Transcript binomial distribution

Chapter 8:
The Binomial and Geometric Distributions
8.1
Binomial Distributions
8.2
Geometric Distributions
1
Let me begin with an example…
• My best friends from Kent School had
three daughters. What is the probability
of having 3 girls?
• We calculate this to be 1/8, assuming
that the probabilities of having a boy
and having a girl are equal.
• But the point in beginning with this
example is to introduce the binomial
setting.
2
The Binomial Setting (p. 439)
•
Let’s begin by defining X as the number of girls.
•
This example is a binomial setting because:
1. Each observation (having a child) falls into just
one of two categories: success or failure of
having a girl.
2. There is a fixed number of observations (n).
3. The n observations are independent.
4. The probability of success (p) is the same for
each observation.
3
Binomial Distributions
• If data are produced in a binomial setting,
then the random variable X=number of
successes is called a binomial random
variable, and the probability distribution of
X is called a binomial distribution.
• X is B(n,p)
• n and p are parameters.
• n is the number of observations.
• p is the probability of success on any
observation.
4
How to tell if we have a binomial
setting …
• Examples 8.1-8.4, pp. 440-441
• Exercise 8.1, p. 441
5
Binomial Probabilities—pdf
• We can calculate the probability of each value
of X occurring, given a particular binomial
probability distribution function.
• 2nd (VARS) 0:binompdf (n, p, X)
• Try it with our opening example:
• What is the probability of having 3 girls?
• Example 8.5, p. 442
• Note conclusion on p. 443
6
Binomial Probabilities—cdf
• Example 8.6, p. 443
• We can use a cumulative distribution
function (cdf) to answer problems like
this one.
• Binomcdf (n, p, X)
7
Problems
• 8.3 and 8.4, p. 445
8
Binomial Formulas
• We can use the TI-83/84 calculator
functions binompdf/cdf to calculate
binomial probabilities, as we have just
discussed.
• We can also use binomial coefficients
(p. 447). Read pp. 446-448 carefully to
be able to use this alternate approach.
9
Homework
• Read through p. 449
• Exercises:
• 8.6-8.8, p. 446
• 8.12-8.13, p. 449
10
Mean and Standard Deviation for
a Binomial Random Variable
• Formulas on p. 451:
  np
  np(1  p)
• Now, let’s go back to problem 8.3, p. 445.
Calculate the mean and standard deviation and
compare to the histogram created.
11
Normal Approximation to
Binomial Distributions
• As the number of trials, n, increases in our binomial
settings, the binomial distribution gets close to the
normal distribution.
• Let’s look at Example 8.12, p. 452, and Figure 8.3.
• I show this because it begins to set the stage for very
important concept in statistical inference—The
Central Limit Theorem.
• Because of the power of our calculators, however, I
recommend exact calculations using binompdf and
binomcdf for binomial distribution calculations.
12
Rules of Thumb
• See box, top of page 454.
13
Problems
• 8.15 and 8.17, p. 454
• 8.19, p. 455
• 8.27, 8.28, and 8.30, p. 461
14
Homework
• Reading, Section 8.2: pp. 464-475
• Review:
• Conditions for binomial setting
• Compare to conditions for geometric setting
• What’s the key difference between the two
distributions?
• Test on Monday
15
8.2 Geometric Distributions
• I am playing a dice game where I will roll one
die. I have to keep rolling the die until I roll a 6,
then the game is over.
• What is the probability that I win right away, on
the first roll?
• What is the probability that it takes me two rolls
to get a 6?
• What is the probability that it takes me three
rolls to get a 6?
16
The Geometric Setting
• This example is a geometric setting
because:
1. Each observation (rolling a die) falls
into just one of two categories: success
or failure.
2.The probability of success (p) is the
same for each observation.
3.The observations are independent.
4.The variable of interest is the number of
trials required to obtain the first
success.
17
Calculating Probability
• For a geometric distribution, the probability that
the first success occurs on the nth trial is:
P( X  n)  (1  p) n1 p
• Let’s look at Example 8.16, p. 465, and then the
example calculations below that example for
Example 8.15.
• Why the name “geometric” for this distribution?
• See middle of page 466.
18
“Calculator Speak”
• Notice that we do not have an “n”
present in the following calculator
commands … that’s the point of a
geometric distribution!
geometpdf ( p, X )
geometcdf ( p, X )
19
Exercises:
8.38, p. 468
8.43 (b&c), p. 474
-- See example 8.15, p. 465
20
Mean and Variance of a
Geometric Random Variable
1
 x  ... Expected Value
p
(1  p)
2
x 
2
p
21
Homework
• Read through the end of the chapter.
• Problems:
• 8.36, p. 463
• 8.39, p. 468
• 8.46, p. 474
• 8.55 and 8.56, p. 479
• 8.1-8.2 Quiz Friday
22
Chapter 8 Review Problems
• pp. 480-482:
• 8.60
• 8.61
• 8.62
• 8.63
23