Transcript StatsChap6

CHAPTER
6
Discrete
Probability
Distributions
Copyright © 2013, 2010 and 2007 Pearson Education, Inc.
Chap 2
2
6.1
Section
Discrete
Random
Variables
Copyright © 2013, 2010 and 2007 Pearson Education, Inc.
A random variable is a numerical
outcome from a probability experiment.
Its value is determined by chance, aka
dumb luck.
Random variables are denoted using
letters such as “x, y, and z”.
6-4
A discrete random variable has countable
number of values.
Its values can be plotted on a number line
in an uninterrupted fashion, i.e., only the
values from 0, 1, 2, 3, and 4.
Think “digital” watch
6-5
A continuous random variable has
infinitely many values. Its values can be
measured, not counted.
Its values can be plotted on a number line
in an uninterrupted fashion, i.e., all the
values from 0 thru 4.
Think “analog” or sweep-second hand
watch
6-6
EXAMPLE
Distinguishing Between Discrete and
Continuous Random Variables
(a) The number of light bulbs that burn out in a room
with 10 light bulbs in the next year.
Discrete; x = 0, 1, 2, …
(b) The number of leaves on a randomly selected oak
tree.
Discrete; x = 0, 1, 2, …
(c) The length of time between calls to 911.
Continuous; t > 0
6-7
A probability distribution provides all
possible values of the outcome random
variable “x” and the corresponding
probabilities of each outcome occurring.
A probability distribution can be in the
form of a table, graph or math formula.
6-8
EXAMPLE
A Discrete Probability Distribution
“x” represents the survey
results of the number of
movies streamed on Netflix
each month at random
homes in the county.
P(x), read “P of x”, is the
probability of each possible
outcome occurring. So, if
you called a random home,
the prob they would say they
streamed 3 movies is 0.10…
x
0
P(x)
0.06
1
2
3
4
0.58
0.22
0.10
0.03
5
0.01
6-9
Rules for a Discrete Probability
Distribution
Σ P(x) = 1: The sum of all Probs = 1
0 ≤ P(x) ≤ 1 : Each individual Prob is
between 0 and 1
5-10
A probability histogram is a graph in
which the horizontal axis plots the
values of “x”, and
the vertical axis plots the probability of
that value of “x” occurring as an
outcome.
6-11
EXAMPLE
Drawing a Probability Histogram
Probability histogram and
probability distribution
representing the number of movies
streamed on Netflix each month.
Movies Streamed from Netflix
0.7
Probability
0.6
0.5
x
0
1
2
3
4
5
P(x)
0.06
0.58
0.22
0.10
0.03
0.01
0.4
0.3
0.2
0.1
0
0
1
2
3
4
5
Number of Movies Streamed
6-12
Mean of a Discrete Random Variable
The mean of a discrete random variable is
given by the formula
 x    x  P x 
where x is the value of the variable and P(x)
is the probability of observing that value x.
5-13
EXAMPLE
Computing the Mean of a Discrete
Random Variable
Compute the mean of this
distribution, which
represents the mean
number of DVDs a random
customer rents from a video
store during a visit.
 x    x  P x 
x
0
1
2
3
4
5
P(x)
0.06
0.58
0.22
0.10
0.03
0.01
 0(0.06)  1(0.58)  2(0.22)  3(0.10)  4(0.03)  5(0.01)
 1.49
6-14
Interpretation of the Mean of a Discrete Random Variable
Suppose an experiment (observing DVD rentals) is repeated a
large number of times (Law Large Numbers).
As the number of repetitions of the experiment increases, the
mean value of the trials will approach μX =1.49 DVDs rented
each time.
So, the first ten customers may rent 5 DVDs each, but after 500
customers have come in, their mean number of DVDs will
approach 1.49 each.
That is what Probability tells us ~ the expected outcome over
time, not instantaneous results.
5-15
The following data represent the actual number of
DVDs rented by 100 customers in a single visit.
Compute the mean number of DVDs rented.
6-16
As the number of trials of the DVD experiment
increases, the mean number of rentals approaches the
mean of the probability distribution.
