Transcript X - cloudfront.net

Probability
Two-way tables, discrete &
continuous random variables; OLI CIS
Module 15 to Module 17
Probability...
• Probability calculations are the basis for
inference (making decisions about a
population based on a sample).
• What we learn in about probability will help
us describe statistics from random samples &
randomized comparative experiments later in
the course.
1-in-6 game…
As a special promotion for its 20-ounce bottles of soda, a soft
drink company printed a message on the inside of each bottle
cap. Some of the caps said, “Please try again!” while others
said, “You’re a winner!” The company advertised the
promotion with the slogan “1 in 6 wins a prize.” The prize is a
free 20-ounce bottle of soda, which comes out of the store
owner’s profits.
Seven friends each buy one 20-ounce bottle at a local
convenience store. The store clerk is surprised when three of
them win a prize. The store owner is concerned about losing
money from giving away too many free sodas. She wonders if
this group of friends is just lucky or if the company’s 1-in-6
claim is inaccurate. In this Activity, you and your classmates
will perform a simulation to help answer this question.
1-in-6 game…
For now, let’s assume that the company is telling
the truth, and that every 20-ounce bottle of
soda it fills has a 1-in-6 chance of getting a cap
that says, “You’re a winner!” We can model the
status of an individual bottle with a six-sided
die: let 1 through 5 represent “Please try again!”
and 6 represent “You’re a winner!”
1-in-6 game…
1. Roll your die seven times to imitate the process
of the seven friends buying their sodas. How many
of them won a prize? Repeat 3 times.
2. Plot your three results on the board to create a
class dot plot displaying the number of prize
winners we got in Step 1 on the graph.
3. What percent of the time did the friends come
away with three or more prizes, just by
chance? Does it seem plausible that the company is
telling the truth or did the friends just get
lucky? Explain.
Whose book is this?
Suppose that four friends (including Ariana Grande) get together to
study at a doughnut shop for their next test in high school statistics.
When they leave their table to go get a doughnut, the doughnut shop
owner decides to mess with them (you know… because of Ariana’s
doughnut scandal) and makes a tower using their textbooks.
Unfortunately, none of the students wrote their name in their book, so
when they leave the doughnut shop, each student takes one of the
books at random.
When the students return the books at the end of the year and the
clerk scans their barcodes, the students are surprised to learn that
none of the four had their own book. How likely is it that none of the
four students ended up with the correct book? … simulation time! 
On four equally-sized slips of paper, write “Student 1,” “Student 2,” “Student
3,” and “Student 4.” Likewise, on four equally-sized slips of paper, write
“Book 1,” “Book 2,” “Book 3,” and “Book 4.”
Place the four papers with the student numbers on your desk. Then shuffle
the papers with book numbers and randomly place one paper on each
‘student.” If the book number matches the student number, this represents a
student choosing his own book from the tower of textbooks.
Count the number of students who get the correct book. Repeat this process
three times. Then plot your results on the board to create a class dot plot.
How likely is it for none of the students to end up with their own book?
What if we were to do this entire simulation again. Would you expect to get
the same exact results? Why or why not?
Investigating Randomness… & More
Simulation
• Pretend that you are flipping a fair coin. Without actually
flipping a coin, imagine the first toss. Write down the
result you see in your mind, heads (H) or tails (T), below.
• Imagine a second coin flip. Write down the result below.
• Keep doing this until you have recorded the results of 25
imaginary flips. Write all 25 of your results in groups of 5
to make them easier to read, like this: HTHTH TTHHT, etc.
Investigating Randomness… & More
Simulation…
• A run is a repetition of the same result. In the
previous example, there is a run of two tails
followed by a run of two heads in the first 10 coin
flips. Read through your 25 imagined coin flips
that you wrote above and find the longest run
(doesn’t matter if it was heads or tails; just your
longest run).
• On the board, plot the length of the longest run
you you got (within your 25 values) to create a
class dot plot.
Investigating Randomness… & More
Simulation
• Now, use a random digits table, technology, or
a coin to generate a similar list of 25 coin flips.
Find the longest run that you have.
• Now lets create another dot plot with this
new data from the class. Plot the length of
the longest run you got above.
