7 Making Sense of Randomness Facilitator Guide

Facilitator Guide
Unit 7
UNIT 07
Making Sense of Randomness
Facilitator Guide
ACTIVITIES
NOTE: At many points in the activities for Mathematics Illuminated, workshop
participants will be asked to explain, either verbally or in written form, the process
they use to answer the questions posed in the activities. This serves two purposes:
for the participant as a student, it helps to solidify any previously unfamiliar
concepts that are addressed; for the participant as a teacher, it helps to develop
the skill of teaching students “why,” not just “how,” when it comes to confronting
mathematical challenges.
NOTE: Instructions, answers, and explanations that are meant for the facilitator
only and not the participant are in grey boxes for easy identification.
UNIT 7
ACTIVITY
Making Sense of Randomness
FACILITATOR GUIDE
1
(30 minutes)
The following questions visit some of the important topics from Unit 7 in
more depth.
A
1. Suppose that 100 people are to flip a fair coin simultaneously. Suppose
further that the 100 people decide to form a traveling show, traveling from city
to city each day and performing their simultaneous coin flip once every night in
front of an enthralled audience. After a year, what would you expect the average
number of flips that end up “heads” to be (per night)?
Answer: 50
2. What is the probability that after 100,000 shows, at least one of the shows
involves 100 simultaneous flips that all come up “heads?”
Answer: The likelihood that a given trial of 100 flips will be all heads is × 1
2
× 21 …100 times, OR 21 100. Multiply this probability by 100,000 chances to
get about 7.9 x 10-26, so there is about one chance in 1025 of all coins coming up
heads in 100,000 tries.
()
()
()
3. Suppose that there are about 1024 grains of sand on Earth. What is more
likely, flipping 100 heads in a row, or randomly selecting a specific grain of sand
on Earth? Compare the two probabilities.
Answer: The probability of finding the grain of sand is about 1 in 1024,
whereas the probability of flipping a coin and getting 100 heads in a row is
about 1 in 1025. So, you are ten times more likely to find a specific grain of sand
on Earth than to have a coin flip come up heads 100 times in a row just once in
100,000 tries.
B
Facilitator’s note: This exercise is designed to be done in groups of three or four.
1. Say that the average height of an American male is 5’ 10”, and the average
height of an American female is 5’ 5”. Furthermore, say that the standard
deviation for men is 3”, and for women it is 2.5”. Find the normalized heights for
the people in your group. Express your answers in terms of standard deviations.
Answer: People should take their height, subtract the mean, and then divide by
the standard deviation—answers should be in units of standard deviation (SD).
Unit 7 | 1
UNIT 7
ACTIVITY
Making Sense of Randomness
FACILITATOR GUIDE
1
2. Compare your normalized height to the normalized heights of the following
famous people:
Yao Ming, U.S.A.(Professional Basketball Player): 7’ 6”
Answer: 6.7 SD
Ekatarina Gamova, (Russian volleyball player and one of the tallest female
athletes as of 2007): 6’ 9”
Answer: 6.4 SD
Leonid Stadnyk, Ukraine (Tallest living man as of 2007): 8’ 5”
Answer: 10.3 SD
De-Fen Yao, China (Tallest living woman in the world as of 2007): 7’ 8”
Answer: 10.8 SD
Robert Wadlow, U.S.A. (Tallest recorded human as of 2007): 8’ 11”
Answer: 12.3 SD
3. Discuss the various assumptions behind this activity.
Answer: This activity assumes that averages and standard deviations are the
same regardless of ethnicity, race, and socioeconomic status. It differentiates
between male and female but not among races, where there is an observed
difference. This activity also assumes that height is normally distributed.
