october 15. in chapter 6: 6.1 binomial random variables 6.2 calculating binomial probabilities 6.3...
TRANSCRIPT
Apr 20, 2023
Chapter 6: Chapter 6: Binomial Probability Binomial Probability
DistributionsDistributions
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
§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
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)
§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:
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:
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
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
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
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
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
Area Under The Curve
Pr(
X =
2)
=.2
109 ×
1.
0
Recall the area under the curve (AUC) concept.
AUC = probability!
§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)
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)
§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
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
§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
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