Randomness…
• The idea of probability is that randomness is
predictable in the long run. Unfortunately, our
intuition about randomness tries to tell us that
random phenomena should also be predictable
in the short run.
• Probability Applet (www.whfreeman.com/tps5e)
Random Phenomenon...
We call an event ‘random’ if individual
outcomes are uncertain but there is
nonetheless a regular distribution of
outcomes in a large number of repetitions.
Big Idea
Chance behavior (random phenomenon) is unpredictable
in the short run, but has a regular and predictable
pattern in the long run.
Individual outcomes are uncertain; but a regular
distribution of outcomes emerges in a large number of
repetitions.
Probability of any outcome of random phenomenon is
the proportion of times an outcome would occur in a
very long series of repetitions. Probability is a longterm relative frequency (simulations very helpful).
Probability vs. Odds
• Probability =
successes
total
successes
• Odds =  failures

Careful...
• It makes no sense to discuss the probability of an
event that has already occurred.
• Meaningless to ask what the probability is of an
already-flipped coin being a tail. It’s already been
decided.
• Probability: future event
• Statistics: past event
Definition: Simulation is...
• the imitation of chance behavior, based on a
model that accurately reflects the
phenomenon under consideration.
• Examples include...
Simulation...
• Why would we want to simulate a situation
(rather than carry the event out in reality)?
• Discuss with a partner for one minute.
Simulation… model must
match situation...
• What model could we use to simulate the
probability of a soon-to-be new-born baby
being a girl or a boy?
What couldn’t be used as a model to simulate
this situation?
Discuss for one minute.
Simulation...
• ... can be an effective tool/method for finding
the likelihood of complex results IF you have a
trustworthy model.
• If not (if model does not correctly describe the
random phenomenon), probabilities derived
from model will also be incorrect/worthless.
Probability Rules ...
1. All probabilities are values between 0 & 1
Consider event A: 0  P(A)  1
2. Sum of probabilities of all outcomes = 1
S sample
P(S) = 1
 space
Momentary detour...
Examples of disjoint/mutually exclusive events
include:
•
•
•
•
miss a bus; catch a bus
play chess; sleep
turn left; turn right
sit down; stand up
Non-examples of disjoint/mutually exclusive
events include:
• listen to music; do homework
• sleep; dream
Mutually Exclusive/Disjoint events
are...
• Events that cannot happen simultaneously
• Other examples of mutually exclusive/disjoint
events?
Another brief detour… “union”
* The union of any collection of events is the
event that at least one of the collection occur.
* Symbol “U”
* P(A or B or C) = P(A U B U C)
Back to the Probability Rules ...
3. If 2 events have no outcomes in common
(disjoint/mutually exclusive) then the probability
of one or the other occurring is the sum of their
individual probabilities.
P(A or B) = P(A) + P(B)
(Addition Rule for Disjoint Events)
Example: P (rolling a 2 or rolling an odd)
Non-example: P (rolling a 4 or rolling an even)
…more examples
P (A or B or C) = P(A U B U C)
= P (A) + P (B) + P (C) only if events are disjoint
A: freshman
B: sophomore
C: junior
D: senior
P(A) = 0.30
P(B) = 0.35
P(C) = 0.20
P(D) = 0.15
All disjoint events.
P(B U C) =
P(A U D) =
P (A U B U C U D) =
Probability Rules ...
4. Probability that an event does not occur is one minus
the probability that the event will occur (complement
rule)
P(Ac)= 1 - P(A)
Example: P (person has brown hair) = 0.53
So, P (person does not have brown hair) = 1 – 0.53 = 0.47
What would AUA c = ?
What would A Ac = ?
Probability Rules ....
... one more probability rule later... Stay tuned ...
Probability Rules Practice
Distance learning courses are rapidly gaining popularity among
college students. The probability of any age group is just the
proportion of all distance learners in that age group. Here is the
probability model:
Are rules 1 & 2 satisfied above?
Are the above groups mutually exclusive events? Why or why not?
P ( 18-23 yr or 30-39 yr) =
P (not being in 18-23 yr category) =
P (24-29 yr & 30-39 yr) =
Caution...
