CH 17
Probability Models
Key Terms
• Bernoulli Trials
• Geometric Probability Model
• Binomial Probability Model
• 10% Condition
• Success/Failure Condition
Bernoulli Trials
An trial is said to be a Bernoulli Trial if
1. There are 2 possible outcomes. 2. The probability of success is constant.
3. The trials are independent.
EXAMPLE Idenfying Bernoulli Trials
Which of the following are binomial experiments?
(a) Flipping a coin.
(b) Rolling a die.
(c) Drawing a card from a standard deck and recording the number of Aces you draw.
(d) Watching planes arrive at an airport and recording if each plane is on me or not.
Geometric Probability Model
A model that counts the number of Bernoulli trials unl the first success.
Probability of Success on the xth trial
p = probability of success
q = probability of failure = 1­p
X = number of trials until the first success occurs
Mean
Standard Deviation
EXAMPLE Using the Geometric Probability Distribuon Funcon
According to a recent survey, 9% of the students at Hayes have $20 or more on their person. If I randomly select students
(a) What is the probability that the 14th student I select is the first student who has $20?
(b) What is the mean of this distribuon?
(c) What is the standard deviaon of this distribuon?
EXAMPLE Construcng Geometric Probability Histograms
(a) Construct a geometric probability histogram p = 0.15.
(b) Construct a binomial probability histogram p = 0. 5.
(c) Construct a binomial probability histogram p =0.85.
(a) Construct a geometric probability histogram p = 0.15.
20x0.5
(b) Construct a binomial probability histogram p = 0. 5.
20x0.5
(c) Construct a binomial probability histogram p =0.85.
20x1
Binomial Probability Model
A model that counts the number of Successes in a fixed number of Bernoulli trials.
Probability of x Success in n trials
n = number of trials
p = probability of success
q = probability of failure = 1­p
X = number of success in n trials
Mean
Standard Deviation
EXAMPLE Using the Binomial Probability Distribuon Funcon
According to the Experian Automove, 35% of all car‐owning households have three or more cars. (a) In a random sample of 20 car‐owning households, what is the probability that exactly 5 have three or more cars?
(b) In a random sample of 20 car‐owning households, what is the probability that less than 4 have three or more cars?
(c) In a random sample of 20 car‐owning households, what is the probability that at least 4 have three or more cars?
EXAMPLE
Finding the Mean and Standard Deviaon of a Binomial Random Variable
According to the Experian Automove, 35% of all car‐owning households have three or more cars. In a simple random sample of 400 car‐owning households, determine the mean and standard deviaon number of car‐owning households that will have three or more cars.
EXAMPLE Construcng Binomial Probability Histograms
(a) Construct a binomial probability histogram with n = 8 and p = 0.15.
(b) Construct a binomial probability histogram with n = 8 and p = 0. 5.
(c) Construct a binomial probability histogram with n = 8 and p = 0.85.
For each histogram, comment on the shape of the distribuon.
Construct a binomial probability histogram with n = 50 and p = 0.8. Comment on the shape of the distribuon.
EXAMPLE Using the Mean, Standard Deviaon and Empirical Rule to Check for Unusual Results in a Binomial Experiment
According to the Experian Automotive, 35% of all car­owning households have three or more cars. A researcher believes this percentage is higher than the percentage reported by Experian Automotive. He conducts a simple random sample of 400 car­owning households and found that 162 had three or more cars. Is this result unusual ?
The result is unusual since 162 > 159.1
10% Condion
As long as the sample is smaller than 10% of the populaon, it is ok to ignore the independence requirement.
Success/Failure Condion
To use a Normal approximaon for a binomial model then,
np≥10 and nq≥10.
For when calculators are too wimpy to do binomials!
EXAMPLE Using the Normal Approximaon
Mr. Gray needs some cash, a lot of it! (Don't ask why). According to a recent survey, 9% of the students at Hayes have $20 or more on their person. In order to get enough money Mr. Gray needs to find 200 students who have $20 so he can use his persuasive skills to get it from them.
What is the probability that Mr. Gray will find 200 or more students who have $20 out of the 1200 students at Hayes?
Problem Set:
pg 401 #1­41 (odd)