6-17
Because the mean of a discrete random
variable represents what we would expect to
happen in the long run, it is also called the
expected value, E(x), of the
random
variable.
6-18
EXAMPLE
Computing the Expected Value of a Discrete
Random Variable
A life insurance policy will pay a discrete sum of money upon the death of the
policy holder. These policies have premiums that must be paid annually.
Suppose a company sells a one-year $250,000 life insurance policy to a
49-year-old woman for $530. The probability she will survive the year is
0.99791. Compute the expected value (distribution mean) of this policy to the
insurance company.
She survives
She does not survive
X ($)
P(x)
+530
0.99791
530 – 250,000 =
-249,470
0.00209
E(X) = 530(0.99791) + (-249,470)(0.00209) = $7.50
What does that EV represent to the company if they sell
100,000 policies – profit or loss?
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Expected Value
At a raffle, 500 tickets are sold for $1 each for two prizes
of $100 and $50. If you buy one ticket, what is the
expected value of your gain?
Gain, x
P ( x)
+$99
1
500
+$49
1
500
–$1
498
500
Winning
no prize
E(x) = Σ[x •P (x) ]
 $99 
1
1
498
 $49 
 ($1) 
500
500
500
 $0.70
Because the expected value is
negative, you can expect to lose
an average of $0.70 for each $1
ticket you buy.
Larson & Farber, Elementary Statistics: Picturing the World, 3e
20
TI-84
To find the Mean (EV) of a Probability Distribution
Go to Stat:Edit – enter the “x” values in List 1
and the “Prob” values in List 2
ló
Go to Stat:Calc:1Var-Stats
Enter: L1,L2
This will give μ, or the Expected Value
of the outcome.
Chap 4
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Section
6.2
The
Binomial
Probability
Distribution
Copyright © 2013, 2010 and 2007 Pearson Education, Inc.
Criteria for a Binomial Probability Experiment
A binomial experiment is:
1. The experiment is performed a fixed number of times.
Each repetition of the experiment is called a trial.
2. The trials are independent. This means the outcome of
one trial will not affect the outcome of the other trials.
3. For each trial, there are only two disjoint outcomes as:
Success/Failure; Pass/Fail; True/False; Yes/No, etc
4. The probability of success “p” is fixed for each trial of
the experiment.
5-23
Binomial Probability Distribution
•There are “n” independent trials of the
experiment.
•Let “p” denote the probability of success so
that “q” = (1 – p) is the probability of failure.
•Let “x” be a binomial discrete variable that
denotes the number of successes in “n” trials
So…. 0 < x < n.
5-24
EXAMPLE
Constructing a Binomial Probability Distribution
According to the Air Travel Consumer Report, the 11
largest air carriers had an on-time arrival percentage
of 79.0% in May, 2008.
Suppose that 4 flights are randomly selected from
May, 2008 and the number of on-time flights “x” is
recorded.
Construct a probability distribution for the random
variable “x” using a tree diagram.
6-25
The probability of obtaining x successes in
n independent trials of a binomial
experiment is given by:
P x   n Cx p 1 p 
x
nx
x  0,1,2,...,n
where “p” is the probability of success for
each trial.
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Phrase
“at least” or “no less than”
“greater than or equal to”
Math Symbol
≥
“more than” “greater than”
“fewer than” “less than”
>
<
“no more than” “at most”
“less than or equal to
“exactly” “equals” “is”
“does not equal” “is not”
≤
=
≠
6-27
EXAMPLE
Using the Binomial Probability Distribution Function
According to the Experian Automotive, 35% of all
car-owning households have three or more cars.
In a random sample of 20 car-owning households,
what is the probability that exactly 5 have three
or more cars?
P(5) 
5
205
C
(0.35)
(1
0.35)
20 5
 0.1272
6-28
TI-83/84

