Binomial Distributions

1
Chapter 12

• Binomial Distributions

2
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

3
Binomial SettingExamples

• 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

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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

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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

6
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.
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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 n10 and p0.1

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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

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Binomial ProbabilitiesExample

• 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.

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Binomial ProbabilitiesExample
(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
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Binomial ProbabilitiesExample
(2) Second, find the number of possible
arrangements for getting two successes and four
failures
SSFFFF SFSFFF SFFSFF SFFFSF
SFFFFS FSSFFF FSFSFF FSFFSF FSFFFS
FFSSFF FFSFSF FFSFFS FFFSSF FFFSFS
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(X2) 15(0.2)2(0.8)4
0.246 .
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Binomial Coefficient

• The number of ways of arranging k successes among
  n observations is given by the binomial
  coefficient

• where n! is n factorial (see next slide).
• the binomial coefficient is read n choose k.

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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! 654321 720,
• and from the previous example

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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

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Case Study
Inspecting Switches
The number X of bad switches has approximately
the binomial distribution with n10 and p0.1.
Find the probability of getting 1 or 2 bad
switches in a sample of 10.
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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

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Case Study
Inspecting Switches
The number X of bad switches has approximately
the binomial distribution with n10 and p0.1.
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

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Case Study
Inspecting Switches
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Normal Approximationto 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

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Normal Approximationto the Binomial

• The Normal distribution that is used to
  approximate the binomial distribution uses the
  same mean and standard deviation

• When n is large, a binomial random variable X
  (with n trials and success probability p) is
  approximately Normal

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Normal Approximationto 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)

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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?
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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 n2500 and p0.60.
• 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
  X1520 to X2500this is not practical.

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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
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Case Study
Shopping Attitudes

• Assuming X has the N(1500, 24.49) distribution
  np and n(1?p) are both 10, we have

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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!