Chapter 12
Binomial Distributions
BPS - 3rd Ed.
Chapter 12
1
Binomial Setting
 Fixed
number n of observations
 The n observations are independent
 Each observation falls into one of just
two categories
– may be labeled “success” and “failure”
 The
probability of success, p, is the
same for each observation
BPS - 3rd Ed.
Chapter 12
2
Binomial Setting
Examples
 In
a shipment of 100 televisions, how
many are defective?
– counting the number of “successes”
(defective televisions) out of 100
 A new
procedure for treating breast
cancer is tried on 25 patients; how
many patients are cured?
– counting the number of “successes”
(cured patients) out of 25
BPS - 3rd Ed.
Chapter 12
3
Binomial Distribution
 Let
X = the count of successes in a
binomial setting. The distribution of X
is the binomial distribution with
parameters n and p.
– n is the number of observations
– p is the probability of a success on any
one observation
– X takes on whole values between 0 and n
BPS - 3rd Ed.
Chapter 12
4
Binomial Distribution
– not all counts have binomial distributions
 trials
(observations) must be independent
 the probability of success, p, must be the same
for each observation
– if the population size is MUCH larger than the
sample size n, then even when the observations
are not independent and p changes from one
observation to the next, the change in p may be
so small that the count of successes (X) has
approximately the binomial distribution
BPS - 3rd Ed.
Chapter 12
5
Case Study
Inspecting Switches
An engineer selects a random sample of
10 switches from a shipment of 10,000
switches. Unknown to the engineer, 10%
of the switches in the full shipment are
bad. The engineer counts the number X of
bad switches in the sample.
BPS - 3rd Ed.
Chapter 12
6
Case Study
Inspecting Switches

X (the number of bad switches) is not quite binomial
– Removing one switch changes the proportion of bad
switches remaining in the shipment (selections are not
independent)

However, removing one switch from a shipment of
10,000 changes the makeup of the remaining 9,999
very little
– the distribution of X is very close to the binomial
distribution with n=10 and p=0.1
BPS - 3rd Ed.
Chapter 12
7
Binomial Probabilities
 Find
the probability that a binomial
random variable takes any particular
value
– P(x successes out of n observations) = ?
– need to add the probabilities for the
different ways of getting exactly x
successes in n observations
BPS - 3rd Ed.
Chapter 12
8
Binomial Probabilities
Example
Each offspring hatched from a particular type
of reptile has probability 0.2 of surviving for at
least one week. If 6 offspring of these reptiles
are hatched, find the probability that exactly 2
of the 6 will survive for at least one week.
Label an offspring that survives with S for
“success” and one that dies with F for “failure”.
P(S) = 0.2 and P(F) = 0.8.
BPS - 3rd Ed.
Chapter 12
9
Binomial Probabilities
Example
(1) First, find probability that the two survivors
are the first two offspring:
Using the Multiplication Rule:
P(SSFFFF) = (0.2)(0.2)(0.8)(0.8)(0.8)(0.8)
= (0.2)2(0.8)4
= 0.0164
BPS - 3rd Ed.
Chapter 12
10
Binomial Probabilities
Example
(2) Second, find the number of possible arrangements
for getting two successes and four failures:
SSFFFF
FSSFFF
FFSFSF
SFSFFF
FSFSFF
FFSFFS
SFFSFF
FSFFSF
FFFSSF
SFFFSF
FSFFFS
FFFSFS
SFFFFS
FFSSFF
FFFFSS
There are 15 of these, and each has the same
probability of occurring: (0.2)2(0.8)4.
Thus, the probability of observing exactly 2 successes
out of 6 is: P(X=2) = 15(0.2)2(0.8)4 = 0.246 .
BPS - 3rd Ed.
Chapter 12
11
Binomial Coefficient
 The
number of ways of arranging k
successes among n observations is
given by the binomial coefficient
n
n!
  
 k  k! (n  k )!
where n! is “n factorial” (see next slide).
– the binomial coefficient is read “n choose k”.
BPS - 3rd Ed.
Chapter 12
12
Factorial Notation
 For
any positive whole number n, its
factorial n! is
n! = n  (n1)  (n2)    3  2  1
– Also, 0! = 1 by definition.
 Example:
6! = 6·5·4·3·2·1 = 720,
and from the previous example:
6
6!
6!
6  5  4  3  2 1
6  5 30
  




