Chapter 6 Discrete Probability Distributions
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CHAPTER 6 DISCRETE PROBABILITY DISTRIBUTIONS
Objectives:
1. The student will be able to distinguish between a discrete and continuous random variable.
2. The student will be able to apply and interpret expected value for discrete variables.
3. The student will be able to use the expected value concept with decision trees.
4. The student will be able to apply expected value to the concepts of risk.
5. The student will be able to use and interpret the binomial probability distribution.
6. The student will be able to use and interpret the poisson probability distribution.
Random variable - a variable whose numerical value is
determined by the outcome of a random trial Pg 188
Discrete random variable is able to take on a countable number of values in an interval. Pg 184
Continuous random variable is assumed able to take any value in
an interval. Pg 184
A probability distribution reports all possible outcomes as well as the corresponding probability. page 184.
Expected value of a discrete random variable is a weighted mean equal to the sum of the products of each value x of the variable and associated probability P(X = x), or Pg 185
μ= xPx
σ2= x-μ2P(x)
AVERAGE SALES/DAY REL. FREQ.
50 .1
150 .2 _
250 .4 X = 200
350 .3 E(X) = 240
σ = 94.33
COMPUTE EXPECTED VALUE AND A SIMPLE AVERAGE
LOTTERY 2,000,000 entries $1 PER TICKET $1,000,000 FIRST PRIZE
COMPUTE THE EXPECTED VALUE OF THE LOTTERY TICKET.
COMPUTE THE AVERAGE PROFIT(LOSS) OF THE LOTTERY TICKET.
DRAW A DECISION TREE FOR THE POSSIBLE OUTCOMES.
RISK TAKER - AN INDIVIDUAL IS WILLING TO PAY TO TAKE A RISK.
[Cyber rebate offers a Fisher Price play phone for $206 with a rebate of $206. Assume the cost for this product to Cyber Rebate of this product is $10. Cyber Rebate also notes that 10% of its
customers do not apply for the rebate. Can Cyber Rebate make a profit with this rebate program?] WSJ 3/5/01 Marketplace
11/25/05 “Beyond Bland: Gift Cards Now Play Music, Record Messages” WSJ B1
10/13/09 Retailers Turn to Gift-Card Promotions to Lure Reluctant Buyers, Boost Spending WSJ B1
11/17/09 Fed Targets Gift-Card Fees WSJ A2
Tennis balls on the Web.
50 free downloads
How should you decide if is a good idea to give something away.
If the benefits exceed the costs
What is the cost? Assume e-music pays royalties of .05 per song
What are the benefits? Subscribe for 9.95 a month but I can cancel anytime in the first 2 weeks and keep 50 songs.
Piracy and formula 6-1
11/19/09 Armed U.S. Ship Repels Attack by Somali Pirates WSJ A12
Piracy and Business decision making
10,000 ships pass near Somilia 40 are taken each year.
What is the probability of being hijacked?
FIRE INSURANCE
Mr. Askil faces a decision to buy or not buy fire insurance
for his home. The cost of the insurance is $150/year, but there is only a 1/1000 chance of a fire destroying their home each year. Draw a decision tree for the outcomes Mr. Askil faces.
RISK AVERSION - An individual must be compensated to take a risk or is willing to pay something to avoid a risk.
Mr. Askil has a business opportunity to purchase 10 acres of
land at $50,000/acre. The land will only produce a profit if Mr.
Askil can get approval from the local government to re-zone the land for residential use. Mr. Askil can buy an option for $10,000/acre to buy in the next month. Mr. Askil believes there is a 50% chance he can convince the local government to rezone the land. If the land is re-zoned, it can be developed for a profit of $30,000/acre. Mr. Askil may decide to seek approval from the local government before making any purchase, but he estimates there will only be a 30% chance the land will still be available.
[Compute the expected value of each decision.]
[Explain which branch of the decision tree you would choose.]
[Compute the standard deviation of each branch.]
DRAW THE OUTCOME TREE FOR THIS PROBLEM. Make a decision.
DRAW OUTCOME TREE FOR A MANUFACTURER REBATE.
RISK NEUTRAL - An individual is indifferent between two assets,
regardless of risk, as long as the expected values are the same.
DISCUSS MARCH MADNESS.
DISCUSS REBATES
TORO START P(S) = .9 100,000 SOLD THIS SUMMER 1,200,000 Attempts
$50 COST/MACHINE
1. What is the average number of times a Toro will not
start on two pulls this summer?
