Announcements
Announcements
Lecture 7: Geometric & Binomial distributions
Due: HW 2 at the beginning of class on Thursday:
Clarification on Exercise 3.4 parts (c) and (d): Mary and Leo’s
percentile mean the proportion of people whose finishing times
are lower than theirs. Note that this doesn’t mean ”% of people
they performed better than” because in a triathlon ”performing
better” means finishing faster.
Statistics 101
Mine Çetinkaya-Rundel
February 7, 2012
Statistics 101 (Mine Çetinkaya-Rundel)
Recap
Review question
Q3: More than three-quarters of the nation’s colleges and universities now
offer online classes, and about 23% of college graduates have taken a
course online. 39% of those who have taken a course online believe that
online courses provide the same educational value as one taken in person,
a view shared by only 27% of those who have not taken an online course. At
a coffee shop you overhear a recent college graduate discussing that she
doesn’t believe that online courses provide the same educational value as
one taken in person. What’s the probability that she has taken an online
course before?
valuable
not valuable
total
Statistics 101 (Mine Çetinkaya-Rundel)
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Qnline quiz 2- Q3
Question 3:
didn’t take
online course
February 7, 2012
Recap
Qnline quiz 2- commonly missed questions
took
online course
L7: Geo & Binom distributions
Which is the correct notation for the following probability?
“At a coffee shop you overhear a recent college graduate discussing that she
doesn’t believe that online courses provide the same educational value as
one taken in person. What’s the probability that she has taken an online
course before?”
(a) P(took online course | not valuable)
(b) P(not valuable | took online course)
(c) P(valuable and took online course)
total
(d) P(took online course and not valuable)
0.23
(e) P(valuable | didn’t take online course)
1
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Statistics 101 (Mine Çetinkaya-Rundel)
L7: Geo & Binom distributions
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Geometric distribution
Bernoulli distribution
Geometric distribution
Milgram experiment
Milgram experiment (cont.)
Stanley Milgram, a Yale University
psychologist, conducted a series of
experiments on obedience to
authority starting in 1963.
These experiments measured the willingness of study
participants to obey an authority figure who instructed them to
perform acts that conflicted with their personal conscience.
Experimenter (E) orders the teacher
(T), the subject of the experiment, to
give severe electric shocks to a
learner (L) each time the learner
answers a question incorrectly.
Milgram found that about 65% of people would obey authority
and give such shocks.
Over the years, additional research suggested this number is
approximately consistent across communities and time.
The learner is actually an actor, and
the electric shocks are not real, but a
prerecorded sound is played each
time the teacher administers an
electric shock.
Statistics 101 (Mine Çetinkaya-Rundel)
Bernoulli distribution
L7: Geo & Binom distributions
Geometric distribution
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Statistics 101 (Mine Çetinkaya-Rundel)
Bernoulli distribution
L7: Geo & Binom distributions
Geometric distribution
Bernouilli random variables
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Geometric distribution
Geometric distribution
Dr. Smith wants to repeat Milgram’s experiments but she only wants to sample people until she finds someone who will not inflict a severe shock. What
is the probability that she stops after the first person?
Each person in Milgram’s experiment can be thought of as a trial.
A person is labeled a success if she refuses to administer a
severe shock, and failure if she administers such shock.
P (1st person refuses ) = 0.35
Since only 35% of people refused to administer a shock,
probability of success is p = 0.35.
... the third person?
When an individual trial has only two possible outcomes, it is
called a Bernoulli random variable.
P (1st and 2nd shock , 3rd refuses ) =
S
S
R
×
×
= 0.652 ×0.35 ≈ 0.15
0.65
0.65
0.35
... the tenth person?
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L7: Geo & Binom distributions
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Statistics 101 (Mine Çetinkaya-Rundel)
L7: Geo & Binom distributions
February 7, 2012
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Geometric distribution
Geometric distribution
Geometric distribution
Geometric distribution
Geometric distribution (cont.)
Geometric distribution describes the waiting time until a success for
independent and identically distributed (iid) Bernouilli random
variables.