Unit 7 | 2
UNIT 7
ACTIVITY
Making Sense of Randomness
FACILITATOR GUIDE
1
C
Suppose that a lottery has the following payoff structure:
Outcome (selected
values) and wildcard
+
+
+
+
+
Prize
Odds
$120,000,000.00
1 in 146,107,962.00
$200,000.00
1 in 3,563, 608.83
$10,000.00
1 in 584,431.85
$100.00
1 in 14,254.44
$100.00
1 in 11,927.18
$7.00
1 in 290.91
$7.00
1 in 745.45
$4.00
1 in 126.88
$3.00
1 in 68.96
1. What is the expected value of a ticket in this lottery if the Grand Prize
is $120,000,000?
) + (200000) × (
)+
(10000) × (
) + (100) × ( ) + (100) × ( ) + (7) × ( ) + (7)
× (
) + (4) × ( ) + (3) × ( )
Answer: Expected value is: (120000000) ×
(
1
1
146107962
3563608.83
1
1
1
1
584431.85
14254.44
11927.18
290.91
1
1
1
745.45
126.88
68.96
Sum of:
0.821310477
0.056122883
0.017110635
0.007015358
0.008384212
0.024062425
0.009390301
0.031525851
0.04350348
Expected Value:
1.018425622
Unit 7 | 3
UNIT 7
ACTIVITY
Making Sense of Randomness
FACILITATOR GUIDE
1
D
Suppose that you must select a three-student team to send to the state math
competition. Out of the 12 students in your math club, how many possible teams
can you select? Assuming that each student is equally likely to be selected,
what is the probability of a particular student being selected for the team?
12!
Answer: The number of possible teams is C(12, 3) = (3!(12-3)!) = 220 possible
three-person teams. The number of ways that a student can be selected out
11!
of a team of three is C(11,2) = (2!(11-2)!) = 55. To obtain this result, imagine one
of the spots on the team as filled by the student in question and look at all the
ways the other two spots can be filled. There are 11 other students, so we need
to consider all the ways to choose 2 from 11, which is the computation that
results in 55. Because there are 55 teams that include the student in question
1
55
and 220 possible teams, any one student has a probability of 220 = 4 of being
selected for the team.
E
In the 1880s, about two out of every thousand couples got divorced each year.
In the 1950s it was about 11 out of every thousand. In the 1980s, it was about
22 out of every thousand. Find the probability that a couple would be together
after 30 years for each of the time periods. Don’t worry about the probability
changing each year.
Answer: In the 1880s each couple had a 0.002 chance of divorcing each year;
that means that there was a 0.998 chance of staying together. Assuming that
each year is independent from the last, the probability of being together after
30 years is given by (0.998) × (0.998) × …(30 times) = (0.998)30 = 94%. In the
1950s, couples experienced a 0.011 chance of divorce, which means there was
a 0.989 chance of staying together. (0.989)30 = 72% In the 1980s, there was a
0.022 chance of divorce each year, which means there was a 0.978 chance of
staying together. (0.978)30 = 51%
Notice that we have tacitly assumed that the probability of getting a divorce
is independent of the number of years that a couple has been married. If time
allows, have the participants identify where they used this assumption, and then
have them discuss whether or not such an assumption of independence
is reasonable.
Note that these numbers, while not entirely accurate, are in the ballpark.
Unit 7 | 4
UNIT 7
ACTIVITY
Making Sense of Randomness
FACILITATOR GUIDE
2
(40 - 50 minutes)
In this activity you will explore some of the many uses of the Central Limit
Theorem (CLT), one of the most important concepts in probability and statistics.
The CLT was described in the text as having to do with the distribution of results
of sets of many trials. It said that even if the results of a single event are not
normally distributed, the results of multiple trials of the single event will be,
at least approximately.
For example, while the outcome of a single coin toss is certainly not normally
distributed, the distribution of outcomes of sets of many coin tosses is
(approximately). The advantage conferred by knowing that something is
approximately normally distributed is that one can then use the concepts of
the mean and standard deviation to make predictions about large populations.
We’ll see how this is done in the following activity.
A
(10 minutes)
The Dishonest Coin
1. Suppose that you have come into possession of a coin that preferentially lands
“heads” up with 0.8 probability. If you were to flip this coin 100 times, how many
“heads” would you expect, on average? Call this result the “mean.”
