1
n 6.1
Discrete and Continuous Random Variables
n 6.2
Transforming and Combining Random Variables
n 6.3
Binomial and Geometric Random Variables
n Binomial
1
2
Settings
When the same chance process is repeated several times, we are often interested
in whether a particular outcome does or doesn’t happen on each repetition. In
some cases, the number of repeated trials is fixed in advance and we are
interested in the number of times a particular event (called a “success”) occurs. If
the trials in these cases are independent and each success has an equal chance
of occurring, we have a binomial setting.
Definition:
A binomial setting arises when we perform several independent trials of the
same chance process and record the number of times that a particular
outcome occurs. The four conditions for a binomial setting are
After this section, you should be able to…
B
• _______________?
The possible outcomes of each trial can be
Binary
classified as “success” or “failure.”
outcomes
y onlya
q
DETERMINE whether the conditions for a binomial setting are met
q
COMPUTE and INTERPRET probabilities involving binomial random
variables
I
q
CALCULATE the mean and standard deviation of a binomial random
variable and INTERPRET these values in context
N
•number
_______________? The number of trials n of the chance process must
be fixed in advance.
q
CALCULATE probabilities involving geometric random variables
S
•success
_______________? On each trial, the probability p of success must be
the same.
n Binomial
The number of heads in n tosses is a binomial random variable X.
The probability distribution of X is called a binomial distribution.
Definition:
The __________ X of successes in a binomial setting is a binomial random
aunt
variable. The probability distribution of X is a binomial distribution with
parameters n and p, where ____ is the number of trials of the chance process
and _____ is the probability of a success on any one trial. The possible values
BCNP.
of X are the whole numbers from 0 to n. As an abbreviation, we say X~
Note: When checking the Binomial condition, be sure to check the BINS
and make sure you’re being asked to count the number of successes in a
certain number of trials!
Binomial and Geometric Random Variables
Consider tossing a coin n times. Each toss gives either heads or tails.
Knowing the outcome of one toss does not change the probability of
an outcome on any other toss. If we define heads as a success, then
p is the probability of a head and is 0.5 on any toss of a fair coin.
p
regarding i nseam
mama
its
• _______________?
Trials must be independent; that
is, knowing
the
Independent
result of one trial must not have any effect on the result of any other trial.
statement
fixes memberoftrials statenumberof
trials
evenprobabilitymusthavethesamesuccess
3
Random Variable
Binomial and Geometric Random Variables
AP Statistics
Chapter 6, Random Variables
4
Binomial Setting-Example
What is the probability that exactly 2 of 4 consumers will have
some frustration with the setup of their new electronic device if
the probability of frustration is 0.2 for any given consumer?
B: Each observation falls into one of _____ categories
“________”
= Frustrated or “_______”
success
failure = Not Frustrated
I: The observations are _______________.
independent
A call is ______ affected by a previous call.
N: __________ number n of observations (in advance)
fixesset
n=4
S: The probability of _______________,
p, is the same for
success
each observation.
P(“Success”) = ______
not
0.2
n Binomial
5
Probabilities
Step 1: P(SSFFF) = (0.25)(0.25)(0.75)(0.75)(0.75) = (0.25)2(0.75)3 = 0.02637
Step 2: However, there are _______ different arrangements in which 2
out of the 5 children have type O blood:
ten
therefore
we must do that
Thus
lox
orcombination
4!
n
3!
n
2!
n
1!
n
0!
in
n
ænö
n!
Ck = C(n, k ) = ç ÷ =
è k ø k !(n - k )!
12
C5 = C (12,5 )
æ12 ö
ç ÷
è7ø
We can generalize this for any setting in which we are interested in k
successes in n trials. That is,
P(X = k) = P(exactly k successes in n trials)
= number of arrangements× p k (1- p) n-k
Definition:
The number of ways of arranging k successes among n observations is
given by the binomial coefficient
æ nö
n!
