than almost all other
measurements.
• z score near 0 or is 0 means the measurement is
located at or near the mean of the sample or
population.
2-30
Interpretation of z–Scores
2-31
Box Plot
1.06
1.09
1.19
1.26
1.27
1.4
1.51
1.72
1.95
2.03
2.05
2.13
2.13
2.16
2.24
2.31
2.41
2.5
2.57
2.64
•The most extreme observation smaller than
upper inner fence(Q3+IQR*1.5=3.4)=2.64
•Q3
•Q2
•Q1
•The most extreme observation bigger than
upper inner fence(Q1-IQR*1.5=-0.2)=1.1
2-32
Box Plot
3. A second pair of fences, the outer fences, are
defined at a distance of 3(IQR) from the hinges.
One symbol (*) represents measurements falling
between the inner and outer fences, and another (0)
represents measurements beyond the outer fences.
4. Symbols that represent the median and extreme data
points vary depending on software used. You may
use your own symbols if you are constructing a box
plot by hand.
2-33
Outlier
An observation (or measurement) that is unusually large
or small relative to the other values in a data set is called
an outlier. Outliers typically are attributable to one of
the following causes:
1. The measurement is observed, recorded, or entered
into the computer incorrectly.
2. The measurement comes from a different
population.
3. The measurement is correct but represents a rare
(chance) event.
2-34
Key Ideas
Rules for Detecting Quantitative Outliers
Method
Box plot:
z-score
Suspect
Values
between inner
and outer
fences
Highly Suspect
Values beyond
outer fences
|z| > 3
2 < |z| < 3
2-35
Experiments & Sample Spaces
1. Experiment
• Process of observation that leads to a single
outcome that cannot be predicted with certainty
2. Sample point
• Most basic outcome of an
experiment
Sample Space
Depends on
Experimenter!
3. Sample space (S)
• Collection of all sample points
2-36
Visualizing
Sample Space
1.
Listing for the experiment of tossing a coin once and
noting up face
S = {Head, Tail}
Sample point
2.
A pictorial method for presenting the sample space
Venn Diagram
H
T
S
2-37
Example
• Experiment: Tossing two coins and
recording up faces:
• Is sample space as below?
S={HH, HT, TT}
2-38
Tree Diagram
1st coin
H
T
2nd coin
H
T
H
T
2-39
Sample Space Examples
•
•
•
•
•
•
•
Experiment
Sample Space
Toss a Coin, Note Face
Toss 2 Coins, Note Faces
Select 1 Card, Note Kind
Select 1 Card, Note Color
Play a Football Game
Inspect a Part, Note Quality
Observe Gender
{Head, Tail}
{HH, HT, TH, TT}
{2♥, 2♠, ..., A♦} (52)
{Red, Black}
{Win, Lose, Tie}
{Defective, Good}
{Male, Female}
2-40
Events
1. Specific collection of sample points
2. Simple Event
• Contains only one sample point
3. Compound Event
• Contains two or more sample points
2-41
What is Probability?
1. Numerical measure of the
likelihood that event will •1
occur
• P(Event)
• P(A)
•.5
• Prob(A)
•Certain
2. Lies between 0 & 1
3. Sum of probabilities for all
•0
sample points in the
sample space is 1
•Impossible
2-42
Equally Likely Probability
P(Event) = X / T
• X = Number of outcomes in the
event
• T = Total number of sample points
in Sample Space
• Each of T sample points is equally
likely
— P(sample point) = 1/T
© 1984-1994 T/Maker Co.
2-43
Thinking Challenge (sol.)
• Consider rolling two fair dice.