Clicker question
Can we calculate the probability of rolling a 6 for the first time on the
6th roll of a die using the geometric distribution?
independence: outcomes of trials don’t affect each other
identical: the probability of success is the same for each trial
(a) no, on the roll of a die there are more than 2 possible outcomes
(b) yes, why not
Geometric probabilities
If p represents probability of success, (1 − p ) represents probability of
failure, and n represents number of independent trials
P (success on the nth trial ) = (1 − p )n−1 p
Statistics 101 (Mine Çetinkaya-Rundel)
L7: Geo & Binom distributions
Geometric distribution
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Statistics 101 (Mine Çetinkaya-Rundel)
Geometric distribution
L7: Geo & Binom distributions
Geometric distribution
Expected value
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Geometric distribution
Expected value and its variability
Mean and standard deviation of geometric distribution
How many people is Dr. Smith expected to test before finding the first
one that refuses to administer the shock?
The expected value, or the mean, of a geometric distribution is
defined as p1 .
1
1
µ= =
= 2.86
p
0.35
s
1
µ=
p
σ=
1−p
p2
Going back to Dr. Smith’s experiment:
s
σ=
She is expected to test 2.86 people before finding the first one that
refuses to administer the shock. But how can she test a non-whole
number of people?
1−p
=
p2
r
1 − 0.35
= 2.3
0.352
Dr. Smith is expected to test 2.86 people before finding the first
one that refuses to administer the shock, give or take 2.3 people.
These values only makes sense in the context of repeating the
experiment many many times.
Statistics 101 (Mine Çetinkaya-Rundel)
L7: Geo & Binom distributions
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Statistics 101 (Mine Çetinkaya-Rundel)
L7: Geo & Binom distributions
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Binomial distribution
Binomial distribution
Suppose we randomly select four individuals to participate in this experiment. What is the probability that exactly 1 of them will refuse to
administer the shock?
Scenario 2:
Scenario 3:
Scenario 4:
0.35
(A) refuse
0.65
(A) shock
0.65
(A) shock
0.65
(A) shock
0.65
(B) shock
0.35
×
(B) refuse
0.65
×
(B) shock
0.65
×
(B) shock
×
0.65
(C) shock
0.65
×
(C) shock
0.35
×
(C) refuse
0.65
×
(C) shock
×
0.65
(D) shock
0.65
×
(D) shock
0.65
×
(D) shock
0.35
×
(D) refuse
×
# of scenarios × P (single scenario )
= 0.0961
= 0.0961
# of scenarios: there is a less tedious way to figure this out,
we’ll get to that shortly...
= 0.0961
P (single scenario ) = p k (1 − p )(n−k )
= 0.0961
probability of success to the power of number of successes, probability of failure to the power of number of failures
The Binomial distribution describes the probability of having exactly k
successes in n independent Bernouilli trials with probability of
success p.
The probability of exactly one 1 of 4 people refusing to administer the
shock is the sum of all of these probabilities.
0.0961 + 0.0961 + 0.0961 + 0.0961 = 4 × 0.0961 = 0.3844
Statistics 101 (Mine Çetinkaya-Rundel)
L7: Geo & Binom distributions
Binomial distribution
Binomial distribution
The question from the prior slide asked for the probability of given
number of successes, k , in a given number of trials, n, (k = 1
success in n = 4 trials), and we calculated this probability as
Let’s call these people Allen (A), Brittany (B), Caroline (C), and
Damian (D). Each one of the four scenarios below will satisfy the
condition of “exactly 1 of them refuses to administer the shock”:
Scenario 1:
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The binomial distribution
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Statistics 101 (Mine Çetinkaya-Rundel)
L7: Geo & Binom distributions
The binomial distribution
Binomial distribution
Counting the # of scenarios
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The binomial distribution
Calculating the # of scenarios
Choose function
The choose function is useful for calculating the number of ways to
choose k successes in n trials.
Earlier we wrote out all possible scenarios that fit the condition of
exactly one person refusing to administer the shock. If n was larger
and/or k was different than 1, writing out the scenarios would get
even more tedious. For example, what if n = 9 and k = 2:
!
n
n!
=
k
k !(n − k )!
RRSSSSSSS
SRRSSSSSS
SSRRSSSSS
···
SSRSSRSSS
k = 1, n = 4:
4
k = 2, n = 9:
9
1
···
2
=
4!
1!(4−1)!
=
4×3×2×1
1×(3×2×1)
=4
=
9!
2!(9−1)!
=
9×8×7!
2×1×7!
72
2
=
= 36
SSSSSSSRR
Note: You can also use R for these calculations:
Writing out all possible scenarios is incredibly tedious and prone to
errors.