ç ÷=
è k ø k!(n - k)!
for k = 0, 1, 2, …, n where
n! = n(n – 1)(n – 2)•…•(3)(2)(1)
and 0! = 1.
æ12 ö
=ç ÷
è5ø
7
n Binomial
8
Probability
The binomial coefficient counts the number of different ways in
which k successes can be arranged among n trials. The
binomial probability P(X = k) is this count multiplied by the
probability of any one specific arrangement of the k successes.
Binomial Probability
If X has the binomial distribution with n trials and probability p of success on
each trial, the possible values of X are 0, 1, 2, …, n. If k is any one of
these values,
æ nö
P(X = k) = ç ÷ p k (1- p) n-k
è kø
Number of
arrangements
of k successes
Probability of k
successes
Probability of
n-k failures
theexponents totalnumberof
trials
Binomial and Geometric Random Variables
Binomial Coefficient
AKA “Combinations”
n
Note, in the previous example, any one arrangement of 2 S’s and 3 F’s
had the same probability. This is true because no matter what
arrangement, we’d multiply together 0.25 twice and 0.75 three times.
Pex
2710883260
7
6
Coefficient
Binomial and Geometric Random Variables
having type O blood. Genetics says that children receive genes from
each of their parents independently. If these parents have 5 children,
the count X of children with type O blood is a binomial random
variable with n = 5 trials and probability p = 0.25 of a success on
each trial. In this setting, a child with type O blood is a “success” (S)
and a child with another blood type is a “failure” (F).
What’s P(X = 2)?
Binomial and Geometric Random Variables
In a binomial setting, we can define a random variable (say, X) as the
number of successes in n independent trials. We are interested in
finding the probability distribution of X.
Example
Each child of a particular pair of parents has probability 0.25 of
n Binomial
9
Finding Binomial Probabilities
9
3 t.IE
n What
is the probability that exactly 2 of 6
consumers will have some frustration with the
setup of their new electronic device if the
probability of frustration is 0.2 for any given
consumer?
nge P
x
2720.87
1510
2
IV
NV45
Finding Binomial Probabilities
10
10
p.389 TPS 4th Ed
n
TI-84: binomialpdf(n, p, k)
n
Can be found at 2nd [DISTR], A:binompdf
n
n = number of trials
n
p = probability of a single trial
n
k = X = number of successful observations
7i
Calculator
Binomppe 6,02 2
24.576 1
Finding Binomial Probabilities
n What
mum
is the probability that no more than 2 of
7 mall shoppers will agree to answer a brief
survey question if the probability of any one
mall shopper agreeing to answer a brief
survey question is 0.40?
1317,04
Co4110.676
79
y Co4
ng
11
11
12
12
Finding Cumulative
Binomial Probabilities
p.389 TPS 4th Ed
n
TI-84: binomialcdf(n, p, k)
n
Can be found at 2nd [DISTR], B:binomcdf
n
n = number of trials
n
p = probability of a single trial
n
k = X = this number and less of successful observations
t
2 It waslooklikethis 04 04
0.675
Onevalue
BinomialPdf
Accumulating values
S BinomialCaf
atleast
n Example:
13
Inheriting Blood Type
We describe the probability distribution of a binomial random variable just like
any other distribution – by looking at the shape, center, and spread.
Consider the probability distribution of X = number of children with type O
blood in a family with 5 children.
(a) Find the probability that exactly 3 of the children have type O blood.
Let X = the number of children with type O blood. We know X has a binomial
distribution with n = 5 and p = 0.25.
successes
or
É
offailures
O
251310.7510
PA3
5o
xi
0
1
2
3
4
5
pi
0.2373
0.3955
0.2637
0.0879
0.0147
0.00098
Shape: The probability distribution of X is skewed to
the right. It is more likely to have 0, 1, or 2 children
with type O blood than a larger value.
0087890625
(b) Should the parents be surprised if more than 3 of their children have
type O blood?