• Let event A=Having the sum of upfaces 6 or
less. So,
• A={ (1,1), (1,2), (1,3), (1,4), (1,5), (2,1),
(2,2), (2,3), (2,4), (3,1), (3,2),(3,3), (4,1),
(4,2), (5,1)}
each with prob.1/36
• P(A)=15/36=5/12
2-44
Combinations Rule
A sample of n elements is to be drawn from a set of N
elements. The, the number of different samples possible
 N
is denoted by   and is equal to
n 
 N
N!
 n   n!N  n !
where the factorial symbol (!) means that
n!=n*(n-1)*…*3*2*1
For example, 5!  5  4  3 2 1
0! is defined to be 1.
2-45
Thinking Challenge
• The price of a european tour includes four
stopovers to be selected from among 10
cities. In how many different ways can one
plan such a tour if the order of the
stopovers does not matter?
2-46
Unions & Intersections
1. Union
•
•
•
Outcomes in either events A or B or both
‘OR’ statement
Denoted by  symbol (i.e., A  B)
2. Intersection
•
•
•
Outcomes in both events A and B
‘AND’ statement
Denoted by  symbol (i.e., A  B)
2-47
• The table displays the probabilities for each of the six
outcomes when rolling a particular unfair die. Suppose
that the die is rolled once. Let A be the event that the
number rolled is less than 4, and let B be the event that
the number rolled is odd. Find P(AB).
Outcome
1
2
3
4
5
6
Probability 0.1
0.1
0.1
0.2
0.2
0.3
A. 0.5
B. 0.2
C. 0.3
D. 0.7
2-48
Event Probability Using
Two–Way Table
Event
Event
B1
B2
Total
A1
P(A 1  B1) P(A 1  B2) P(A 1)
A2
P(A 2  B1) P(A 2  B2) P(A 2)
Total
Joint Probability
P(B 1)
P(B 2)
1
Marginal (Simple) Probability
2-49
Thinking Challenge
What’s the Probability?
1. P(A) =
2. P(D) =
Event
C
D
4
2
3. P(C  B) =
Event
A
Total
6
4. P(A  D) =
B
1
3
4
5. P(B  D) =
Total
5
5
10
2-50
Solution*
The Probabilities Are:
1. P(A) = 6/10
2. P(D) = 5/10
Event
C
D
4
2
3. P(C  B) = 1/10
Event
A
Total
6
4. P(A  D) = 9/10
B
1
3
4
5. P(B  D) = 3/10 Total
5
5
10
2-51
Complementary Events
Complement of Event A
• The event that A does not occur
• All events not in A
• Denote complement of A by AC
AC
A
S
2-52
Rule of Complements
The sum of the probabilities of complementary events
equals 1:
P(A) + P(AC) = 1
AC
A
S
2-53
3.4
The Additive Rule and
Mutually Exclusive Events
2-54
Mutually Exclusive
Events Example
Experiment: Draw 1 Card. Note Kind & Suit.
Sample
Space:
2, 2,
2, ..., A


Event Spade:
2, 3, 4, ..., A
S
Outcomes
in Event
Heart:
2, 3, 4
, ..., A
Events  and are Mutually Exclusive
2-55
Additive 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)
2-56
Thinking Challenge
• Let P(A)=0.25 and P(BC)=0.4. If P(A ∪
B)=0.85. Are the two events A, B mutually
exclusive events?
A. True
B. False
2-57
Thinking Challenge
Using the additive rule, what is the probability?
1. P(A  D) =
2. P(B  C) =
Event
A
Event
C
D
4
2
Total
6
B
1
3
4
Total
5
5
10
2-58
Solution*
Using the additive rule, the probabilities are:
1. P(A  D) = P(A) + P(D) – P(A  D)
6
5
2
9
=
+
–
=
10 10 10 10
2. P(B  C) = P(B) + P(C) – P(B  C)
4
5
1
8
=
+
–
=
10 10 10 10
2-59
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(A  B
P(B)
P(B)
2-60
Thinking Challenge
Using the table then the formula, what’s the
probability?
