Download 6: Binomial Probability Distributions

Chapter 6:
Binomial Probability
Distributions
April 17
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
6.6 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
parameters
n  number of trials
p  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
Formula for binomial probabilities:
Pr( X  x) n C x p q
x
n x
where
nCx
≡ the binomial coefficient (next slide)
p ≡ probability of success for each trial
q ≡ probability of failure = 1 – p
Binomial Coefficient
Formula for the binomial coefficient:
n!
n Cx 
x!(n  x)!
where ! represents the factorial function, calculated:
x! = x  (x – 1)  (x – 2)  …  1
For example, 4! = 4  3  2  1 = 24
By definition 1! = 1 and 0! = 1
For example:
4!
4!
4  3  2 1


6
4 C2 
(2!)( 4  2)! (2!)( 2)! (2 1)( 2 1)
Binomial Coefficient
n!
n Cx 
x!(n  x)!
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?
Pr ( X  0)  n C x p x q n  x
 4 C0  0.75 0  0.25 40
4!
0
4

 0.75  0.25
0!4!
 1 1  .0039
 .0039
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
Recall the
area under
the curve
(AUC)
concept.
AUC =
probability!
Pr(X = 2)
=.2109 × 1.0
Area Under The Curve
§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
  npq
2
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)
x=2
p = .75
n=4
Pr(X = 2) = .2109
Pr(X ≤ 2) = .2617