DISTR:0 is binompdf for “prob density
function”. It gives you individual discrete
probabilities.

DISTR:A is binomcdf for “cum density
function”. It gives you cumulative
(starting from zero) discrete probabilities.
Chap 4
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binompdf… (n , p , x)
If you wanted to know the prob that
exactly 5 households have three or
more cars,
 … and the prob of “success” on each
trial was 0.350, then …


Binompdf (20,0.350,5)= 0.1272
Chap 4
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EXAMPLE
Using the Binomial Probability Distribution Function
According to the Experian Automotive, 35% of all
car-owning households have three or more cars.
In a random sample of 20 car-owning households,
what is the probability that less than 4 have three or
more cars?
P(X  4)  P(X  3)
 P(0)  P(1)  P(2)  P(3)
 0.0444
6-31
x
binomcdf (n,p,x)

This function accumulates prob starting with
x=0

For P(x = 3 or less)
binomcdf (20,0.35,3)=0.0444

So, the prob of having 4 or more cars is:
1 - 0.0444 = 0.9556
Chap 4
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Mean (Expected Value) and Standard
Deviation of a Binomial Random Variable
A binomial experiment with “n” independent
trials and probability of success “p” has a
mean / standard deviation given by:
 X  np and  X  np 1 p 
5-33
EXAMPLE
Finding the Mean and Standard Deviation of a
Binomial Random Variable
According to Experian Automotive, 35% of all car-owning
households have three or more cars. In a simple random
sample of 400 car-owning households, determine the mean
and standard deviation of car-owning households that will
have three or more cars.
 X  np
 (400)(0.35)
 140
 X  np(1  p)
 (400)(0.35)(1  0.35)
 9.54
So, if the data were normally distributed, 68% of these households
would own between 130.46 and 149.54 cars
6-34
EXAMPLE
Constructing Binomial Probability
Histograms
Construct a binomial probability histograms with
n = 8 and varying “p”
In other words, you flip a coin 8 times.
Success is (approx) :
a) getting a 1
p ≈ 0.15
b) getting a 3 or less
p ≈ 0.50
c) getting a 2 or more p ≈ 0.85
For each histogram, comment on the shape of the
distribution.
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In other words, in 8 coin flips, you got 2 successes (rolling a 1) about 24% of
the time.
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In other words, in 8 coin flips, you got 5 successes (rolling 3 or less)
about 22% of the time.
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In other words, in 8 coin flips, you got 7 successes
(rolling 2 or more) almost 40% of the time.
μ 7/8 = 0.875
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μ 21/25 = 0.840
6-39
μ 58/70 = 0.829
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For a fixed probability of success, p, as the
number of trials n in a binomial experiment
increase, the probability distribution of the
random variable “x” becomes bell-shaped.
As a general rule, if σ > 3 then the binomial
probability distribution will be
approximately bell-shaped.
(called a “normal” distribution).
5-41
Use The Empirical Rule (for normal
distributions) to identify “unusual”
observations in an experiment.
95% of the observations lie within ± 2σ
from the μ
Any observation that lies outside this
interval may be considered “unusual”
because it occurs less than 5% of the time.
5-42
Section
6.3
The
Poisson
Probability
Distribution
Copyright © 2013, 2010 and 2007 Pearson Education, Inc.
Poisson Distribution
The Poisson (Fr: b 1781) distribution satisfies the following
conditions.
1. The experiment consists of counting the number of times an
event, x, occurs in a given interval. The interval can be an
interval of time, area, or volume.
2. The probability of the event occurring is the same for each
interval.
3. The number of occurrences in one interval is independent of the
number of occurrences in other intervals.
The probability of exactly x occurrences in an interval is
x μ
μ
P (x )  e
x!
where e  2.71818 and μ is the mean number of occurrences.
Larson & Farber, Elementary Statistics: Picturing the World, 3e
44
Poisson Distribution
The mean number of power outages in the city of Gulf Breeze is
4 per week. Find the probability that in any given week….
a.) there are exactly 3 outages b.) there are more than 3 outages.
Note: λ = μ in the Poisson formula
a.)   4, x  3
43(2.71828)-4
P (3) 
3!
 0.195
b.) P (more than 3)
 1  P (x  3)
 1  [P (3)  P (2) + P (1) + P (0)]
 1  (0.195  0.147  0.073  0.018)
 0.567
Larson & Farber, Elementary Statistics: Picturing the World, 3e
45
Poisson Distribution
Statistics show that, on average, sharks kill 10 (λ)
people each year worldwide. Find the probability
that….
a) 3 people are killed by sharks this year
b) Two or three people are killed by sharks this
year
P(3) = 0.0076
P(2 or 3) = 0.0023 + 0.0076 = 0.0099
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DISTR:B:poissonpdf ( λ ,x)