PA3
P e4
4
P xs
0.0146s
10.25 0.757
g 0.2550
becausethe
yes
probability is less
tort
co
than tuextremes
Center: The median number of children with type O
blood is _____. Based on our formula for the mean:
p
µ X = å x i pi = (0)(0.2373) + 1(0.39551) + ...+ (5)(0.00098)
= 1.25
this is
Binomial and Geometric Random Variables
and Standard Deviation of a Binomial
Distribution
Each child of a particular pair of parents has probability 0.25 of having blood
type O. Suppose the parents have 5 children
25,37 3
Binomialpor15.0
14
n Mean
assoriasSpread: The variance of X is s 2 = (x - µ ) 2 p = (0 -1.25) 2 (0.2373) + (1-1.25) 2 (0.3955) + ...+
requites
å i X i
rare
X
reactionfrom imparents
(5 -1.25) 2 (0.00098) = 0.9375
É
The standard deviation of X is s X = 0.9375 = 0.968
oasis
arcs
Binomial
aneroid
n Mean
Notice, the mean µX = 1.25 can be found another way. Since each
child has a 0.25 chance of inheriting type O blood, we’d expect
one-fourth of the 5 children to have this blood type.
That is, µX = 5(0.25) = 1.25. This method can be used to find the
mean of any binomial random variable with parameters n and p.
Mean and Standard Deviation of a Binomial Random Variable
If a count X has the binomial distribution with number of trials n and
probability of success p, the mean and standard deviation of X are
µ X = np
s X = np(1- p)
Note: These formulas work ONLY for binomial distributions.
They can’t be used for other distributions!
15
Binomial and Geometric Random Variables
and Standard Deviation of a Binomial
Distribution
n Example:
16
Inheriting Blood Type
Each child of a particular pair of parents has probability 0.25 of having blood type O. If
these parents have 5 children, let X = the number of children with type O Blood.
Find the mean and standard deviation of X.
Since X is a binomial random variable with parameters n = 5 and p = 1/3, we can
use the formulas for the mean and standard deviation of a binomial random
variable. Interpret each.
µ X = np
= __________
506251 = 1.25
In the long run
s X = np(1 - p)
50.75
= __________ = 0.968
we’d expect about 1.25 out of 5
_______ ______________, we’d expect
on
average
the number of children out of a family of
children to have type O blood.
5 to vary by __________ 0.968 children
_______ _______ _______ _______,
about
from the mean of 1.25 children.
n Binomial
Distributions in Statistical Sampling
The actual probability is
9000 8999 8998
8991
×
×
× ...×
= 0.3485
10000 9999 9998
9991
replacement
æ10ö
P(X = 0) = ç ÷(0.10) 0 (0.90)10 = 0.3487
è0ø
P(no defectives) =
no
Using the binomial distribution,
In practice, the binomial distribution gives a good approximation as long as we don’t
sample more than 10% of the population.
Sampling Without Replacement Condition
When taking an SRS of size n from a population of size N, we can use a
binomial distribution to model the count of successes in the sample as
1
long as
n£ N
in other terms: Population ≥ ________
longto
10
n Normal
Approximation for Binomial Distributions
As n gets larger, something interesting happens to the shape of a
binomial distribution. The figures below show histograms of
binomial distributions for different values of n and p. What do
you notice as n gets larger?
18
Binomial and Geometric Random Variables
Suppose 10% of CDs have defective copy-protection schemes that can harm
computers. A music distributor inspects an SRS of 10 CDs from a shipment of
10,000. Let X = number of defective CDs. What is P(X = 0)? Note, this is not
quite a binomial setting. Why?
Binomial and Geometric Random Variables
The binomial distributions are important in statistics when we want to
make inferences about the proportion p of successes in a population.
17
assample
size
increases
symmetry
Normal Approximation for Binomial Distributions
Suppose that X has the binomial distribution with n trials and success
probability p. When n is large, the distribution of X is approximately
Normal with mean and standard deviation
µ X = np
s X = np(1- p)
As a rule of thumb, we will use the Normal approximation when n is so
large that np ≥ 10 and n(1 – p) ≥ 10. That is, the expected number of
successes and failures are both at least 10.
popmess
KamEst
The Normal Approximation to
Binomial Distributions
19
19
p. 526 TPS 3rd Ed
n
As the number of trials n gets larger, the binomial distribution
gets close to a normal distribution.
