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 questionsQ1

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(YO)=P(Y)+P(O)-P(YO)=0.20+0.1+0=0.30

3) It is not green?

P(Gc)=1-P(G)=0.9

Review questionsQ1(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(BrBr)P(BrBrBr)=(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(RB)P(RBR)=(10/100)(20/99)(19/98)

P(BrRR)=P(Br)P(RBr)P(RBrR)=(30/100)(20/99)(19/98)

P(YRR)=P(Y)P(RY)P(RYR)=(20/100)(20/99)(19/98)

P(ORR)=P(O)P(RO)P(ROR)=(10/100)(20/99)(19/98)

P(GRR)=P(G)P(RG)P(RGR)=(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 thatP(B1) + P(B2) + … + P(Bk) = 1,and an observed event A, then

P(Bi| A)

P(Bi A)

P( A)

P(B

i)P( A | B

i)

P(B1)P( A | B

1) P(B

2)P( A | B

2) ... P(B

k)P( A | B

k)

•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 •RandomVariable

•PossibleValues

•Count Cars at Toll•Between 11:00 & 1:00

•# Cars•Arriving

•0, 1, 2, ..., ∞

•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

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

•Measure Time•Between Arrivals

•Inter-Arrival•Time

•0, 1.3, 2.78, ...

•Experiment •RandomVariable

•PossibleValues

•Weigh 100 People •Weight •45.1, 78, ...

•Measure Part Life •Hours •900, 875.9, ...

•Amount spent on food •$ amount •54.12, 42, ...

Ex. 4.1Which 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. 4.1Which 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

Probability Distribution

Values, x Probabilities, p(x)

0 1/4 = .25

1 2/4 = .50

2 1/4 = .25

•Experiment: Toss 2 coins. Count number of tails.

•© 1984-1994 T/Maker Co.

Visualizing Discrete Probability Distributions

Listing Table

Formula

# Tailsf(x)

Countp(x)

0 1 .251 2 .502 1 .25

P(x)n

x!(n – x)!!

= px(1 – p)n – x

Graph

.00

.25

.50

0 1 2x

p(x)

{ (0, .25), (1, .50), (2, .25) }

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

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 2p(x)

Summary Measures

3. Standard Deviation2

Summary Measures Calculation Table

x p(x) x p(x) x –

Total (x 2p(x)

(x – 2 (x – 2p(x)

xp(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*

0 .25 –1.00 1.00

1 .50 0 0

2 .25 1.00 1.00

0

.50

.50

=1.0

x p(x) x p(x) x – (x – 2 (x – 2p(x)

.25

0

.25

2.

.

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 Empirical RuleP x x µ 0 .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 DistributionNumber 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 ProbabilityCharacteristics 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

3 5 3

!( ) (1 )

!( )!

5!(3) .5 (1 .5)

3!(5 3)!

.3125

x n xnp x p p

x n x

p

Experiment: Toss 1 coin 5 times in a row. Note number of tails. What’s the probability of 3 tails?

•© 1984-1994 T/Maker Co.

Binomial Probability Distribution

!( ) (1 )

! ( )!x n x x n xn n

p x p q p px x n 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)

n = 5 •p

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

.0

.5

1.0

0 1 2 3 4 5

X

P(X)

.0

.2

.4

.6

0 1 2 3 4 5

X

P(X)

n = 5 p = 0.1

n = 5 p = 0.5

E (x ) np•Mean

•Standard Deviation

npq

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 = .20A. p(0) = .0687 B. p(2) = .2835C. 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

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

c) 0.35

d) 0.2522

e) 0.2616