1 chapter 4 probability distributions 4-1 random variables 4-2 binomial probability distributions...
TRANSCRIPT
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Chapter 4
Probability Distributions
4-1 Random Variables
4-2 Binomial Probability Distributions
4-3 Mean, Variance, Standard Deviation for the Binomial Distribution
4-4 Other Discrete Probability Distributions
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OverviewThis chapter will deal with the construction of
probability distributions
by combining the methods of Chapter 2 with the those of Chapter 3.
Probability Distributions will describe what will probably happen instead of what
actually did happen.
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Combining Descriptive Statistics Methods and Probabilities to Form a Theoretical Model of
Behavior
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4-1
Random Variables
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Definitions Random Variable
a variable (typically represented by x) that has a single numerical value, determined by chance, for each outcome of a procedure
Probability Distribution
a graph, table, or formula that gives the probability for each value of the random variable
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Probability DistributionNumber of Girls Among Fourteen Newborn Babies
0 1 2 3 4 5 6 7 8 91011121314
0.0000.0010.0060.0220.0610.1220.1830.2090.1830.1220.0610.0220.0060.0010.000
x P(x)
Questions:
Let X = # of girls
1. P(X=3) = ?
2. P(X ≤ 2) = ?
3. P(X ≥ 1) = ?
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Probability Histogram
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DefinitionsDiscrete random variable
has either a finite number of values or countable number of values, where ‘countable’ refers to the fact that there might be infinitely many values, but they result from a counting process.
Continuous random variable
has infinitely many values, and those values can be associated with measurements on a continuous scale with no gaps or interruptions.
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Requirements for Probability Distribution
P(x) = 1 where x assumes all possible values
0 P(x) 1 for every value of x
See #3 on hw
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Formula (Mean)
µ = [x • P(x)]
Formula (Variance)
2 = [(x - µ)2 • P(x)]
Formula (Standard Deviation)
= [(x - µ)2 • P(x)]
Mean, Variance and Standard Deviation of a Probability Distribution
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Roundoff Rule for µ, 2, and
Round results by carrying one more decimal place than the number of decimal places used for the random variable x. If the values of x are integers, round µ, 2, and to one decimal place.
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TI-83 Calculator
Calculate Mean and Std. Dev from a Probability Distribution
1. Press Stat
2. Press “1” Edit
3. Enter values of random variable (x) in L1
4. Enter probability P(x) in L2
5. Press Stat
6. Cursor over to CALC
7. Choose the 1-Var stats option
8. Enter 1-Var stats L1,L2
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Using Excel
See Probability Distribution Worksheet
Examples:1. See Introduction to probability distributions handout
2. Go to Excel (dice example – class assignment)
3. Find missing probability, mean, SD and unusual values (test question see #6 on hw)
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Unusual and Unlikely Values
1. Unusual if greater than 2 standard deviations from the mean, that is x + 2and x – 2
2. Unlikely if probability is very small, usually less than .05. This is consistent with the 2 idea associated with the empirical rule. #6f and 7c on HW
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Definition
Expected Value The average value of outcomes
E = [x • P(x)]
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E = [x • P(x)]
Event
Win
Lose
x
$5
- $5
P(x)
244/495
251/495
x • P(x)
2.465
- 2.535
E = -$.07
Example: #9 on hw
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4-2
Binomial Random Variables
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Binomial Random Variables
Facts:Discrete (we can count the outcomes)
Have to do with random variables having 2 outcomes. Examples: heads/tails, boy/girl, yes/no,
defective/not defective, etc.
A binomial distribution is the sum of several trials. Example: # of heads when a coin is tossed three times
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Notation for Binomial Probability Distributions
n = fixed number of trials
x = specific number of successes in n trials
p = probability of success in one of n trials
q = probability of failure in one of n trials
(q = 1 - p ) P(x) = probability of getting exactly x successes among n
trials
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P(x) = • px • qn-x (n - x )! x!
Binomial Probability Formula
n !
Method 1
P(x) = nCx • px • qn-x
for calculators use nCr key, where r = x
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P(x) = • px • qn-xn ! (n - x )! x!
Number of outcomes with
exactly x successes
among n trials
Probability of x successes
among n trials for any one
particular order
Binomial Probability Formula
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This is a binomial experiment so you need to know 4 things p, q, n and x.
p=.5
q=.5
n=3
a) x = 2 b) x = 0 then 1 then 2
Example: Toss a coin 3 times. Let x = number of heads and find
a) P(2) = b) P(at least 2)
On the test you will have to construct the entire probability distribution for tossing a coin “n” times and observing the number of heads. Should do this before the test.
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This is a binomial experiment where:
n = 5
x = 2
p = 0.80
q = 0.20
Using the binomial probability formula to solve:
P(2) = 5C2 • 0.8 • 0.2 = 0.0512
2 3
Example: Find the probability of getting exactly 2 correct responses among 5 different requests from directory assistance. Assume in general, they are correct 80% of the time.