In other words,
As long as _________
and __________
nu p 210
MPs lo
lot
Attitudes Toward Shopping
rule
20
Sample surveys show that fewer people enjoy shopping than in the past. A survey asked a
nationwide random sample of 2500 adults if they agreed or disagreed that “I like buying
new clothes, but shopping is often frustrating and time-consuming.” Suppose that
exactly 60% of all adult US residents would say “Agree” if asked the same question. Let
X = the number in the sample who agree. Estimate the probability that 1520 or more
of the sample agree.
1) Verify that X is approximately a binomial random variable.
___: There are exactly 2 outcomes.
Success = agree, Failure = don’t agree
___: Since we are sampling without replacement, we
must check that
independent
B(n, p ) ® N (np, np (1 - p ))
Check the
n Example:
it is
___: n = 2500 trials of the chance process
___: The probability of selecting an adult, who agrees is
p = 0.60
n Example:
21
Attitudes Toward Shopping
n Example:
Sample surveys show that fewer people enjoy shopping than in the past. A survey asked a nationwide
random sample of 2500 adults if they agreed or disagreed that “I like buying new clothes, but
shopping is often frustrating and time-consuming.” Suppose that exactly 60% of all adult US
residents would say “Agree” if asked the same question. Let X = the number in the sample who
agree. Estimate the probability that 1520 or more of the sample agree.
2) Check the conditions for using a Normal approximation.
np
2500o.o
1500
10
22
Attitudes Toward Shopping
nup 250060.4
4) Calculate P(X ≥ 1520) using a Normal approximation.
h
3) Calculate the mean and standard deviation.
Neup 2500 oo soo 24.40
O Ftp Acog
Ia
IIe 139ua
P as
t
The Normal Approximation to
Binomial Distributions (POD p.422)
The Normal Approximation to
Binomial Distributions (POD p.422)
23
Premature babies are those born more than 3 weeks early.
Newsweek reported that 10% of the live births in the US are
premature. Suppose that 250 live births are randomly selected.
n
Find the probability that the number of “preemies” is less than 20.
n
State: What is the probability that the number of “preemies” is
less than 20?
n
Plan:
n
Sample Randomly Selected? Randomlyselected
n nso P o.io
Population
n
Normality?
n
Since we’re sampling without replacement, can we consider the
sample independent?
sirens
Do:
births
A totalamount ofareeve
soolivebirthsintheU.sper
year
safeto assumetherearemorethan a
n
24
nasim
mmht.im
TotalPop ofalllive
n
t
25001.6
St 2
0 s 0.82
nom
re
Conclude:
i
P 2C
There isabout a 14.687 that the
number of U.s premiebirths in a sample
of 250 is lessthan20
q.gg
1.05
1468
trials
a
get
to
how
outcome
tans
itsuccessful
many
n Geometric
Settings
25
Definition:
A geometric setting arises when we perform independent trials of the same
chance process and record the number of trials until a particular outcome
occurs. The four conditions for a geometric setting are
B
• Binary? The possible outcomes of each trial can be classified as
“success” or “failure.”
I
• Independent? Trials must be independent; that is, knowing the result
of one trial must not have any effect on the result of any other trial.
T
• __________? The goal is to count the number of trials until the first
trials
success occurs.
S
• Success? On each trial, the probability p of success must be the
same.
26
Random Variable
In a geometric setting, if we define the random variable Y to be the
number of trials needed to get the first success, then Y is called a
geometric random variable. The probability distribution of Y is
called a geometric distribution.