2. What is the average cost per lawn mower to Toro of this plan?
Binomial and poisson probability distributions are used to
determine the probability of events that can occur in more than
one way. In the previous chapters we computed probability of events occurring in a specific manner.
Binomial distribution represents the probabilities of various
numerical outcomes over several identical, independent trials,
where there are two possible outcomes for each trial. Pg 194
ASSUMPTIONS
1. MUTUALLY EXCLUSIVE, ONLY TWO OUTCOMES ARE POSSIBLE
2. PROBABILITY OF SUCCESS AND FAILURE REMAIN CONSTANT
3. TRIALS ARE INDEPENDENT
TOSS A COIN THREE TIMES
OUTCOMES PROBABILITY BINOMIAL FORMULA
0 .125
1 .375
2 .375
3 .125
Binomial formula=nCx πx1-πn-x
x = number of successes, n = number of trials, n-x = number of failures
10/10/06 “More Fliers Forced To Give Up Seats” WSJ D1
9/17/09 Bumped Passengers Learn a Cruel Flying Lesson WSJ D1
1/07/09 An Airline Report Card: Fewer Delays, Hassles Last Year, but Bumpy
Times May Be Ahead WSJ D1
[Askil is a small commuter airline with a capacity of 14 passengers per plane. Askil also determines that 95% of people, who have reservations, take the flight. On Reserve
A. What percent of the time will this airplane fly with at least one empty seat? 5 pts
B. The airline decides to over book this flight by one person(take 15 reservations). What is the probability that 15 people show up for the flight. 5 pts
C. The FAA(Federal Aviation Association) requires compensation if you are bumped. Assume this compensation cost the airline $100 for each individual that is bumped(can not get on the plane because more people have tickets than seats on the aircraft). If this airline consistently over books by one person, determine the average cost of over booking this flight. Label and interpret your answer. 10 pts] Fall 01
Excel - top menu - formula paste function fx - binomial
MEAN AND STANDARD DEVIATION
E(X) = n∏ π = Probability of event A
(1-p) = Probability of the complement of event A.
σx= nπ(1-π)
Function key > Probability Distribution>Binomial
In the dialog box select Probability, set the number of trials to 6, the probability of success to .5 and the input column to c1
trials (n=3) and the probability of success is .5 (p=.5)
Poisson distribution represents the random arrival of events per
unit of time, distance or area.
= the mean number of Poisson distributed events over the sampling medium that is being examined.
x = the number of occurrences over the sampling medium
ASSUMPTIONS
1. All the assumptions relating to the binomial plus
2. A precise maximum does not exist in the sample space.
Compare the average number of occurrences to the actual number of
occurrences to determine the probability.
Determine the poisson approximation to the binomial distribution
for tossing a coin 25 times and getting 12 heads.
Excel - fx - poisson
Chapter 7 CONTINUOUS PROBABILITY DISTRIBUTIONS
Objectives:
1. The student will be able to use and interpret the normal distribution.
2. The student will be able to compute and interpret a value for a continuous variable if given a probability of occurrence.
3. The student will be able to know when it is appropriate to approximate the binomial distribution with the normal distribution.
4. The student will be able to determine when it is appropriate to use the binomial, poisson, and normal distributions.
A PROBABILITY DISTRIBUTION IS CONTINUOUS WHEN THE RANDOM
VARIABLE MAY ASSUME ANY VALUE WITHIN SOME SPECIFIED RANGE.
THE NORMAL DISTRIBUTION - Pg 227
PROPERTIES OF THE NORMAL CURVE
1. Only the mean (μ) and standard deviation (σ) need to
be known to compute probabilities for the normal distribution.
2. The graph of a normal distribution is bell shaped and
symmetrical around the mean.
3. Since the normal curve is measured on a continuous
scale the probability of obtaining a precise value is
approximately 0.
4. The probability that a random variable will have a
value between any two points is equal to the area under the normal curve between those two points.
5. The area under the normal curve between the mean and
any other point can be determined by knowing how many
standard deviations this data value is from the mean.
Z value= x-μσ
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Excel - fx - standardize - for Z-values
fx - normdist - Returns the probability of less than a specified value for a given mean and standard deviation. For less than probability cumulative probability put 1 in the cumulative window.
X
Mean
S Dev
Cum
fx - normsdist converts z-values to less than probabilities. Z
fx - normsinv - Gives a Z-value for a given probability.
fx - norminv - Gives a data value. Only a designated percent of the data will be less than or equal to this value given a mean and standard deviation.
[Your product requires ball bearings that have an average diameter 1 cm. You are looking for a supplier of ball bearings. The quality control department determines that a ±.008 cm variation around the mean is acceptable.
