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The Binomial Probability
Distribution and
Related Topics
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Copyright © Cengage Learning. All rights reserved.
Section
6.2
Binomial
Probabilities
Copyright © Cengage Learning. All rights reserved.
Binomial Experiment
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Binomial Experiment
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Computing Probabilities for a Binomial Experiment Using the Binomial
Distribution Formula
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Using a Binomial Distribution
Table
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Using a Binomial Distribution Table
In many cases we will be interested in the probability of a
range of successes. In such cases, we need to use the
addition rule for mutually exclusive events. For instance, for
n = 6 and p = 0.50,
P(4 or fewer successes) = P(r  4)
= P(r = 4 or 3 or 2 or 1 or 0)
= P(4) + P(3) + P(2) + P(1) + P(0)
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Using a Binomial Distribution Table
It would be a bit of a chore to use the binomial distribution
formula to compute all the required probabilities.
Table 2 of Appendix gives values of P(r) for selected
p values and values of n through 20.
To use the table, find the appropriate section for n, and
then use the entries in the columns headed by the p values
and the rows headed by the r values.
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Using a Binomial Distribution Table
Table 6-10 is an excerpt from Table 2 of Appendix showing
the section for n = 6. Notice that all possible r values
between 0 and 6 are given as row headers.
Excerpt from Table 2 of Appendix for n = 6
Table 6-10
The value p = 0.50 is one of the column headers. For n = 6
and p = 0.50, you can find the value of P(4) by looking at the
entry in the row headed by 4 and the column headed by
0.50. Notice that P(4) = 0.234 (to three digits).
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Using a Binomial Distribution Table
Likewise, you can find other values of P(r) from the table. In
fact, for n = 6 and p = 0.50,
P(r  4) = P(4) + P(3) + P(2) + P(1) + P(0)
= 0.234 + 0.312 + 0.234 + 0.094 + 0.016 = 0.890
Alternatively, to compute P(r  4) for n = 6 , you can use
the fact that the total of all P(r) values for r between 0 and 6
is 1 and the complement rule.
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Using a Binomial Distribution Table
Since the complement of the event r  4 is the event r  5,
we have
P(r  4) = 1 – P(5) – P(6)
= 1 – 0.094 – 0.016
= 0.890
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Example 5 – Using the binomial distribution table to find P(r)
A biologist is studying a new hybrid tomato. It is known that
the seeds of this hybrid tomato have probability 0.70 of
germinating. The biologist plants six seeds.
a. What is the probability that exactly four seeds will
germinate?
Solution:
This is a binomial experiment with n = 6 trials. Each seed
planted represents an independent trial.
We’ll say germination is success, so the probability for
success on each trial is 0.70.
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Example 5 – Solution
n=6
p = 0.70
q = 0.30
cont’d
r=4
We wish to find P(4), the probability of exactly four
successes. In Table 2, Appendix, find the section with
n = 6 (excerpt is given in Table 6-10).
Excerpt from Table 2 of Appendix for n = 6
Table 6-10
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Example 5 – Solution
cont’d
Then find the entry in the column headed by and the row
headed by p = 0.70 and the row headed by r = 4.
This entry is 0.324.
P(4) = 0.324
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Example 5 – Using the binomial distribution table to find P(r)
cont’d
b. What is the probability that at least four seeds will
germinate?
Solution:
In this case, we are interested in the probability of four or
more seeds germinating. This means we are to compute
P(r  4). Since the events are mutually exclusive, we can
use the addition rule
P(r  4) = P(r = 4 or r = 5 or r = 6)
= P(4) + P(5) + P(6)
We already know the value of P(4). We need to find P(5)
and P(6).
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Example 5 – Solution
cont’d
Use the same part of the table but find the entries in the
row headed by the r value 5 and then the r value 6. Be sure
to use the column headed by the value of p, 0.70.
P(5) = 0.303
and
P(6) = 0.118
Now we have all the parts necessary to compute P(r  4).
P(r  4) = P(4) + P(5) + P(6)
= 0.324 + 0.303 + 0.118
= 0.745
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Using Technology to Compute Binomial
Probabilities
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Using Technology to Compute Binomial Probabilities
Some calculators and computer-software packages support
the binomial distribution. In general, these technologies will
provide both the probability P(r) for an exact number of
successes r and the cumulative probability P(r  k), where
k is a specified value less than or equal to the number of
trials n.
Note that most of the technologies use the letter x instead
of r for the random variable denoting the number of
successes out of n trials.
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Sampling Without Replacement: Use of the
Hypergeometric Probability Distribution
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Sampling Without Replacement: Use of the Hypergeometric Probability Distribution
If the population is relatively small and we draw samples
without replacement, the assumption of independent trials
is not valid and we should not use the binomial distribution.
The hypergeometric distribution is a probability distribution
of a random variable that has two outcomes when sampling
is done without replacement.
This is the distribution that is appropriate when the sample
size is so small that sampling without replacement results
in trials that are not even approximately independent.
A discussion of the hypergeometric distribution can be
found in Appendix I of Understandable Statistics, 10th
edition.
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