Be careful to apply the addition rule only to
disjoint/mutually exclusive events
P (queen or heart) =
4/52 + 13/52 (??) .... not disjoint... this
probability rule would not be correct in this
case; more on this in a minute...
Review/Preview...
Mutually Exclusive/Disjoint
• sleeping; playing chess
• walking; riding a bike
Overlapping Events (not mutually exclusive)
• roll an even; roll a prime
• select 12th grader; select athlete
• choose hard-cover book; choose fiction
What if ...
• What if events are not disjoint/mutually
exclusive? i.e., they can occur simultaneously
(overlapping events)
• How do we calculate P(A or B)?
Data Collection Time…
Which do you use?
Facebook
Twitter
No Twitter
Total
No Facebook
Total
More Two-Way Table Practice..
• Go to my website, Math 140 data, copy &
paste hair color & need glasses/contacts into
StatCrunch
• Create a two-way table in StatCrunch
• Stat, Tables, Contingency, With Data
General Addition Rule (disjoint or overlapping)
P (A or B) = P (A) + P (B) – P (A and B)
P (A U B) = P (A) + P (B) – P (A∩ B)
Pierced ears, anyone?
Find the probability that a given student:
• has pierced ears
• is a male
• is male and has pierced ears
• is male or has pierced ears
Morale of the story?
Be careful to apply the addition rule for mutually
exclusive events only to disjoint/mutually
exclusive events
P (queen or heart) =
4/52 + 13/52 .... not disjoint... counted queen
of hearts twice
P (queen or heart) = 4/52 + 13/52 – 1/52
(think of a Venn diagram; overlap)
Venn Diagrams…
(a) Event A and 𝐴𝑐
(b) A, B mutually exclusive/disjoint
Venn Diagrams
(a) Intersection of A & B (and)
(b) Union of A, B (or)
Venn Diagrams with our Facebook &
Twitter Class Data...
Facebook
Twitter
Conditional Probability...
Remember... Probability assigned to an event can
change if we know that some other event has
occurred (“given”)
Conditional Probability...
P (A | B) is read “the probability of A given B”
P (female) =
versus
P (female | 15-17 years) =
Conditional Probability... caution
P (male | 18-24 yr) =
P (18-24 yr | male) =
Formula…
To find the conditional probability P (A | B)
𝑃(𝐴 ∩ 𝐵)
𝑃 𝐴𝐵 =
𝑃(𝐵)
The conditional probability P (B | A) is given by
𝑃(𝐵 ∩ 𝐴)
𝑃 𝐵𝐴 =
𝑃(𝐴)
General Multiplication Rule for Any Two Events
The joint probability that events A and B both
happen is P (A ∩ B) = P (A) P (B|A)
P (female and 15-17yr) =
89/16,639
P(A ∩ B) = P(A) P(B|A)
A = female
B = 15-17 years
= (9,321/16,639) x (89/9,321)
= 89/16,639 ✓
Facebook & Twitter... Conditional
probabilities...
• Table
• Venn Diagram
Tree diagram…
About 27% of adult Internet users are 18 to 29
years old, another 45% are 30 to 49 years
old, and the remaining 28% are 50 and over.
The Pew Internet and American Life Project finds
that 70% of Internet users aged 18 to 29 have
visited a video-sharing site, along with 51% of
those aged 30 to 49 and 26% of those 50 or
older.
Let’s try a tree diagram now...
Yes
Are you
married?
No
Have a FT
job
Do not have
a FT job
Have a FT
job
Do not have
a FT job
Review/Preview ....
Two events A & B are independent if knowing that one occurs
does not change the probability that the other occurs.
Examples:
- Roll a die twice. What I roll the first time does not change
the probability of what I will roll the second time.
- Win at chess; win the lottery
- Student on debate team; student on swim team
So, if events A and B are independent, then P (A|B) = P(A) and
likewise P (B|A) = P (B).
Are these events independent?
Event A: Honors student
Honors
Student
Basketball
Event B: Basketball
Non-Honors
Student
Total
450
1,800
1,500
6,000
Non-BB Player
Total
P (B) =
1800/6000 = 0.3
P (B|A) =
450/1500 = 0.3
P (A) =
1500/6000 = 0.25
P (A|B) =
450/1,800 = 0.25
So remember… Independent Events
Two events A and B that both have positive
probability are independent if
P (B|A) = P (B)
or
P (A|B) = P (A)
Last Probability Rule ....
