1. Basic Practice of Statistics - 3rd Edition Binomial Distributions
Chapter 13
Binomial Distributions
BPS - 5th Ed.
Chapter 13
Chapter 12

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
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Chapter 13

3. 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
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4. 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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Chapter 13

5. 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
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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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Chapter 13

7. 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
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Chapter 13

8. 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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9. 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.
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10. 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 =
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Chapter 13

11. Binomial Probabilities Example
(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(X=2) = 15(0.2)2(0.8)4 =
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Chapter 13

12. 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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Chapter 13

13. 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:
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Chapter 13

14. 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
BPS - 5th Ed.
Chapter 13

15. 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.
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Chapter 13

16. 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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Chapter 13

17. Case Study Inspecting Switches µ = np = (10)(0.1) = 1
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
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18. Probability Histogram
Case Study
Inspecting Switches
Probability Histogram
n=10, p=0.1
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Chapter 13

19. 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
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20. Normal Approximation to 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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Chapter 13

21. 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)
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22. 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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23. 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.
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Chapter 13

24. Case Study Shopping Attitudes Chapter 13
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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25. 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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Chapter 13

26. 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 - 5th Ed.
Chapter 13