Geometric
lowest
is
Definition:
Binomial and Geometric Random Variables
havetothisyourself
Binomial and Geometric Random Variables
In a binomial setting, the number of trials n is fixed and the binomial random variable
X counts the number of successes. In other situations, the goal is to repeat a
chance behavior until a success occurs. These situations are called geometric
settings.
n Geometric
successes of
of
o forbinomial
1
The number of trials Y that it takes to get a success in a geometric setting is
a geometric random variable. The probability distribution of Y is a
geometric distribution with parameter p, the probability of a success on
any trial. The possible values of Y are 1, 2, 3, ….
Note: Like binomial random variables, it is important to be able to
distinguish situations in which the geometric distribution does and
doesn’t apply!
untilis a keywon
Scuba Diving
this
remember
p.520 TPS
3rd
27
27
What is the probability that the first female is reached on the
fourth call?
afemalebting chosen
let a
failure 0.7
g
PCX 4
28
28
(POD p.389)
Suppose that 40% of students who drive to school carry jumper cables
(and would be willing to help you!). Find the probability that the
number of students that need to be stopped to find jumper cables is:
Ed
The mailing list of an agency that markets scuba-diving trips to
the Florida Keys contains 70% males and 30% females. The
agency calls 30 people chosen at random from its list.
BIsuccess0.3
Geometric Probabilities
P (Exactly 1) =
P (Exactly 2) =
P (Exactly 3) =
1029
x 0.4
X
ha Of
Of
P (Exactly 3 or less) =
P (Exactly k) =
g
EG
failures
X
0.24
gu
pay pay
x
success
Geometric Probability
If Y has the geometric distribution with probability p of
success on each trial, the possible values of Y are
1, 2, 3, … . If k is any one of these values,
P(Y = k) = (1 - p) k -1 p
0.144
o.gg
n TI-84:
n Can
29
29
n Mean
The table below shows part of the probability distribution of Y. We can’t show the
entire distribution because the number of trials it takes to get the first success
could be an incredibly large number.
geometpdf(p, x)
be found at 2nd [DISTR], E:geometpdf
np
= probability of a single trial
nx
= number of trials until a success is observed
yi
1
2
3
4
5
6
pi
0.143
0.122
0.105
0.090
0.077
0.066
n Can
Center: The mean of Y is µY = _______. We’d expect
it to take _______ guesses to get our first success.
our
be found at 2nd [DISTR], F: geometcdf
= probability of a single trial
nx
= number of trials or less until a success is observed
n
Spread: The standard deviation of Y is σY = _______. If the class played the Birth Day
game many times, the number of homework problems the students receive would differ
from 7 by an average of _______.
geometcdf(p, x)
np
…
Shape: The heavily right-skewed shape is
characteristic of any geometric distribution. That’s
because the most likely value is _______.
7
n TI-84:
30
of a Geometric Distribution
oÉÉI
Mean (Expected Value) of Geometric Random Variable
If Y is a geometric random variable with probability p of success on
each trial, then its mean (expected value) is E(Y) = µY = 1/p.
31
GEOMETRIC Mean & Standard Deviation
•
•
It’s also on the AP Statistics Exam Formula Sheet!!
Again, you just have to know when and how to apply it.
32
P(X > n)
TPS 3rd Ed p.546
The probability that it takes more than n
trials to see the first success is:
P(X > n) = (1 – p)n
tiffs
Binomial and Geometric Random Variables
Finding Geometric Probabilities
a of students
Example: Suppose that 40%
who drive to school carry jumper
cables. Find the probability that you
have to ask more than 5 people before
someone helps you?
s ump
Jumper cables ask morethan speoplenotcarry
1 2
34 s 6 78
X
t
x
Quiz 6.3B
AP Statistics
Name:
Nishi
3. Suppose that 20% of a herd of cows is infected with a particular disease.
(a) What is the probability that the first diseased cow is the 3rd cow tested?
1. Determine whether each random variable described below satisfies the conditions for a
binomial setting, a geometric setting, or neither. Support your conclusion in each case.
