Both the quizzes and exams are closed book.
However,
•For quizzes:
Formulas will be provided with quiz papers if there
is any need.
•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.
Statistics for Business and
Economics
Chapter 3
Probability
Review questions
Q1
The masterfoods company says that before the introduction of purple,
in a pack of 100 candies, yellow candies made up 20% of their plain
M&M’s, red another 20%, and orange, blue and green each made up
10%. The rest were brown.
a) If you pick an M&M at random, what is the probability that
1) It is brown?
0.30
2) It is yellow or orange?
P(YO)=P(Y)+P(O)-P(YO)=0.20+0.1+0=0.30
3) It is not green?
P(Gc)=1-P(G)=0.9
Review questions
Q1(cont.)
b) If you pick three M&M’s in a row, without replacement, what is the
probability that
1) They are all brown?
P(BrBrBr)=P(Br)P(BrBr)P(BrBrBr)=(30/100)(29/99)(28/98)
2) Both of the last two are red?
P(BRR)+P(BrRR)+P(YRR)+P(ORR)+P(GRR)
P(BRR)=P(B)P(RB)P(RBR)=(10/100)(20/99)(19/98)
P(BrRR)=P(Br)P(RBr)P(RBrR)=(30/100)(20/99)(19/98)
P(YRR)=P(Y)P(RY)P(RYR)=(20/100)(20/99)(19/98)
P(ORR)=P(O)P(RO)P(ROR)=(10/100)(20/99)(19/98)
P(GRR)=P(G)P(RG)P(RGR)=(10/100)(20/99)(19/98)
Q2
• A certain bowler can bowl a strike 70% of the time. What
is the probability that she
A) goes three consecutive frames without a strike?
0.3*0.3*0.3=0.027
B) makes her first strike in the third frame?
0.3*0.3*0.7=0.063
C) Has at least one strike in the first three games?
3*(0.3*0.3*0.7)+3*(0.3*0.7*0.7)+0.7*0.7*0.7)=0.973
D) Bowls a perfect game (12 consecutive strikes)?
(0.7)12=0.0138
a) 45/57
b) 50/111
c) No.
3.7
Bayes’s Rule
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.
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?
Let events be
•U+=Athlete uses testosterone
•U- = Athlete do not use testosterone
•T+=Test is positive
•T- = Test is negative
We are given;
•P(U+)=100/1000=0.1
•P(T+ U+)=50/100=0.5
•P(T+ U-)=9/900=0.01
a) P(T+ U+)=0.5 sensitivity of
the drug test
b) P(T- U-)=1-P(T+ U-)
=1-0.01=0.99
specificity of th e drug test
Ex. 3.84, cont. (sol.)
c)
Statistics for Business and
Economics
Chapter 4
Random Variables &
Probability Distributions
Content
1. Two Types of Random Variables
2. Probability Distributions for Discrete
Random Variables
3. The Binomial Distribution
4. Other Discrete Distributions: Poisson
and Hypergeometric Distributions
5. Probability Distributions for Continuous
Random Variables
6. The Normal Distribution
Content (continued)
7. Descriptive Methods for Assessing
Normality
8. Other Continuous Distributions: Uniform
and Exponential
Learning Objectives
1. Develop the notion of a random variable
2. Learn that numerical data are observed
values of either discrete or continuous
random variables
3. Study two important types of random
variables and their probability models: the
binomial and normal model
4. Present some additional discrete and
continuous random variables
Thinking Challenge
• You’re taking a 33 question
multiple choice test. Each
question has 4 choices.
Clueless on 1 question, you
decide to guess. What’s the
chance you’ll get it right?
• If you guessed on all 33
questions, what would be
your grade? Would you
pass?
4.1
Two Types of Random
Variables
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.
Discrete
Random Variable
Random variables that can assume a
countable number (finite or infinite) of values
are called discrete.
Discrete Random Variable
Examples
•Experiment
•Random
Variable
•Possible
Values
•Make 100 Sales Calls
•# Sales
•0, 1, 2, ..., 100
•Inspect 70 Radios
•# Defective •0, 1, 2, ..., 70
•Answer 33 Questions
•# Correct
•0, 1, 2, ..., 33
•Count Cars at Toll
•Between 11:00 & 1:00
•# Cars
•Arriving
•0, 1, 2, ...,
∞
Continuous
Random Variable
Random variables that can assume values
corresponding to any of the points
contained in one or more intervals (i.e.,
values that are infinite and uncountable) are
called continuous.
Continuous Random Variable
Examples
•Experiment
•Random
Variable
•Possible
Values
•Weigh 100 People
•Weight
•45.1, 78, ...
•Measure Part Life
•Hours
•900, 875.9, ...