Answer: The expected value, which is the same as the mean in this case, is the
possible maximum payoff (in terms of number of “heads”) times probability:
100 × 0.8 = 80 “heads.”
2. The outcome of any particular set of 100 flips is by no means guaranteed to
be what you just calculated. You can use the CLT, however, to make predictions
about how close the results of a specific set of 100 flips will be to the mean. To
do this, we first need to find the standard deviation, which is the average number
of flips by which a given set of 100 flips should differ from the mean. If the
standard deviation is defined as:
σ = ((# of flips) × (probability of heads) × (probability of tails))
what is the standard deviation for sets of 100 flips of your “dishonest” coin?
Answer: σ = (100)× (0.8)× (0.2) = 4 flips
Unit 7 | 5
UNIT 7
ACTIVITY
Making Sense of Randomness
FACILITATOR GUIDE
2
3. Using the 68-95-99.7 rule from the text (the probability that a given trial falls
within 1, 2, or 3 standard deviations of the mean, respectively), if you perform 100
coin flips of your dishonest coin, what is the probability that between 84 and 76 of
the results will be heads?
Answer: There is a 68% chance that heads will be the result between 84 and 76
times; these numbers represent one standard deviation above and below the
mean respectively.
4. Give the range in which you expect the number of heads to fall 99.7% of the time.
Answer: You should expect that 99.7% of the time the number of heads will fall
between 92 and 68.
5. If you perform 1,000 sets of 100 flips, how many would you expect to have more
than 92 or fewer than 68 “heads”?
Answer: If 99.7% of the results have between 92 and 68 “heads,” then 0.3% will
have either more or fewer. 0.3% of 1000 is 3. You should expect that only 3 out of
your 1000 trials would have more than 92 heads or fewer than 68.
Convene the large group and discuss results. (5 minutes)
B
(10 minutes)
Suppose that you work for a company that makes frozen hamburger patties.
One of your beef suppliers reports that all of his cows have been diagnosed
with mad cow disease, and he is afraid that his most recent shipment of cattle
to you was infected as well. This supplier’s cattle make up 1 out of every 10 of
the cattle you process into hamburger patties. Because you process cattle from
each supplier separately into hamburger patties, yet mix patties randomly from
different suppliers when packaging, potentially 1 out of every 10 of your frozen
hamburger patties is contaminated.
1. If you package patties in bags of 50, how many patties per package are,
on average, contaminated?
Answer: 5
2. What is the standard deviation for bags of 50 patties, 1 in 10 of which
are contaminated?
Answer: σ = (50) × (0.1) × (0.9) ≈ 2.12 patties
Unit 7 | 6
UNIT 7
ACTIVITY
Making Sense of Randomness
FACILITATOR GUIDE
2
3. If you shipped 10,000 bags of 50 patties, how many bags would you expect to
have between 3 and 7 contaminated patties?
Answer: The range is approximately within 1 standard deviation, so 0.68 x 10,000
= 6,800 bags
4. How many bags would have between 0 and 12 contaminated patties?
How many bags would fall outside of this range?
Answer: 9,970 bags = 0.997 × 10,000; 30 bags fall outside this range.
5. What are the chances that a bag contains no contaminated patties?
Answer: Each patty has a 109 chance of being uncontaminated, so the probability
that a bag of 50 patties is completely safe is (0.9) × (0.9) × …(50 times) =
(0.9)50 = 0.0051.
6. Using the probability you just found, how many uncontaminated bags would
you expect to find out of 10,000?
Answer: 51 bags
7. How would the above result change if 1 out of 100 patties were contaminated
instead of 1 out of 10?
99
Answer: In this case, each patty has a 100 chance of being uncontaminated, so
the probability that a bag of 50 patties is completely safe is (0.9) × (0.9) × …
(50 times) = (0.99)50 = 0.605. Multiplying this probability by 10,000 yields 6,050
uncontaminated bags.