Statistics 101 (Mine Çetinkaya-Rundel)
L7: Geo & Binom distributions
February 7, 2012
> choose(9,2)
[1] 36
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L7: Geo & Binom distributions
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Binomial distribution
The binomial distribution
Binomial distribution
Properties of the choose function
Binomial distribution (cont.)
If k = 1, only 1 of the n trials result in a success, it could be the
first, the second, · · · , or the nth trial, so there are n ways this can
happen:
Binomial probabilities
If p represents probability of success, (1 − p ) represents probability of
failure, n represents number of independent trials, and k represents
number of successes
!
n
=n
1
If k = n, all n trials result in a success, and there’s only one way
this can happen:
!
n k
P (k successes in n trials ) =
p (1 − p )(n−k )
k
We can use the binomial distribution to calculate the probability of k
successes in n trials, as long as
!
n
=1
n
If k = 0, all n trials result in a failure, and there’s only one way
this can happen as well:
!
n
=1
0
Statistics 101 (Mine Çetinkaya-Rundel)
L7: Geo & Binom distributions
Binomial distribution
The binomial distribution
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1
the trials are independent
2
the number of trials, n, is fixed
3
each trial outcome can be classified as a success or a failure
4
the probability of success, p, is the same for each trial
Statistics 101 (Mine Çetinkaya-Rundel)
The binomial distribution
L7: Geo & Binom distributions
Binomial distribution
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The binomial distribution
Clicker question
Clicker question
A January 27, 2012 Gallup survey suggests that 48% of Americans
would vote for Obama over Romney if the presidential election was
held that day. Among a random sample of 10 Americans what is the
probability that exactly 8 would vote for Obama over Romney?
A January 27, 2012 Gallup survey suggests that 48% of Americans
would vote for Obama over Romney if the presidential election was
held that day. Among a random sample of 10 Americans what is the
probability that exactly 8 would vote for Obama over Romney?
(a) pretty low
(a) 0.488 × 0.522
(b) pretty high
(b)
8
× 0.488 × 0.522
10
8
2
(c) 10
8 × 0.48 × 0.52
10
(d) 8 × 0.482 × 0.528
http:// www.gallup.com/ poll/ election.aspx , February 6, 2012.
Statistics 101 (Mine Çetinkaya-Rundel)
L7: Geo & Binom distributions
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Statistics 101 (Mine Çetinkaya-Rundel)
L7: Geo & Binom distributions
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Binomial distribution
The binomial distribution
Binomial distribution
Expected value
The binomial distribution
Expected value and its variability
Mean and standard deviation of binomial distribution
A January 27, 2012 Gallup survey suggests that 48% of Americans
would vote for Obama over Romney if the presidential election was
held that day. Among a random sample of 100 people, how many
would you expect to vote for Obama?
µ = np
σ=
q
np (1 − p )
Going back to the voters:
Easy enough, 100 × 0.48 = 48.
σ=
Or more formally, µ = np = 100 × 0.48 = 48.
But this doesn’t mean in every random sample of 100 people
exactly 48 will vote for Obama. In some samples this value will
be less, and in others more. How much would we expect this
value to vary?
√
q
np (1 − p ) =
100 × 0.48 × 0.52 ≈ 5
We would expect 48 out of 100 randomly sampled voters, give or
take 5.
Note: Mean and standard deviation of a Binomial might not always be whole numbers, and that
is alright, these values represent what we would expect to see on average.
Statistics 101 (Mine Çetinkaya-Rundel)
L7: Geo & Binom distributions
Binomial distribution
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Statistics 101 (Mine Çetinkaya-Rundel)
The binomial distribution
L7: Geo & Binom distributions
Binomial distribution
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The binomial distribution
Unusual observations
Clicker question
Using the notion that observations that are more than 2 standard
deviations away from the mean are considered unusual and the mean
and the standard deviation we just computed, we can calculate a
range for how many people we should expect to find in random
samples of 100 that are planning to vote for Obama in the 2012
presidential election.
A January 2012 Gallup report suggests that 7.5% of 18-29 year old
Americans have been diagnosed with high blood pressure. Would a
random sample of 1,000 adults in this age range where only 60 of
them have high blood pressure be considered unusual?