If events A & B are independent, then
P (A & B) = P (A) P(B)
(this is the multiplication rule for independent events)
Example: Consider the following probabilities.
P( student has 4.0 GPA) = 0.15
P(student miss bus) = 0.30
If these two events are independent, then P (4.0 GPA &
missing bus) = (0.15)(0.30) = 0.045
Caution...
P ( heart & 3) -- without replacement – is not
independent; knowing outcome of first pick changes
outcome of second pick
Independent is not mutually exclusive/disjoint.
Mutually exclusive/disjoint is not independent.
(remember... mutually exclusive/disjoint events can’t
happen at same time; independent events can)
Draw Venn diagrams for...
•
•
•
•
Mutually exclusive/disjoint events
Independent events
Dependent events
Overlapping events
Let’s Make a Deal!
• Go to New York Times website and do
simulation
• Discuss which strategy is best.
Top 3 Concepts on Probability...
• With a partner, come to a consensus about
the top 3 concepts on probability
• Write 2-3 sentences on each concept
explaining what the concept is and why you
believe it should be on your ‘top 3’ list.
• Print and turn in.
Continuous & Discrete Random
Variables: Modeling Random
Events... Normal & Binomial Models
OLI CIS Modules 16 & 17
Models...
• Help us predict what is likely to happen
• Remember LSRLs (a model for linear, bi-variate data); not
exact (some points above line, some below); but best
model we have
• Two more models we will discuss are for Normal (or ≈
Normal) and for Binomial distributions
• Both of these describe/model numeric, uni-variate data
• Again, models are not perfect... But in many cases, they are
good, effective, or “good enough”
Two types of numeric data...
• Discrete random variables/data (we will use
binomial distribution with discrete data)
• Continuous random variables/data (we will
often use Normal distribution with continuous
data)
How are these the same? How are
they different? Don’t look ahead!
Discrete Random Variable Examples
Continuous Random Variable
Examples
Stars in the sky
Hours of sleep you got last night
Number of burgers you buy at In-NOut
Gallons of gas you put in your car last
time you filled up
How many coins you have in your
pocket
Your current weight
Times you have gone bowling this
month
Burgers you ate at In-N-Out
Number of pets your parents have
A friend’s age
Number of social media platforms you A co-worker’s height
use regularly
Olympic athlete’s ranking in his/her
sport/event
Olympic athlete’s best time in his/her
sport/event
Discrete Random Variables
• Discrete random variables have a “countable”
number of possible positive outcomes and
must satisfy two requirements. Note:
‘countable’ is not the same as finite.
• (1) Every probability is a # between 0 and 1;
• (2) The sum of the probabilities is 1
World-Wide 2015 High School AP Statistics Score
Distribution
• This a discrete random variable probability
distribution? Why? Explain.
1
2
3
4
5
.238
.189
.252
.189
.132
Discrete Random Variables...
• Discuss other examples of discrete random
variables. 1 minute.
Other Examples of Discrete Random
Variables...
• Number of times people have seen Shawn
Mendes in concert
• Number of gifts we get on our birthday
• Number of taco’s sold at Taco Bell per day
All are whole, countable numbers
Non-examples of Discrete Random
Variables...
• Your height
• Weight of a candy bar
• Time it takes to run a mile
Continuous Random Variables...
• are usually measurements
• heights, weights, time
• amount of sugar in a granny smith apple, time
to finish the New York marathon, height of Mt.
Whitney
How can we distinguish between
continuous and discrete?
• Discuss in your groups for a few minutes.
How can we distinguish between
continuous and discrete?
• Discuss in your groups for a few minutes.
• Ask yourself ‘How many? How much? Are you
sure?’
• For example, # of children, pounds of Captain Crunch
produced each year, # of skittles, ounces in a bag of
skittles
Continuous Random Variables ...
• take on all values in an interval of numbers
• probability distribution is described by a density curve
• probability of event is area under the density curve and above
the values of X that make up the event
• total area under (density) curve = 1
Continuous Random Variables...