 15
 2  2! (6  2)! 2!4! (2 1)  (4  3  2 1) 2 1 2
BPS - 3rd Ed.
Chapter 12
13
Binomial Probabilities
 If
X has the binomial distribution with n
observations and probability p of
success on each observation, the
possible values of X are 0, 1, 2, …, n.
If k is any one of these values, then
n k
n k
P( X  k )    p (1 p)
k 
BPS - 3rd Ed.
Chapter 12
14
Case Study
Inspecting Switches
The number X of bad switches has approximately the
binomial distribution with n=10 and p=0.1. Find the
probability of getting 1 or 2 bad switches in a sample of 10.
P( X  1 or 2)  P( X  1)  P( X  2)
10 
1
10-1  10 
  (0.1) (1- 0.1)   (0.1) 2 (1- 0.1)10-2
1
2
10!
10!
1
9

(0.1) (0.9) 
(0.1) 2 (0.9) 8
1!9!
2!8!
 (10)(0.1)( 0.3874)  (45)(0.01) (0.4305)
 0.3874  0.1937  0.5811
BPS - 3rd Ed.
Chapter 12
15
Mean and Standard Deviation
 If
X has the binomial distribution with n
observations and probability p of
success on each observation, then the
mean and standard deviation of X are
μ  np
σ  np(1 p)
BPS - 3rd Ed.
Chapter 12
16
Case Study
Inspecting Switches
The number X of bad switches has approximately the
binomial distribution with n=10 and p=0.1. Find the mean
and standard deviation of this distribution.
µ = np = (10)(0.1) = 1
 the probability of each being bad is one tenth; so we
expect (on average) to get 1 bad one out of the 10
sampled
σ  np(1 p)  (10)(0.1)( 1 0.1)  0.9  0.9487
BPS - 3rd Ed.
Chapter 12
17
Case Study
Inspecting Switches
Probability Histogram
n=10, p=0.1
BPS - 3rd Ed.
Chapter 12
18
Normal Approximation
to the Binomial
 The
formula for binomial probabilities
becomes cumbersome as the number of
trials n increases
 As the number of trials n increases, the
binomial distribution gets close to a Normal
distribution
– when n is large, Normal probability
calculations can be used to approximate
binomial probabilities
BPS - 3rd Ed.
Chapter 12
19
Normal Approximation
to the Binomial
 The
Normal distribution that is used to
approximate the binomial distribution uses
the same mean and standard deviation:
μ  np and σ  np(1 p)
 When
n is large, a binomial random variable
X (with n trials and success probability p) is
approximately Normal:

X is approx. N np, np(1 p)
BPS - 3rd Ed.
Chapter 12

20
Normal Approximation
to the Binomial (Sample Size)
 As
a rule of thumb, we will use the
Normal approximation to the Binomial
when n is large enough to satisfy the
following:
np ≥ 10 and n(1p) ≥ 10
– Note that these conditions also depend
on the value of p (and not just on n)
BPS - 3rd Ed.
Chapter 12
21
Case Study
Shopping Attitudes
Hall, Trish. “Shop? Many say ‘Only if I must’,” New York
Times, November 28, 1990.
Nationwide random sample of 2500 adults were
asked if they agreed or disagreed with the
statement “I like buying clothes, but shopping is
often frustrating and time-consuming.” Suppose
that in fact 60% of the population of all adult U.S.
residents would say “Agree” if asked this
question. What is the probability that 1520 or
more of the sample agree?
BPS - 3rd Ed.
Chapter 12
22
Case Study
Shopping Attitudes
The responses of the 2500 randomly chosen
adults (from over 210 million adults) can be taken
to be independent.
 The number X in the sample who agree has a
binomial distribution with n=2500 and p=0.60.
 To find the probability that at least 1520 people in
the sample agree, we would need to add the
binomial probabilities of all outcomes from
X=1520 to X=2500…this is not practical.

BPS - 3rd Ed.
Chapter 12
23
Case Study
Shopping Attitudes
Histogram of 1000
simulated values of the
binomial variable X,
and the density curve of
the Normal distribution
with the same mean
and standard deviation:
Find
probability
of getting at
least 1520:
µ = np = 2500(0.6) = 1500
σ  np(1  p )
 (2500)(0.6 )(0.4)
 600  24.49
BPS - 3rd Ed.
Chapter 12
24
Case Study
Shopping Attitudes
Assuming X has the N(1500, 24.49) distribution
[np and n(1p) are both ≥ 10], we have
 X  μ 1520  1500 
P( X  1520)  P


24.49
 σ

 P(Z  0.82)
 1 0.7939 (from Standard Normal Table)
 0.2061
BPS - 3rd Ed.
Chapter 12
25
Case Study
Shopping Attitudes
The probability of observing 1520 or more adults
in the sample who agree with the statement has
been calculated as 20.61% using the Normal
approximation to the Binomial.
 Using a computer program to calculate the actual
Binomial probabilities for all values from 1520 to
2500, the true probability of observing 1520 or
more who agree is 21.31%
 This is a very good approximation!

BPS - 3rd Ed.
Chapter 12
26