The mean diameter of the ball bearings is 1 cm. What percent of parts will be acceptable if the standard deviation is .008?
The quality control department wants to establish a failure rate of .0026 or .26%. What standard deviation should the quality control department demand from its suppliers?
This is a 3-sigma quality control level.]
[4.5-sigma .49996599x2 = .999993198 1 - .999993198=.000003198]
Incentive Pay
(How would you design a pay incentive plan for cashiers?)
NORMAL APPROXIMATION TO THE BINOMIAL DISTRIBUTION
Z= X-nπnπ1-π
Rule of thumb np(1-p) 5 Pg 238
1. Determine the probability of 3 defectives in a box of 5.
2. Determine the probability of 5 or more heads in 20 tosses.
3. Determine the probability of 3 defects in a yard of cloth.
4. Determine the probability of 4 phone calls in two hours.
5. Determine the probability of an individual's income, picked at random, being equal to $36,000.
6. Determine the probability of an individual's income, picked at random, being greater than $36,000.
7. Determine the probability of 400 heads in 1000 tosses.
Continuity correction is an adjustment that we make by adding or subtracting ½ to a discrete value when we use a continuous distribution to approximate a discrete distribution. Pg 238
Normal approximation of a binomial proportion of successes.
Z= p-pp1-pn
Sometimes, an appropriate procedure is to work not with the number X of successes in n trials but, instead with the proportion or fraction of successes.
[What is the probability that a data value is more than 1.96 standard deviations from its mean?]
[Assume 1=0, what is the probability that a population slope is more than 1.96 standard deviations from 0?]
Chapter 8 Sampling Distributions
Objectives:
1. The student will be able to understand the importance of random sampling in the process of inference.
2. The student will be able to distinguish between a population and a sample. Pg 264-265
3. The student will be able to choose the appropriate sampling method (i.e. systematic, stratified, cluster or simple random samples). Pg 265-270.
SAMPLING TECHNIQUES
1. sample versus census
why use a sample?
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2. Sample should be representative of the population -
random
If samples are random we can use the normal distribution. There are N!/[n!(N - n)!] samples of size n in a finite population of size, N, and random sampling means choosing one in such a way that all are equally likely to be choose.
POPULATION SAMPLE IS THIS RANDOM
1. ALL MEN MEN IN THIS CLASS
2. COLLEGE STUDENTS UW-PLATTEVILLE
3. VOTERS PHONE SURVEY BEFORE THE ELECTION
4. UNEMPLOYED PEOPLE WHO RECEIVE U.E. COMP.
5. T.V. VIEWERS 5500 VIEWERS RANDOMLY SELECTED
9/6/07 “New Way to Count Listeners Shakes Up Radio” WSJ B1 survey
1/10/08 Thomas E. Obama WSJ A14
Sampling Computations Results Inference
Why Sample? 1. Impractical (destructive testing) 2.Time and Money Pg 264-265
NON-RANDOM SAMPLING
1. JUDGEMENT SAMPLING
2. QUOTA OR CONVENIENCE SAMPLING
RANDOM SAMPLING
I. Complete list of the population is available - random
numbers with minitab.
Calc-random data.
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2/27/10 Lights, Camera, Calculator! The New Celebrity Math WSJ A2 Sampling
II. Population list is not available
Non-human subjects
1. Systematic sampling population does not
follow a pattern - Data set homogeneous-Usually non-human -Assembly line
Human Subjects
2. Stratified sample from some subgroup - Data set is not homogenous, such as human subjects - effects of drugs on people, advertising dollars based on return on equity
3. Cluster sample- Sampling a geographic area - divide sample into small units called primary units - entire subgroups- Pg 270
10/18/06 “655,00 War Dead? WSJ A20 Cluster sampling
Good samples are ones that are representative of the
population.
In general if the sample is random it should be representative of the population.
QUESTIONNAIRE CONSTRUCTION
1. NON-RESPONSE BIAS
PROPER STEPS FOR CONSTRUCTING QUESTIONNAIRES
1. DEFINE YOUR OBJECTIVES
2. FORMULATE THE QUESTIONS
3. DETERMINE THE TABULATION METHOD
MAKE SURE YOU CAN QUANTIFY THE DATA
4. PREPARE THE INSTRUMENT
5. PRETEST THE INSTRUMENT
AER papers and Proceedings May 2001 ‘Do People Mean What they Say? Implications for Subjective Survey Data’ Pg 67
Cognitive problems
1. Ordering of questions
How happy are you with life in general?