• Probability distribution is area under the
density curve, within an interval, above x-axis
Continuous Random Variables...
Random Variables...
Consider a six-sided die... What is the probability...
P ( roll less than or equal to a 2) is
P ( roll less than a 2) is
Different probabilities; discrete random variable
Note: possible outcomes are 1, 2, 3, 4, 5, 6; but probabilities of
those outcomes are (often) fractions/decimals
Continuous random variables ...
• All continuous random variables assign
probabilities to intervals
• All continuous random
variables assign a probability of zero to every
individual outcome. Why?
Continuous Random Variables...
• There is no area under a vertical line (sketch)
• Consider...
0.7900 to 0.8100
P = 0.02
0.7990 to 0.8010
P = 0.002
0.7999 to 0.8001
P = 0.0002
P (an exact value –vs. an interval--) = 0
Density Curves & Continuous RV’s..
• Can use ANY density curve to assign probabilities/model
continuous distributions/RV’s; many models
• Most familiar density curves are the Normal (bell) density
curves
• Based on Empirical Rule, 68-95-99.7, symmetric, uni-modal,
chapter 3)
• Many distributions/events are considered Normal & can be
modeled by Normal density curves, such as ... cholesterol
levels in boys, heights of 3-year-old females, Tiger Woods’
distance golf ball travels on driving range, basic skills
vocabulary test scores for 7th graders, etc.
Mean & Standard Deviation of Normal
Distributions...
• μx for continuous random variables lies at the
center of a symmetrical (or fairly symmetrical)
density curve (Normal or approximately Normal)
• N (μ, σ)
• Remember... μ and σ are population parameters;
and s are sample statistics.
x
• Calculating σ and/or σ2 for continuous random
variables…. beyond the scope of this course… will
be given this information if needed
Normal distributions/density curves...
Four possibilities that we need to
know how to calculate...
• When calculating probabilities for Normal
distributions, there are four possibilities that we
might be asked to calculate.
• Let’s draw some pictures and match up some
questions to each drawing
• Remember for continuous random variables (like
Normal distributions) there is no difference
between ‘less than’ and ‘less than or equal to;’
likewise no difference between ‘greater than’ or
‘greater than or equal to.’
Female heights... N(64.5, 2.5)
Female heights... N(64.5, 2.5)
• Calculate the probability that a randomly chosen female is:
– shorter than 62 inches
– no more than 64.5 inches
– taller than 57 inches
– 67 inches or taller
– between 62 inches and 67 inches
– between 57 inches and 72 inches
– Shorter than 57 inches or taller than 72 inches
Female heights... N(64.5, 2.5)
• Now, what about these? Can we calculate the probability that a
randomly chosen female is:
– shorter than 63 inches
– no more than 61 inches
– taller than 66 inches
– 59 inches or taller
– between 63.5 inches and 64 inches
– between 5 inches and 72 inches
– Shorter than 58 inches or taller than 70 inches
Female heights... N(64.5, 2.5)
• Sometimes we want to turn this around... Sometimes
we are given a probability & we need to find the value
that corresponds to that probability.
• What is the height of a randomly chosen woman if she
is in the 20th percentile? Let’s draw a picture...
• What is the height of a randomly chosen woman if she
is in the 85th percentile?
Normal model...
• Very helpful, but one size does not fit all
• Good first choice if data is continuous,
uni-modal, symmetric, 68-95-99.7
Another important, “special” type of
distribution...
• If certain criteria is met, easier to calculate probabilities in
specific situations
• Next types of distributions we will examine are situations
where there are only two outcomes
• Win or lose; make a basket or not; boy or girl ...
Discuss situations where there are
only two outcomes...
•
•
•
•
•
•
Yes or no
Open or closed
Patient has a disease or doesn’t
Something is alive or dead
Person has a job or doesn’t
A part is defective or not
that is what this section is all about...
• ... a class of distributions that are concerned about events
that can only have 2 outcomes
• The Binomial Distribution
• Binomial Distributions are a special type of discrete random
variables
Binary; Independent; fixed Number;
probability of Successes
The Binomial setting is:
1. Each observation is either a success or a
failure (i.e., it’s binary)
2. All n observations are independent
3. Fixed # (n) of observations
4. Probability of success, p, is the same for each
observation
“BINS”
Binomial Distribution: practice…
I roll a die 3 times and observe each roll to see if it is even or
odd. Is:
1. each observation is either a success or a failure?
