chapter 6: binomial probability distributions
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Chapter 6: Binomial Probability Distributions. In Chapter 6:. 6.1 Binomial Random Variables 6.2 Calculating Binomial Probabilities 6.3 Cumulative Probabilities 6.4 Probability Calculators 6.5 Expected Value and Variance of Binomial Random Variables - PowerPoint PPT PresentationTRANSCRIPT
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Apr 20, 2023
Chapter 6: Chapter 6: Binomial Probability Binomial Probability
DistributionsDistributions
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In Chapter 6:
6.1 Binomial Random Variables6.2 Calculating Binomial Probabilities6.3 Cumulative Probabilities6.4 Probability Calculators6.5 Expected Value and Variance of Binomial Random Variables6.6 Using the Binomial Distribution to Help Make Judgments
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§6.1 Binomial Random Variables• Binomial = a family of discrete random
variables
• Binomial random variable ≡ the random number of successes in n independent Bernoulli trials (a Bernoulli trials has two possible outcomes: “success” or “failure”
• Binomials random variables have two parametersn number of trialsp probability of success of each trial
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Binomial Example• Consider the random number of successful
treatments when treating four patients
• Suppose the probability of success in each instance is 75%
• The random number of successes can vary from 0 to 4
• The random number of successes is a binomial with parameters n = 4 and p = 0.75
• Notation: Let X ~b(n,p) represent a binomial random variable with parameters n and p. The illustration variable is X ~ b(4, .75)
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§6.2 Calculating Binomial Probabilities
xnxxn qpCxX )Pr(
where
nCx ≡ the binomial coefficient (next slide)
p ≡ probability of success for each trial
q ≡ probability of failure = 1 – p
Formula for binomial probabilities:
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Binomial Coefficient
)!(!
!
xnx
nCxn
where ! represents the factorial function, calculated:x! = x (x – 1) (x – 2) … 1For example, 4! = 4 3 2 1 = 24By definition 1! = 1 and 0! = 1
6)12)(12(
1234
)!2)(!2(
!4
)!24)(!2(
!424
C
For example:
Formula for the binomial coefficient:
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Binomial Coefficient
)!(!
!
xnx
nCxn
The binomial coefficient is called the “choose function” because it tells you the number of ways you could choose x items out of n
nCx the number of ways to choose x items out of n
For example, 4C2 = 6 means there are six ways to choose two items out of four
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Binomial Calculation – Example Recall the “Four patients example”. Four patients; probability of success of each treatment = .75. The number of success is the binomial random variable X ~ b(4,.75). Note q = 1 −.75 = .25. What is the probability of observing 0 successes under these circumstances?
0039.
0039.11
250750!4!0
!4
250750
)0(Pr
40
04004
..
..C
qpCX xnxxn
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X~b(4,0.75), continued
Pr(X = 1) = 4C1 · 0.751 · 0.254–1
= 4 · 0.75 · 0.0156
= 0.0469
Pr(X = 2) = 4C2 · 0.752 · 0.254–2
= 6 · 0.5625 · 0.0625
= 0.2106
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X~b(4, 0.75) continued
Pr(X = 3) = 4C3 · 0.753 · 0.254–3
= 4 · 0.4219 · 0.25
= 0.4219
Pr(X = 4) = 4C4 · 0.754 · 0.254–4
= 1 · 0.3164 · 1
= 0.3164
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pmf for X~b(4, 0.75)Tabular and graphical forms
x Pr(X = x)
0 0.0039
1 0.0469
2 0.2109
3 0.4210
4 0.3164
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Area Under The Curve
Pr(
X =
2)
=.2
109 ×
1.
0
Recall the area under the curve (AUC) concept.
AUC = probability!
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§6.3: Cumulative Probability
• Recall the cumulative probability concept
• Cumulative probability ≡ the probability of that value or less
• Pr(X x) • Corresponds to left
tail of pmf
Pr(X 2) on X ~b(4,.75)
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Cumulative Probability Function
• Cumulative probability function lists cumulative probabilities for all possible outcome
• Example: The cumulative probability function for X~b(4, 0.75)Pr(X 0) = 0.0039
Pr(X 1) = 0.0508
Pr(X 2) = 0.2617
Pr(X 3) = 0.6836
Pr(X 4) = 1.0000
Pr(X 1) = Pr(X = 0) + Pr(X = 1) = .0039 + .0469 = 0.0508
Pr(X 2) = Pr(X = 0) + Pr(X = 1) + Pr(X = 2)
Pr(X 4) = Pr(X = 0) + Pr(X = 1) + Pr(X = 2) + Pr(X = 3)
Pr(X 4) = Pr(X = 0) + Pr(X = 1) + … + Pr(X = 4)
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§6.5: Expected Value and Variance for Binomials
• The expected value (mean) μ of a binomial pmf is its “balancing point”
• The variance σ2 is its spread
• Shortcut formulas:
np npq2
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Expected Value and Variance, Binomials, Illustration
For the “Four patients” pmf of X~b(4,.75)
μ = n∙p = (4)(.75) = 3
σ2 = n∙p∙q = (4)(.75)(.25) = 0.75
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§6.6 Using the Binomial• Suppose we observe 2
successes in the “Four patients” example
• Note μ = 3, suggesting we should see 3 success on average
• Does the observation of 2 successes cast doubt on p = 0.75?
• No, because Pr(X 2) = 0.2617 is not too unusual
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StaTable Probability Calculator• Calculates
probabilities for many types of random variables
• This figure shows probabilities for X~b(4,0.75)
• Available in Java, Windows, and Palm versions (download from website)
Pr(X = 2) = .2109
Pr(X ≤ 2) = .2617
x = 2
p = .75
n = 4