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Binomial Table
Method 2
Two tables are available on website
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Example: Using Table for n = 5 and p = 0.80, find the following: a) The probability of exactly 2 successesb) The probability of at most 2 successesc) The probability of at least 1 success
a) P(2) = 0.0512
b) P(at most 2) = P(0) or P(1) or P(2)
= 0.0003 + 0.0064 + 0.0512
= 0.0579
c) P(at least 1) = 1 – P(0) = 1 – .0003 = .9997
Test Question
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Using Technology
Calculator function (TI-83)
See binomial distribution worksheet
See also coin example worksheet
Method 3
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TI-83 Calculator
Finding Binomial Probabilities (complete distribution)
1. Press 2nd Distr
2. Choose binopdf
3. Enter binopdf(n,p)
4. Press STO L2 (stores probabilities in column L2)
5. Press Stat
6. Choose Edit (to view probabilities)
7. Optional: enter the values of the random variable in L1
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TI-83 Calculator
Finding Binomial Probabilities (individual value)
1. Press 2nd Distr
2. Choose binopdf
3. Enter binopdf(n,p,x)
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TI-83 Calculator
Finding Binomial Probabilities (cummulative)
1. Press 2nd Distr
2. Choose binocdf
3. Enter binocdf(n,p,x)
This yields the sum of the probabilities from 0 to x.
Example:
Let n=6 and p=0.2
P(X<3) = binocdf(6,.2,3)
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4.3 Mean, variance and standard deviation of a Binomial Probability Distribution
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For Any Discrete Probability Distribution the general formulas
are:
• µ = [x • P(x)]
• 2= [(x - µ) 2 • P(x) ]
= [(x - µ) 2 • P(x)]
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= n • p • q
For Binomial Distributions:
• µ = n • p
• 2= n • p • q
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• We previously discovered that this scenario could be considered a binomial experiment where:
• n = 20
• p = 0.2
• q = 0.8
• Using the binomial distribution formulas:
µ = (20)(0.2) = 4 correct answers
= (20)(0.2)(0.8) = 1.8 answers (rounded)
Example: Find the mean and standard deviation for students that guess answers on a multiple choice test with 5 answers and 20 questions.
Test question
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Reminder
• Maximum usual values = µ + 2 • Minimum usual values = µ - 2
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For this binomial distribution,
• µ = 4 answers
• = 1.8 answers
• µ + 2 = 4 + 2(1.8) = 7.6
• µ - 2 = 4 - 2(1.8) = .4
The usual number of correct answers would be from .4 to 7.6, so guessing 7 correct answers would not be an unusual result.
Example: Determine whether guessing 7 correct answers is unusual.
Test question
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4.4 Other Discrete Probability Distributions
• Poisson
• Geometric
• Hypergeometric
• Negative Binomial• And more
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Poisson Distribution
Definition a discrete probability distribution that
applies to occurrences of some event over a specific interval.
Will be a question on the test for you to differentiate between a binomial and a poisson distribution
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DefinitionPoisson Distribution
a discrete probability distribution that applies to occurrences of some event over a specific interval.
Probability Formula
P(x) = where e 2.71828µ x • e -µ
x!
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• Why is this a poisson distribution?
• µ = 5
• We need to find various probabilities using
• Let’s find P(7)
• Look at the Poisson function in Excel
Example: Look at #1
µ x • e -µ P(x) =
x!
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Geometric Distribution
Definition a discrete probability distribution of
the number of trials needed to get one success.
Will be a question on the test for you to differentiate between a binomial, poisson and a geometric distribution
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Geometric Distribution
Example: Roll a die 5 times. What is the
probability of getting your first 2 on the 5th roll.
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Negative Binomial Distribution
Definition a discrete probability distribution of
the number of trials needed to get a get a specified number of successes.
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Negative Binomial Distribution
Example a basketball player has a 70% chance
of making a free throw, what is the probability of making his 3rd free throw on his 5th shot.
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Hypergeometric Distribution
Hypergeometric Experiment• A sample of size n is randomly selected
without replacement from a population of N items.
. In the population, k items can be classified as successes, and N - k items can be classified as failures.
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Hypergeometric Distribution
Notation• N: The number of items in the population.
• k: The number of items in the population that are classified as successes.
• n: The number of items in the sample.
• x: The number of items in the sample that are classified as successes.
• kCx: The number of combinations of k things, taken x at a time.
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Hypergeometric Distribution
Example
Suppose we randomly select 5 cards without replacement from an ordinary deck of playing cards. What is the probability of getting exactly 2 red cards (i.e., hearts or diamonds)?
P = [ kCx ] [ N-kCn-x ] / [ NCn ] = [ 26C2 ] [ 26C3 ] / [ 52C5 ] = [ 325 ] [ 2600 ] / [ 2,598,960 ] = 0.32513