2. all n observations are independent?
3. fixed # (n) of observations?
4. probability of success, p, is the same for each observation ?
BINS
Binomial Distribution
• If BINS is satisfied, then the distribution can be described as B
(n, p)
• B
binomial
• n
the fixed number of observations
• p
probability of success
• Note: This is a discrete probability distribution.
• Remember N (μ, σ)… is that discrete??
Binomial Distribution
• Most important: being able to recognize situations and then
use appropriate tools for that situation
• Let’s practice...
Are these binomial distributions? Why
or why not?
• Toss a coin 20 times to see how many tails occur.
• Asking people if they watch ABC news
• Rolling a die until a 6 appears
• Asking 20 people how old they are
• Drawing 5 cards from a deck for a poker hand
Side note: Binomial Distribution... Is the
situation ‘independent enough’?
An engineer chooses a SRS of 20 switches from a
shipment of 10,000 switches. Suppose (unknown
to the engineer) 12% of switches in the shipment
are bad.
Not quite a binomial setting. Why?
For practical purposes, this behaves like a binomial
setting; ‘close enough’ to independence; as long
as sample size is small compared to population.
Rule of thumb: sample ≤ 10% of population size
Practice...
Each child born to a particular set of parents has
probability 0.25 of having blood type O. If these
parents have 5 children, what is the probability
that exactly 2 of them have type O blood?
Binomial setting? Check for BINS.
p = 0.25
n=5
X=2
Practice...
Each child born to a particular set of parents has
probability 0.25 of having blood type O. If these
parents have 5 children, what is the probability
that exactly 2 of them have type O blood?
Binomial setting? Check for BINS.
p = 0.25
n=5
= 0.2636; context, always!
X=2
Practice...
Each child born to a particular set of parents has
probability 0.25 of having blood type O. If these
parents have 5 children, what is the probability
that exactly 4 of them have type O blood?
Binomial setting? Check for BINS.
p = 0.25
n=5
X=4
Practice...
Each child born to a particular set of parents has
probability 0.25 of having blood type O. If
these parents have 5 children, what is the
probability that exactly 4 of them have type O
blood?
Binomial setting? Check for BINS.
p = 0.25
n=5
X=4
= 0.0146; context, always
Practice...
Each child born to a particular set of parents has
probability 0.25 of having blood type O. If these
parents have 5 children, what is the probability
that exactly 1 of them have type O blood?
Binomial setting? Check for BINS.
p = 0.25
n=5
X=1
Practice...
Each child born to a particular set of parents has
probability 0.25 of having blood type O. If these
parents have 5 children, what is the probability
that exactly 1 of them have type O blood?
Binomial setting? Check for BINS.
p = 0.25
n=5
X=1
= 0.3955; context, always
Practice...
Each child born to a particular set of parents has
probability 0.25 of having blood type O. If these
parents have 5 children, what is the probability
that at most 2 of them have type O blood?
Binomial setting? Check for BINS.
p = 0.25
n=5
X=2
Practice...
Each child born to a particular set of parents has
probability 0.25 of having blood type O. If these
parents have 5 children, what is the probability
that at most 2 of them have type O blood?
Binomial setting? Check for BINS.
p = 0.25
n=5
X=2
= 0.8965; context, always
Practice...
Each child born to a particular set of parents has
probability 0.25 of having blood type O. If
these parents have 5 children, what is the
probability that at most 4 of them have type O
blood?
Binomial setting? Check for BINS.
p = 0.25
n=5
X=4
Practice...
Each child born to a particular set of parents has
probability 0.25 of having blood type O. If these
parents have 5 children, what is the probability
that at most 4 of them have type O blood?
Binomial setting? Check for BINS.
p = 0.25
n=5
X=4
= 0.9990, context, always
Practice...
Each child born to a particular set of parents has
probability 0.25 of having blood type O. If these
parents have 5 children, what is the probability
that at most 1 of them have type O blood?