1. P(A|D) =
2. P(C|B) =
Event
A
Event
C
D
4
2
Total
6
B
1
3
4
Total
5
5
10
2-61
Solution*
Using the formula, the probabilities are:
P A  B  2 5 2
P A D  


5
P D 
5
10
P(D)=P(AD)+P(BD)=2/10+3/10
P C  B  110 1
P C B  


4
P B 
4
10
P(B)=P(BD)+P(BC)=3/10+1/10
2-62
Multiplicative Rule
1. Used to get compound probabilities for
intersection of events
2. P(A and B) = P(A  B)
= P(A)  P(B|A)
= P(B)  P(A|B)
3. The key words both and and in the statement
imply and intersection of two events, which in
turn we should multiply probabilities to obtain the
probability of interest.
2-63
Thinking Challenge
• Suppose that 23% of adults smoke cigarettes. Given a
selected adult is a smoker, the probability that he/she
has a lung condition before the age of 60 is 57%. What
is the probability that a randomly selected person is a
smoker and has a lung condition before the age of 60.
A. 0.57
B. 0.99
C. 0.77
D. 0.13
2-64
Multiplicative Rule Example
Experiment: Draw 1 Card. Note Kind & Color.
Color
Type
Ace
Red
Black
2
2
Total
4
Non-Ace
24
24
48
Total
26
26
52
P(Ace  Black) = P(Ace)∙P(Black | Ace)
 4  2  2
    
 52  4  52
2-65
Statistical Independence
1. Event occurrence does not affect probability of
another event
• Toss 1 coin twice
2. Causality not implied
3. Tests for independence
• P(A | B) = P(A)
• P(B | A) = P(B)
• P(A  B) = P(A)  P(B)
2-66
Thinking Challenge
• Consider a regular deck of 52 cards with two black suits i.e.
♠(13),♣(13) and two red suits i.e. ♥(13) ♦(13). Given that you
have a red card what is the probability that it is a queen? Also
are the events getting a red card and getting a queen
independent?
A. 1/13, No (i.e. P(Queen)≠ P(Queen|Red))
B. 1/13, Yes (i.e. P(Queen)=P(Queen|Red)
C. 2/13, Yes (i.e. P(Queen)≠ P(Queen|Red)
D. 2/13, No (i.e. P(Queen)=P(Queen|Red)
2-67
Bayes’s Rule
Given k mutually exclusive and exhaustive events B1,
B1, . . . Bk , such that
P(B1) + P(B2) + … + P(Bk) = 1,
and an observed event A, then
P(Bi | A) 

P(Bi  A)
P( A)
P(Bi )P( A | Bi )
P(B1 )P( A | B1 )  P(B2 )P( A | B2 )  ...  P(Bk )P( A | Bk )
•Bayes’s rule is useful for finding one conditional probability
when other conditional probabilities are already known.
2-68
Bayes’s Rule Example
A company manufactures MP3 players at two
factories. Factory I produces 60% of the MP3
players and Factory II produces 40%. Two
percent of the MP3 players produced at Factory
I are defective, while 1% of Factory II’s are
defective. An MP3 player is selected at random
and found to be defective. What is the
probability it came from Factory I?
2-69
Bayes’s Rule Example
0 .6
0 .4
Factory
I
Factory
II
0.02
Defective
0.98
Good
0.01
Defective
0.99
Good
P(I | D) 
P(I )P(D | I )
0.6  0.02

 0.75
P(I )P(D | I )  P(II )P(D | II ) 0.6  0.02  0.4  0.01
2-70
Random Variable
A random variable is a variable that
assumes numerical values associated with
the random outcomes of an experiment,
where one (and only one) numerical value is
assigned to each sample point.
2-71
Random Variable (cont.)
• There are two types of random variables:
– Discrete random variables can take one of a
finite number of distinct outcomes.
• Example: Number of credit hours
– Continuous random variables can take any
numeric value within a range of values.