(a) A high school principal goes to 10 different classrooms and randomly selects one student
from each class. X = the number of female students in his group of 10 students.
non orten above as the
It is different
each trial as the
is
is limy to be diff each
time
success
gg.m.in
0.128 o
w
of
wsuccessunfortunatesuccess
a st
pain por
probability
of femalestunts (b)your
What is the probability that 4 or more cows would need to be tested until a diseased cow
was found?
(b) You are on Interstate 80 in Pennsylvania, counting the occupants in every fifth car you
pass. Let Z = the number of cars you pass before you see one with more than two
occupants.
transpno diseased cows
outan ortreason
thisgeometric as we aretryingtofigure
in
3 trims
morethantwooccupants z
F
2. A manufacturer produces a large number of toasters. From past experience, the manufacturer
knows that approximately 2% are defective. In a quality control procedure, we randomly
select 20 toasters for testing.
in
y
(a) Determine the probability that exactly one of the toasters is defective.
Binomial
F
I.to
P oonor at
failure
coonIIagiarain0.27orsat
(b) Find the probability that at most two of the toasters are defective.
y
binomcsf28102727
0.0270.8
no
t
0.9929
4 Cooncoas 4 6.025Coast
(c) Let X = the number of defective toasters in the sample of 20. Find the mean and standard
deviation of X.
My
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© 2011 BFW Publishers
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as
Tomo
0.63 or 63 1
The Practice of Statistics, 4/e- Chapter 6
271
272
The Practice of Statistics, 4/e- Chapter 6
© 2011 BFW Publishers
Quiz 6.3C
3. Suppose there are 1100 students in your high school, and 28% of them take Spanish. You
select a sample of 50 student in the school, and you want to calculate the probability that 15
or more of the students in your sample take Spanish. Which condition for the binomial
setting has been violated here, and why does the binomial distribution do a good job of
estimating this probability anyway?
Name:Nishi
AP Statistics
1. Determine whether each random variable described below satisfies the conditions for a
binomial setting, a geometric setting, or neither. Support your conclusion in each case.
(a) Draw a card from a standard deck of 52 playing cards, observe the card, return the card to
the deck, and shuffle. Count the number of times you draw a card in this manner until
you observe a jack.
successfailure
hasbeenviolated as tu porsensing amo
Independence
studentwhotans spanish changes
storms take Spanish
as the is
seatiara
are returningthecartothe anam on
asyou
acar soonimma a e orswing
ita year that he
gens
4.
aof.at
of times in saIEfEE
(b) Joey buys a Virginia lottery ticket
Itevery week. XIis the number
wins a prize.
failure
isnot
binary as success is winninga prizep
B binary
Geometric because
it on or
sawing
Binomialbecause
B
A fair coin is flipped 20 times.
(a) Determine the probability that the coin comes up tails exactly 15 times.
winning
not
once
or
as winning
I winning
next
week
losing
eat ofor
there
defines
is a fixernumberor
as
trialsis
N
the
or
weanssecessions
u.tngtonaticYEe
losing
independent
does
for as
2. When a computerized generator is used to generate random digits, the probability that any
particular digit in the set {0, 1, 2, . . . , 9} is generated on any individual trial is 1/10 = 0.1.
Suppose that we are generating digits one at a time and are interested in tracking occurrences
of the digit 0.
getting firstzero
ofthat the first 0 occurs as the fifthtie
(a) Determine the probability
random digit generated.
numbers sweater
on
5
Pix5 Geompsfons o.ot P
(b) How many random digits would you expect to have to generate in order to observe the
5 0.57 o575 7
(b) Let X = the number of tails in the 20 flips. Find the mean and standard deviation of X.
Mx 2010.5
first 0?
(c) Let X = number of digits selected until first zero is
encountered. Construct a probability distribution
histogram for X = 1 through X = 5. Use the grid
provided.
it
of X
Probability
si
coacoaco1 o.o
© 2011 BFW Publishers
so
0.01478571
I 2.231
(c) Find the probability that X takes a value within 1 standard deviation of its mean.
ff
The Practice of Statistics, 4/e- Chapter 6
N is 1 or a
s
10 112.2367 77 764
10 12363 7
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