Binomial setting? Check for BINS.
p = 0.25
n=5
X=1
Practice...
Each child born to a particular set of parents has
probability 0.25 of having blood type O. If these
parents have 5 children, what is the probability
that at most 1 of them have type O blood?
Binomial setting? Check for BINS.
p = 0.25
n=5
X=1
= 0.6328, context, always
Practice...
... probability 0.25 of having blood type O. If these
parents have 5 children, what is the probability
that at least 2 (meaning 2, 3, 4, or 5) of them
have type O blood?
p = 0.25
n=5
X = 2, 3, 4, or 5
Practice...
... probability 0.25 of having blood type O. If these
parents have 5 children, what is the probability
that at least 2 (meaning 2, 3, 4, or 5) of them
have type O blood?
p = 0.25
n=5
X = 2, 3, 4, or 5
0.3672; context, always
Practice...
... probability 0.25 of having blood type O. If these
parents have 5 children, what is the probability
that at least 3 (meaning 3, 4, or 5) of them
have type O blood?
p = 0.25
n=5
X = 3, 4, or 5
Practice...
... probability 0.25 of having blood type O. If these
parents have 5 children, what is the probability
that at least 3 (meaning 3, 4, or 5) of them
have type O blood?
p = 0.25
n=5
X = 3, 4, or 5
0.1035; context, always
Practice...
... probability 0.25 of having blood type O. If
these parents have 5 children, what is the
probability that more than 3 (meaning 4 or or
5) of them have type O blood?
p = 0.25
n=5
X = 4 or 5
Practice...
... probability 0.25 of having blood type O. If these
parents have 5 children, what is the probability
that more than 3 (meaning 4 or or 5) of them
have type O blood?
p = 0.25
n=5
X = 4 or 5
0.0156; context, always
Practice...
... probability 0.25 of having blood type O. If these
parents have 5 children, what is the probability
that … of them have type O blood?
p = 0.25
n=5
a) …at most 3 of them…
b) …at least 4 of them…
c) …more than 1 of them…
d) …exceeds 3 of them…
e) … below 2 of them…
f) … 0 of them…
More practice with binomials?
• See my website, COC Math 140 Binomial
Practice
Binomial Distribution: Mean and
Standard Deviation
• If a basketball player makes 75% (“p” of her
free throws, what do you think the mean
number of baskets made will be in 12 tries?
Binomial Distribution: Mean and
Standard Deviation
• If a basketball player makes 75% (“p” of her
free throws, what do you think the mean
number of baskets made will be in 12 tries?
• (0.75) (12) = 9; we expect she should make 9
baskets in 12 tries.
• Her mean number of baskets made should be
9.
• We expect her x = 9 baskets; E(x) = 9
Binomial Mean & Standard Deviation
• If a count X is a binomial distribution with number of
observations n and probability of success p, then
• μ= np
and
σ=
np(1  p)
• Only for use with binomial distributions; remember criteria...
BINS

Practice...
• If a basketball player makes 75% (“p”) of her free throws, we
expect her to make 9 baskets in 12 tries.
• What is the SD of this distribution?
Practice...
• If a basketball player makes 75% (“p”) of her free throws, we
expect her to make 9 baskets in 12 tries.
• What is the SD of this distribution?
• SD = np(1  p) =
= 1.5


(12)(0.75)(1 0.75)
Practice...
• Each child born to a particular set of parents has
probability 0.25 of having blood type O. If these
parents have 5 children, what is the mean and
standard deviation for this distribution?
Practice...
• Each child born to a particular set of parents has
probability 0.25 of having blood type O. If these parents
have 5 children, what is the mean and standard
deviation for this distribution?
• mean = np = (5)(.25) = 1.25; the family should expect to
have 1.25 children with type O blood
• SD = np(1  p) =
= 0.9682
(5)(0.25)(1 0.25)
Remember & Caution...
• Binomial distribution is a special case of a probability
distribution for a discrete random variable
• All binomials distributions are discrete random variable
distributions BUT not all discrete random variable
distributions are binomial distributions
• Don’t assume; don’t apply binomial tools to all discrete
RV distributions; must meet binomial distribution
criteria (BINS)
Next test... Exam #2 ...