• Example: Cost of books this term
2-72
Discrete
Probability Distribution
The probability distribution of a discrete
random variable is a graph, table, or
formula that specifies the probability
associated with each possible value the
random variable can assume.
2-73
Requirements for the
Probability Distribution of a
Discrete Random Variable x
1. p(x) ≥ 0 for all values of x
2.  p(x) = 1
where the summation of p(x) is over all
possible values of x.
2-74
a) It is not valid
b) It is valid
c) It is not valid
d) It is not valid
2-75
a) {HHH,HTT,THT,TTH,THH,HTH,HHT,TTT}
{0,1,2,3}
b) {1/8,3/8,3/8,1/8}
d) P(x=2 or x=3)= P(x=2)+P(x=3)=3/8+1/8=1/2
2-76
Summary Measures
1. Expected Value (Mean of probability
distribution)
• Weighted average of all possible values
•  = E(x) = x p(x)
2. Variance
• Weighted average of squared deviation
about mean
• 2 = E[(x 2(x 2 p(x)=x2p(x)-2
3.
Standard Deviation
  2
2-77
Thinking challenge
• For the probability model given below,
what is the value of P and E(X)?
X2 3
4
5
P(x) 0.2 0.3 0.1 P
A.0.1, 3.0
B.0.2, 4.7
C. 0.3, 1.7
D. 0.4, 3.7
2-78
Binomial Probability
Characteristics of a Binomial Experiment
1.The experiment consists of n identical trials.
2.There are only two possible outcomes on each trial. We
will denote one outcome by S (for success) and the
other by F (for failure).
3.The probability of S remains the same from trial to trial.
This probability is denoted by p, and the probability of
F is denoted by q. Note that q = 1 – p.
4.The trials are independent.
5.The binomial random variable x is the number of S’s in n
trials.
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Binomial Probability
Distribution
 n  x n x
n!
x
nx
p( x)    p q 
p (1  p )
x ! ( n  x )!
 x
p(x) = Probability of x ‘Successes’
p = Probability of a ‘Success’ on a single trial
q = 1–p
n = Number of trials
x = Number of ‘Successes’ in n trials
(x = 0, 1, 2, ..., n)
n – x = Number of failures in n trials
2-80
Binomial Probability
Distribution Example
Experiment: Toss 1 coin 5 times in a row. Note
number of tails. What’s the probability of 3
n!
tails?
p( x) 
p x (1  p ) n  x
x !(n  x)!
•© 1984-1994 T/Maker Co.
5!
p (3) 
.53 (1  .5)53
3!(5  3)!
 .3125
2-81
Binomial Distribution
Characteristics
n = 5 p = 0.1
•Mean
  E ( x )  np
P(X)
1.0
.5
.0
X
•Standard Deviation
  npq
0
1
2
3
4
5
n = 5 p = 0.5
.6
.4
.2
.0
P(X)
X
0
1
2
3
4
5
2-82
Binomial Distribution
Thinking Challenge
You’re a telemarketer selling
service contracts for Macy’s.
You’ve sold 20 in your last 100
calls (p = .20). If you call 12
people tonight, what’s the
probability of
A. No sales?
B. Exactly 2 sales?
C. At most 2 sales?
D. At least 2 sales?
2-83
Binomial Distribution Solution*
n = 12, p = .20
E(X)=n*p=12*0.2=2.4
=(np(1-p))1/2=(12*0.2*0.8)1/2 =1.38
A. p(0) = .0687
B. p(2) = .2835
C. p(at most 2)
= p(0) + p(1) + p(2)
= .0687 + .2062 + .2835
= .5584
D. p(at least 2)
= p(2) + p(3)...+ p(12)
= 1 – [p(0) + p(1)]
= 1 – .0687 – .2062
= .7251
2-84
By using TI-84:
• B.P(X = 2) = p(2) = P(X ≤ 2) – P(X ≤ 1)
• binomcdf(10,.20,2) - binomcdf(10,.20,2)
2-85