Binomial
Probability
Distribution
Permutations
Number of possible arrangements
n
Pn  n( n  1( n  2)...(1)  n !
n( n  1)...( n  r )( r ) r  1)...(1)
n!

n Pr 
( r )( r  1)...(1)
( n  r )!
5  4  3  2  1 5!
  5  4  3  60
5 P3 
2 1
2!
Combinations
Number of sets or groups
 n  n Pr
n!

n Cr    
 r  r ! r !( n  r )!
# Arrangements of n things taken r at a time

# Arrangements of r things
5!
5  4  3 60


 10
5 C3 
3!(2)!  3  2  1 6
Binomial Theorem
n


n


n
k nk
k nk

(a  b)     a b    nCk a b
1  k 
 1
n
 3 3 0  3 2 1  3 1 2  3 0 3
(x  2)    x 2    x 2    x 2    x 2
3
2
 1
 0
3
(x  2)  x  3x  2   3x  4   8
3
3
2
 x  6x  12x  8
3
2
Binomial Distribution
The binomial distribution is the probability distribution
that results from sampling a binomial population.
Binomial samples have the following properties:
1. Fixed number of trials, represented as n.
2. Each trial has two possible outcomes, a “success” and a
“failure”.
3. P(success)=p (and thus: P(failure)=1–p), for all trials.
4. The trials are independent, which means that the
outcome of one trial does not affect the outcomes of
any other trials. May be population is large enough that
sampling without replacement does not change values.
Success and Failure…
…are just labels for a binomial experiment, there is no value
judgment implied.
For example a coin flip will result in either heads or tails. If
we define “heads” as success then necessarily “tails” is
considered a failure (in as much as we attempting to have
the coin lands heads up).
Other binomial experiment notions:
An election candidate wins or loses
An employee is male or female
Example
Pat Statsdud is a (not good) student taking a statistics
course. Pat’s exam strategy is to rely on luck for the next
quiz. The quiz consists of 10 multiple-choice questions. Each
question has five possible answers, only one of which is
correct. Pat plans to guess the answer to each question.
• What is the probability that Pat gets no answers correct?
•
What is the probability that Pat gets two answers correct?
Pat Statsdud…
Algebraically then: n=10, and
P(success)= 1/5 = .20
Pat plans to guess the answer to each question.
Is this a binomial experiment? Check the conditions:
There is a fixed finite number of trials (n=10).
An answer can be either correct or incorrect.
The probability of a correct answer (P(success)=.20)
does not change from question to question.
 Each answer is independent of the others.
Pat Statsdud…
n=10, and P(success) = .20
a. What is the probability that Pat gets no answers correct?
i.e. # success, x, = 0; hence we want to know P(x=0)
Pat has about an 11% chance of getting no answers correct
using the guessing strategy.
Pat Statsdud…
n=10, and P(success) = .20
b. What is the probability that Pat gets two answers
correct?
I.e. # success, x, = 2; hence we want to know P(x=2)
Pat has about a 30% chance of getting exactly two answers
correct using the guessing strategy.
Binomial Probability Distribution
The binomial distribution with n trials and success
probability p has:



Mean =
np
Variance =
  np 1  p 
2
Standard deviation =
    np 1  p 
2
Cumulative Probability…
Thus far, we have been using the binomial probability
distribution to find probabilities for individual values of
x. To answer the question: (Example 10)
“Find the probability that Pat fails the quiz”
requires a cumulative probability, that is, P(X ≤ x)
If a grade on the quiz is less than 50% (i.e. 5
questions out of 10), that’s considered a failed quiz.
Thus, we want to know what is: P(X ≤ 4) to answer
Pat Statsdud
P(X ≤ 4) = P(0) + P(1) + P(2) + P(3) + P(4)
We already know P(0) = .1074 and P(2) = .3020.
Using the binomial formula to calculate the others:
P(1) = .2684 , P(3) = .2013, and P(4) = .0881
We have P(X ≤ 4) = .1074 + .2684 + … + .0881 =
.9672
Thus, its about 97% probable that Pat will fail the test
using the luck strategy and guessing at answers…
BinomialPDF & …CDF
Pat Statsdud Has Been Cloned
Suppose that a professor has a class full of students like
Pat. What is the mean?
What is the standard deviation?
The mean = μ = np = 10(0.2) = 2
The standard deviation is
σ = √ np ( 1 – p ) = √ 10( .2)( 1 - .2)
= 1.26
Lock Section 6.1
Sampling Distribution
of the
Sample Proportion
Binomial Population
Two Choices:
Success
Failure
Fixed Probability
Independence
Sampling Distribution
P-Hat
p
P-Hat Definition
pˆ  X / n
Properties of p-hat



When sample sizes are fairly large, the shape of
the p-hat distribution will be normal.
The mean of the distribution is the value of the
population parameter p.
The standard deviation of this distribution is
the square root of p(1-p)/n.
sd ( pˆ ) 
p(1  p)
n
Sampling Distribution of p-hat
For the sampling distribution to be normal, you must have:
np  10 and n(1- p)  10
Calculate Probabilities