(a) Yes
(b) No
48 ± 2 × 5 = (38, 58)
http:// www.gallup.com/ poll/ 152108/ Key-Chronic-Diseases-Decline.aspx
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L7: Geo & Binom distributions
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Statistics 101 (Mine Çetinkaya-Rundel)
L7: Geo & Binom distributions
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Binomial distribution
Normal approximation to the binomial
Binomial distribution
A recent study found that “Facebook users get more than they give”.
For example:
40% of Facebook users in our sample made a friend request, but
63% received at least one request
Users in our sample pressed the like button next to friends’
content an average of 14 times, but had their content “liked” an
average of 20 times
Normal approximation to the binomial
This study also found that approximately %25 of Facebook users are
considered power users. The same study found that the average Facebook user has 245 friends. What is the probability that the average
Facebook user with 245 friends has 70 or more friends who would be
considered power users?
We are given that n = 245, p = 0.25, and we are asked for the
probability P (K ≥ 70).
Users sent 9 personal messages, but received 12
12% of users tagged a friend in a photo, but 35% were
themselves tagged in a photo
P (X ≥ 70) = P (K = 70 or K = 71 or K = 72 or · · · or K = 245)
Any guesses for how this pattern can be explained?
= P (K = 70) + P (K = 71) + P (K = 72) + · · · + P (K = 245)
This seems like an awful lot of work...
http:// www.pewinternet.org/ Reports/ 2012/ Facebook-users/ Summary.aspx
Statistics 101 (Mine Çetinkaya-Rundel)
L7: Geo & Binom distributions
Binomial distribution
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Statistics 101 (Mine Çetinkaya-Rundel)
Normal approximation to the binomial
Binomial distribution
Normal approximation to the binomial
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Normal approximation to the binomial
P (K ≥ 70) = P (K = 70) + P (K = 71) + P (K = 72) + · · · + P (K = 245)
= P (X > 70 − 0.5)
In the case of the Facebook power users, n = 245 and p = 0.25.
σ=
February 7, 2012
Correction for the normal approximation
When the sample size is large enough, the binomial distribution with
parameters n and p can be approximated
by the normal model with
p
parameters µ = np and σ = np (1 − p ).
µ = 245 × 0.25 = 61.25
L7: Geo & Binom distributions
√
245 × 0.25 × 0.75 = 6.78
0.06
Bin(n = 245, p = 0.25) ≈ N (µ = 61.25, σ = 6.78).
Bin(245,0.25)
N(61.5,6.78)
0.05
0.06
Bin(245,0.25)
N(61.5,6.78)
0.05
0.04
0.04
We apply a 0.5
correction in order to
account for the
probability of exactly
70 “successes”.
0.03
0.03
0.02
0.02
0.01
0.01
0.00
0.00
20
40
60
80
100
20
k
40
60
80
100
k
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L7: Geo & Binom distributions
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Statistics 101 (Mine Çetinkaya-Rundel)
L7: Geo & Binom distributions
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Binomial distribution
Normal approximation to the binomial
Binomial distribution
Normal approximation to the binomial
Histograms of number of successes
Hollow histograms of samples from the binomial model where
p = 0.10 and n = 10, 30, 100, and 300.
Clicker question
What is the probability that the average Facebook user with 245 friends
has 70 or more friends who would be considered power users?
0
(a) 0.0984
(c) 0.8888
(b) 0.1112
(d) 0.9016
2
4
6
0
2
n = 10
0
5
10
4
6
8
10
n = 30
15
20
10
20
n = 100
30
40
50
n = 300
Try it yourself at http:// www.socr.ucla.edu/ htmls/ SOCR Distributions.html .
Statistics 101 (Mine Çetinkaya-Rundel)
L7: Geo & Binom distributions
Binomial distribution
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Statistics 101 (Mine Çetinkaya-Rundel)
Normal approximation to the binomial
L7: Geo & Binom distributions
Binomial distribution
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Normal approximation to the binomial
Low large is large enough?
Clicker question
Below are four pairs of Binomial distribution parameters. Which distribution can be approximated by the normal distribution? There are two
correct answers.
The sample size is considered large enough if the expected number
of successes and failures are both at least 10.
np ≥ 10
Statistics 101 (Mine Çetinkaya-Rundel)
and
n(1 − p ) ≥ 10
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(a) n = 100, p = 0.8
(c) n = 150, p = 0.05
(b) n = 25, p = 0.6
(d) n = 500, p = 0.015
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L7: Geo & Binom distributions
February 7, 2012
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