•Amount spent on food
•$ amount
•54.12, 42, ...
•Measure Time
•Between Arrivals
•Inter-Arrival •0, 1.3, 2.78, ...
•Time
Ex. 4.1
Which of the following describe continuous random
variables? Which describe discrete random variables?
a) The number of newspapers sold by the New York
Times each month
b) The amount of ink used in printing a Sunday edition of
the New York Times.
c) The actual number of ounces in a 1-gallon bottle of
laundry detergent.
d) The number of defective parts in ashipment of nuts
and bolts.
e) The number of people collecting unemployment
insurance each month.
Ex. 4.1
Which of the following describe continuous random
variables? Which describe discrete random variables?
a) The number of newspapers sold by the New York
Times each month (D)
b) The amount of ink used in printing a Sunday edition of
the New York Times. (C)
c) The actual number of ounces in a 1-gallon bottle of
laundry detergent. (C )
d) The number of defective parts in ashipment of nuts
and bolts. (D)
e) The number of people collecting unemployment
insurance each month. (D)
4.2
Probability Distributions for
Discrete Random Variables
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.
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.
Discrete Probability
Distribution Example
•Experiment: Toss 2 coins. Count number of
tails.
Probability Distribution
Values, x Probabilities, p(x)
•© 1984-1994 T/Maker Co.
0
1/4 = .25
1
2/4 = .50
2
1/4 = .25
Visualizing Discrete
Probability Distributions
Listing
Table
{ (0, .25), (1, .50), (2, .25) }
# Tails
f(x)
Count
p(x)
0
1
2
1
2
1
.25
.50
.25
Graph
p(x)
.50
.25
.00
Formula
x
0
1
2
P(x) =
n!
px(1 – p)n – x
x!(n – x)!
a) It is not valid
b) It is valid
c) It is not valid
d) It is not valid
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
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)
3.
Standard Deviation
  2
Summary Measures
Calculation Table
x
p(x)
Total
x p(x)
x p(x)
x–
(x – 2
(x – 2p(x)
(x 2 p(x)
Thinking Challenge
You toss 2 coins. You’re
interested in the number
of tails. What are the
expected value,
variance, and standard
deviation of this random
variable, number of tails?
•© 1984-1994 T/Maker Co.
Expected Value & Variance
Solution*
x
p(x)
x p(x)
x–
0
.25
0
–1.00
1.00
.25
1
.50
.50
0
0
0
2
.25
.50
1.00
1.00
.25
 =1.0
(x –  2 (x –  2p(x)
2 .50
 .71
Probability Rules for Discrete
Random Variables
Let x be a discrete random variable with probability
distribution p(x), mean µ, and standard deviation .
Then, depending on the shape of p(x), the
following probability statements can be made:
•Chebyshev’s Rule
P x    x  µ    Rule  0
Empirical
 .68
P x  2  x  µ  2 
 34
 .95
P x  3  x  µ  3 
 89
 1.00
a) 34.5, 174.75,13.219
c) 34.5 ±2*13.219=8.062,60.938
The probability that x will fall in this interval is 1.00.
Under the empirical rule we expect that this value would be
0.95.
4.3
The Binomial Distribution
Binomial Distribution
Number of ‘successes’ in a sample of n
observations (trials)
• Number of reds in 15 spins of roulette wheel
• Number of defective items in a batch of 5
items
• Number correct on a 33 question exam
• Number of customers who purchase out of
100 customers who enter store (each
customer is equally likely to purchase)
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.
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
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
Binomial Probability Table
(Portion)
•p
n=5
k
.01
…
0.50
…
.99
0
.951
…
.031
…
.000
1
.999
…
.188
…
.000
2
1.000
…
.500
…
.000
3
1.000
…
.812
…
.001
4
1.000
…
.969
…
.049
Cumulative Probabilities
P(x=3)=P(x ≤ 3) – P(x ≤ 2) = .812 – .500 = .312
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
Thinking Challenge
• You’re taking a 33 question
multiple choice test. Each
question has 4 choices.
Clueless on 1 question, you
decide to guess. What’s the
chance you’ll get it right?
• If you guessed on all 33
questions, what would be
your grade? Would you
pass?
X= number of correct answers
•Binomial random variable since
we just guess the answer.
•Total number of trials=33
•p= 0.25
•Expected number of correct
answers=33*0.25=8.25
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?
Binomial Distribution Solution*
n = 12, p = .20
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
c) 0.35
d) 0.2522
e) 0.2616
a) Adult not working during summer vacation.
b)
•The experiment consist of 10 identical trials. A trial for this experiment
is an individual.
•There are only two possible outcomes: work or do not work
•The probability remains same for each individual (trial)
•Individuals are independent