The mean number of major hurricanes
that strike Florida in any given year is
0.7 (7 storms every 10 years)

What is the probability that 2 major
hurricanes will strike Florida this year?

poissonpdf(0.7,2) = 0.1217
Chap 4
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TI-83/84
You can also use
functions for Poisson
distributions.
CUMULATIVE
This is similar to Binomcdf, where the calculator
will count up all probabilities up to, and including
the count that you specify.
Ex: you want to calc the prob that 2 or less major
hurricanes will strike Florida this year.
DISTR:D poissoncdf (0.7,2) = 0.9659 (not 0.1217)
Chap 4
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The following Sections on
Geometric and Hypergeometric
Probabilities are Optional
They are for your academic
interest. They will not be
covered on exams.
Chap 4
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Geometric Distributions
Chap 4
50
Geometric Distribution
A geometric distribution satisfies the following
conditions.
1. A trial is repeated until a: success occurs.
2. The repeated trials are independent of each other.
3. The probability of a success p is constant for each
trial.
The probability that the first success
will occur on trial x is
P (x) = p (q)x – 1 where q = 1 – p.
Larson & Farber, Elementary Statistics: Picturing the World, 3e
51
Geometric Distribution
A study has found that the probability that a
person who enters a store will actually make
a purchase is 0.30
Therefore, the probability that the first purchase
will be made by customer number 4 is:
P(4) = (.30)(.70)3 = 0.1029
52
DISTR:D:geometpdf… (p,x)
You own a clothing store. Suppose the prob
you will make a sale to any given customer is
0.30
To find the prob that your first sale occurs
with the 4th customer…
(F,F,F,S) = (0.70^3)(0.30^1)

geometpdf(0.30,4) = 0.1029
Chap 4
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Hypergeometric Distributions
Chap 4
54
Hypergeometric Probability Distribution
The probability of obtaining “x” successes
from a sample size “n” is:

k C x  N k Cn x 
P x  
N
Cn
where “k” is the number of successes in the
population size “N” .
5-55
EXAMPLE
A Hypergeometric Probability Distribution
The Dow Jones Industrial Average (DJIA) is a
collection of 30 traded companies on the US Stock
Market that are meant to be representative of the
entire United States economy. In one certain month
18 of the 30 stocks (60%) in the DJIA increased in
value.
What is the probability that an investor who
randomly invests in 4 of these 30 stocks at the
beginning of that month found that 3 of his 4 stocks
increased in value?
6-56
EXAMPLE
A Hypergeometric Probability Distribution
k = 18 N = 30
and
x=3 n=4

k C x  N k Cn x 
P(x) 
N
Cn

18 C 3  3018 C 43 

30
C4
816 12 


27, 405
 0.3573
6-57
EXAMPLE
A Hypergeometric Probability Distribution
Try this Hypergeometric problem:
Women make up 54% of the adult population of the
US.
What is the probability that, if you selected 17 US
adults, that exactly 13 of them would be women?
Ans: 0.0325 or: if you picked 17 adults from the
population 10,000 times, 325 of those times there
would be 13 women out of the 17.
6-58
Jury Selection Problem
There are 60 women and 40 men waiting
outside the courtroom in the jury pool.
Assuming the probability of being
selected for the jury is the same for
each person, what is the probability
that the 12 person jury selected will
consist of 7 men and 5 women?
59
End Chap 6 !
Are you having fun yet??
Don’t worry, it gets even better!
60
Chap 2
61