Transcript p.p chapter 6.3

Binomial & Geometric
Random Variables
Section 6.3
Reference Text:
The Practice of Statistics, Fourth Edition.
Starnes, Yates, Moore
Objectives
1. Binomial Random Variables and Binomial
Distribution
1.
2.
3.
4.
5.
Requirements to be Binomial- B.I.N.S
Binomial Coefficient: by formula and TI-83
Binomial Probability: by formula and TI-83
Mean and Standard Deviation
“10% condition” sampling w/o replacement
2. Geometric Random Variables and Geometric
Distribution
1. Requirements to be Geometric- B.I.T.S
2. Geometric Probability: by formula and TI-83
3. Mean (expected value)
Consider this:
• Toss a coin 5 times, count the number of heads.
• Spin a roulette wheel 8 times. Record how many
times a ball lands in a red slot.
• Take a random sample of 100 babies born in U.S
hospitals today. Count the number of females.
In each case, we’re preforming repeated trials of the same
chance process. The number of trials is fixed in advance.
(number of trials n) In addition, the outcome of one trial has no
effect on the outcome of any other trial, that is, the trials are
independent. We’re interested in the number of times a specific
event occurs, (we’ll call it a “success”). Our chances of getting a
“success” are the same on each trial. (probability p is the same)
Binomial Random Variables and
Binomial Distribution
• The count X of successes in a binomial
setting is a binomial random variable.
The probability distribution of X is a
binomial distribution with parameters n
and p, where n is the number of trials of
the chance process and p is the probability
of a success on any one trial. The possible
values of X are the whole numbers from 0
to n.
Requirements to be Binomial- B.I.N.S
• A binomial setting arises when we preform 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:
• Binary? The possible outcomes of each trial can be classified as
“success” or “failure”
• 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
• Number? The number of trials n of the chance process must be
fixed in advance.
• Success? On each trial, the probability p of success must be the
same.
Examples:
Here are 3 scenarios
• In each case, determine whether the given random
variable has a binomial distribution. Justify your
answer. B.I.N.S
1. Genetics say that children receive genes from each of their parents
independently. Each child of a particular pair of parents have a probability
0.25 of having type O blood. Suppose these parents have 5 children. Let X
= the number of children with type O.
2. Shuffle a deck of cards. Turn over the first 10 cards, one at a time. Let Y=
the number of Aces you observe.
3. Shuffle a deck of cards. Turn over the top card. Put the card back in the
deck, and shuffle again. Repeat this process until you get an ace. Let W=
the number of cards required.
Solutions to 3 Scenarios:
Genetics say that children receive genes from each of their parents
independently. Each child of a particular pair of parents have a
probability 0.25 of having type O blood. Suppose these parents have 5
children. Let X = the number of children with type O.
Binary? “success” = has type O blood, “failure” = does
not have type O
Independent? – children inherit genes are
independent of each parent.
Number? There are n=5 number of trials
Success? The probability of a success is p=0.25
This is a binomial setting, n = 5 and p = 0.25
Solutions to 3 Scenarios:
Shuffle a deck of cards. Turn over the first 10 cards, one at a time. Let Y=
the number of Aces you observe.
Binary? “success” = is an ace, “failure” = is not an ace
Independent? – if the first card that gets turned over is an Ace then there is
less of a probability to get another ace, not independent
This is not a binomial setting since they are not independent.
Shuffle a deck of cards. Turn over the top card. Put the card back in the deck, and
shuffle again. Repeat this process until you get an ace. Let W= the number of
cards required.
Binary? “success” = is an ace, “failure” = is not an ace
Independent? –since the card is being replaced, the probability is not affected.
Number? There has not been a fix number of trials in advance. You could get
an ace the first time, or many times later.
This is not a binomial setting since there was not a fixed number
of trials.
Check Your Understanding
• Determine whether the given random variables has a
binomial distribution. Justify your answer.
1. Shuffle a deck of cards. Turn over the top card. Put
the card back in the deck, and shuffle again. Repeat
this process 10 times. Let X = the number of aces
you observe.
2. Chose three students at random from your class,
without replacement. let Y = the number who are
over 6 feet tall.
3. Flip a coin. If heads, roll a 6-sided die. If its tails, roll
an 8-sided die. Repeat this process 5 times. Let W=
the number of 5’s you roll.
Binomial Coefficient
Binomial Coefficient: With TI-83
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,
• With our formula in hand, we can now
calculate any binomial probability!
• Lets take a look at those 5 kids and blood
type O!
Inheriting Blood Type
Each child of a particular pair of parents has
probability 0.25 of having type O blood.
Suppose the parents have 5 children.
a) Find the probability that exactly 3 of the
children have type O blood.
b) Should the parents be surprised if more than
3 of their children have type O blood? Justify
your answer.
Binomial Probability with TI-83
Binomial Probability with TI-83
For example we want to find probability of exactly 3
children with blood type O.
P(X = 3), n = 5, p =0.25, k=3
TI-83:
• 2nd > VARS >binompdf(> fill in the
information> binompdf(5,0.25,3)
TI-89:
In the CATALOG under Flash Apps
What if I wanted to find out P(X>3)???
What if I wanted to find out
P(X>3)???