Because the shape of the distribution is
normal, we can standardize the variable p-hat
to a Z standard normal distribution. Use Ztransform:
Z
pˆ  p
p(1  p )
n
Lock Section 6.2
Confidence Intervals
for
Population Proportions
AP Statistics
Chap 10-25
Confidence Intervals
Confidence
Intervals
Population
Mean
σ Known
AP Statistics
Population
Proportion
σ Unknown
Chap 10-26
Confidence Intervals for the
Population Proportion, p

Recall that the distribution of the sample
proportion is approximately normal if the
sample size is large, with standard deviation
p(1  p)
σp 
n

We will estimate this with sample data:
sp 
AP Statistics
p(1  p )
n
Chap 10-27
How Big????

Rule of Thumb 1: Formula for standard
deviation of p-hat only when the population,
N, is at least 10 times the sample size: N ≥
10 n

Rule of Thumb 2: The sampling distribution
of p-hat is approximately Normal when

np ≥ 10 and n(1-p) ≥ 10
Confidence interval endpoints

Upper and lower confidence limits for the
population proportion are calculated with the
formula
p  z /2

where



AP Statistics
p(1  p )
n
z is the standard normal value for the level of confidence desired
p is the sample proportion
n is the sample size
Chap 10-29
Example
AP Statistics

A random sample of 100 people
shows that 25 are left-handed.

Form a 95% confidence interval for
the true proportion of left-handers
Chap 10-30
Example
(continued)

1.
A random sample of 100 people shows
that 25 are left-handed. Form a 95%
confidence interval for the true proportion
of left-handers.
p  25/100  .25
2. Sp 
3.
p(1  p )/100  .25(.75)/100  .0433
.25  1.96 (.0433)
0.1651 . . . . . 0.3349
AP Statistics
Chap 10-31
1-PropZInterval
Interpretation
AP Statistics

We are 95% confident that the true
percentage of left-handers in the population
is between
16.51% and 33.49%.

Although this range may or may not contain
the true proportion, 95% of intervals formed
from samples of size 100 in this manner will
contain the true proportion.
Chap 10-33
Changing the sample size

Increases in the sample size reduce
the width of the confidence interval.
Example:

If the sample size in the above example is
doubled to 200, and if 50 are left-handed in the
sample, then the interval is still centered at .25,
but the width shrinks to
.19 …… .31
AP Statistics
Chap 10-34
Example 2
A random sample of 400 graduates showed
32 went to grad school. Set up a 95%
confidence interval estimate for p.

ˆ ×(1- p)ˆ
ˆ p) ˆ
p
p
×
(1pˆ - Z α/2 ×
≤ p ≤ pˆ + Z α/2 ×
n
n
.08 ×(1- .08)
.08 ×(1- .08)
.08 -1.96 ×
≤ p ≤ .08 +1.96 ×
400
400
.053  p  .107
Example 2
A random sample of 400 graduates showed
32 went to grad school. Set up a 95%
confidence interval estimate for p.

Thinking Challenge
You’re a production
manager for a
newspaper. You want to
find the % defective. Of
200 newspapers, 35 had
defects. What is the
90% confidence interval
estimate of the
population proportion
defective?

Confidence Interval Solution
ˆ ×(1- p)ˆ
ˆ p) ˆ
p
p
×
(1pˆ - z a/2 ×
≤ p ≤ pˆ + z a/2 ×
n
n
.175 ×(.825)
.175 ×(.825)
.175 -1.645 ×
≤ p ≤ .175 +1.645 ×
200
200
.1308 ≤ p ≤ .2192
Required Sample Size
Define the
margin of error:
Solve for n:
e  z/2
n
z
2
/2
p(1  p)
n
p (1  p)
2
e
p can be estimated with a pilot sample, if
necessary (or conservatively use p = .50)
AP Statistics
Chap 10-39
What sample size...?

How large a sample would be necessary to
estimate the true proportion defective in a
large population within 3%, with 95%
confidence?
(Assume a pilot sample yields p-hat = .12)
AP Statistics
Chap 10-40
What sample size...?
(continued)
Solution:
For 95% confidence, use Z = 1.96
E = .03
P-hat = .12, so use this to estimate p
z2 /2 p (1  p) (1.96)2 (.12)(1  .12)
n

 450.74
2
2
e
(.03)
So use n = 451
AP Statistics
Chap 10-41
Lock Section 6.3
HYPOTHESIS TESTS
FOR SINGLE
PROPORTIONS
Test of Proportions


Old drug cures 80 percent of the time. New
drug for heartworm disease in dogs cures
90 percent of n=1000 dogs, so p-hat=.9.
What is the probability that this results
occurred by random chance?
po 1- po 
.8 .2 
SE =
=
= .0126
n
1000
Z=
p - value = 0
p - po
po 1- po 
n

.9 - .8
.8 .2 
1000
 7.9
Proportion Test