Check Your Understanding
• To introduce her class to binomial distributions, Mrs. Desai
gives a 10-item, multiple choice quiz. The catch is, students
must simply guess an answer (A through E) for each question.
Mrs. Desai uses her computer’s random number generator to
produce the answer key, so that each possible answer as an
equal chance to be chosen. Patti is one of the students in this
class. Let X= the number of Patti’s correct guesses
1) Show that X is a binomial random variable.
2) Find P(X=3), explain what this result means.
3) To get a passing score on the quiz, a student must guess
correctly at least 6 times. Would you be surprised if Patti
earned a passing score? Compute an appropriate probability
to support your answer.
Check Your Understanding
1) Show that X is a binomial random variable.
1) Binary?
Independent?
Number?
Success?
2) Find P(X=3), explain what this result means.
1) Binompdf(10, .2, 3) = .2013. there is a 20.13% chance that
Patti will answer exactly 3 questions correctly.
3) To get a passing score on the quiz, a student must
guess correctly at least 6 times. Would you be
surprised if Patti earned a passing score? Compute
an appropriate probability to support your answer.
1) 1 – binomcdf(10, .2, 5) = .0064, since there is only a .64%
chance that a student will pass, we would be quite
surprised if Patti was able to pass.
Mean and Standard Deviation of
Binomial Dist.
“10% condition” sampling w/o replacement
Geometric Random Variables
• In a binomial setting, the number of trials n
is fixed in advance, and the binomial
random variable X counts the number of
successes. The possible values of X are
0,1,2,….,n. In other situations, the goal is
to repeat a chance process until success
occurs.
Consider this:
• Roll a pair of dice until you get doubles.
• In basketball, attempt a three-point shot until you
make one.
• Keep placing $1 bet on the number 15 in roulette
until you win.
These are all examples of a geometric setting. Although the
number of trials isn’t fixed in advance, the trials are independent
and the probability of success remains constant.
Geometric Random Variables and
Geometric Distribution
• The number of trials Y that it takes to get a
success in geometric setting is a
geometric random variable. The probability
distribution of Y is a geometric distribution
with parameters p, the probability of a
success on any trial. The possible values
of Y are 1,2,3…
Requirements to be Geometric- B.I.T.S
• A geometric setting arises when we preform 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:
• Binary? The possible outcomes of each trial can be classified as
“success” or “failure”
• 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
• Trials? The goal is the count the number of trials until the first
success occurs.
• Success? On each trial, the probability p of success must be the
same.
The Birth Day Game
• Your teacher is planning to give you 10 problems for homework. As
an alternative, you can agree to play the Birth Day Game. Here’s
how it works. A student will be selected at random from your class
and asked to guess the day of the week (for instance, Thursday) on
which one of your teacher's friends was born. If the student guesses
correctly, then the class will have only one homework problem. If the
student guesses the wrong day of the week, your teacher will once
again select a student from the class at random. The chosen student
will try to guess the day of the week on which a different one of your
teacher’s friends was born. If this student gets it right, the class will
have two homework problems. The game continues until a student
correctly guesses the day on which one of your teacher's many
friends was born. Your teacher will assign a number of homework
problems that is equal to the total number of guesses made by
members of your class. Are you ready to play the Birth Day Game?
Birthday Game Geometric?
Binary? “success” = correct guess, “failure” = incorrect
guess
Independent? – student guess has no influence on
another guess, since it’s a different person’s birthday.
Trials? We are counting the number of trials up and
including the first correct guess.
Success? The probability of a success is p=1/7
This is a geometric setting
Geometric Probability
The Birth Day Game
Geometric Probability with TI-83
Geometric Probability with TI-83
For example we want to find probability that the class
gets exactly 10 homework problems
P(Y = 10), p =1/7, k=10
TI-83:
• 2nd > VARS >geometpdf(> fill in the
information> geometpdf(1/7, 10)
TI-89:
In the CATALOG under Flash Apps
What if I wanted to find out P(Y<10)???
Mean of Geometric Dist.
Objectives
1. Binomial Random Variables and Binomial
Distribution
1.
2.
3.
4.
5.
Requirements to be Binomial- B.I.N.S
Binomial Coefficient: by formula and TI-83
Binomial Probability: by formula and TI-83
Mean and Standard Deviation
“10% condition” sampling w/o replacement
2. Geometric Random Variables and Geometric
Distribution
1. Requirements to be Geometric- B.I.T.S
2. Geometric Probability: by formula and TI-83
3. Mean (expected value)
Test Results!
•
•
•
•
•
•
Grade:
Amount: Marginal %
……A......……....6.……….31%
…….B…………...5……...26% 73% Passed
…….C…………..3..……...16%
…….D…………..3.……...16%
…….F…..............2………..11% 27% Failed
• Mean: 79% Max: 100% Min: 50% No Outliers
Tracking AP Stats
• 2014-2015 (WHS)
•
•
•
•
•
•
Ch. 1 Test Ch. 2 Test
A5
A5
B5
B6
C6
C4
D2
D1
F1
F2
Ch. 3 Test Ch. 4 Test Ch. 5 Test
A3
A1
A6
B5
B9
B5
C6
C3
C3
D2
D5
D3
F2
F1
F2
Homework
Worksheet