8. Say that you ship 100 shipments of 10,000 bags, with 1 out of 100 patties
contaminated. Give the range of expected numbers of safe bags per shipment
for 99 of the shipments.
Answer: If, on average, 6,050 bags are expected to be uncontaminated, this is
the mean for shipments of 10,000 bags. This leads to a standard deviation of
49 bags, using the formulas from before. Consequently, 99 of the shipments
should have approximately between 5,900 and 6,200 uncontaminated bags
(representing values within three standard deviations of the mean).
Unit 7 | 7
UNIT 7
ACTIVITY
Making Sense of Randomness
FACILITATOR GUIDE
2
9. How many shipments would you have to send before you would expect to
ship a completely uncontaminated batch? (feel free to approximate liberally)
Answer: Each bag has a 60% chance of being uncontaminated, and each
shipment consists of 10,000 bags. The probability of a shipment being
completely uncontaminated is (0.6)10,000 ≈ 10-10,000, so you would expect about
one shipment out of a number consisting of 1 followed by 10,000 zeros to be
completely safe. This is a statistical impossibility.
Convene the large group and discuss results. (5 minutes)
C
(10 minutes)
Say that you are conducting an exit poll for an election. You interview an
appropriate (unbiased) sample of 1,200 people as they leave the voting area
and ask them who they voted for. You find that 54% of people say they voted for
candidate A and 46% say they voted for candidate B. You can view each person
polled as a coin that comes up heads 54% of the time and tails 46% of the time.
1. What is the standard deviation of your sample?
Answer: ((1200) × (0.54) × (0.46)) ≈ 17 voters
2. Express this deviation as a percent of the sample size.
17
Answer: 1200 ≈1.4%
The percent you just found (actually, twice the percent you just found) is what is
commonly known as the “margin of error” in polling. It involves the assumption
that the standard deviation of your sample is very close to the standard deviation
of the pool of all people voting, which is usually a reasonable assumption.
3. How likely is it that candidate A garners between 52.6% and 55.4% of the
overall vote?
Answer: 68% chance
4. How likely is it that candidate A garners between 51.2% and 56.8% of the
overall vote?
Answer: 95% chance
Unit 7 | 8
UNIT 7
ACTIVITY
Making Sense of Randomness
FACILITATOR GUIDE
2
5. What is the chance that candidate A won the overall election outright?
Assume that a candidate needs 51% of the vote to win outright. Assume
further that the difference in percent certainty between the 2nd and 3rd
standard deviation (the difference between 95% and 99.7% probability) can be
apportioned evenly with portions of a whole standard deviation—that is, half of
a standard deviation yields a percent certainty halfway between 95% and 99.7%.
(Note: because the normal distribution is a curve, this isn’t actually the case,
but for a rough estimate, it will work.)
Answer: 51% is 2.14 standard deviations from the measured sample mean of
54%. 2 standard deviations gets us to 95% certainty. The difference between
95% and 99.7% (corresponding to the third standard deviation) is 4.7%. Find
0.14 of 4.7% to find the portion of percent certainty that 0.14 of a standard
deviation buys us between the 2nd and 3rd standard deviations according to the
assumptions. 0.14 x 4.7% ≈ 0.66%, so we go 0.66% beyond the 95% certainty of
the second standard deviation: 95% + 0.66% = 95.66% certainty that candidate A
won the overall election outright.
6. If you did the same poll of 1200 people for 100 elections and got 54% for
candidate A in every one of them, how many of those elections would you expect
ending up going to candidate B?
Answer: 95.66% of 100 elections is about 96 wins for candidate A, which leaves
about 4 wins for candidate B.
7. How does this make you feel about the reliability of exit poll data?
Answer: Answers will vary.
Convene the large group and review/discuss the various ways that you have
used the Central Limit Theorem. (5 minutes)
Unit 7 | 9
UNIT 7
ACTIVITY
Making Sense of Randomness
FACILITATOR GUIDE
3
(30 minutes)
A
(15 minutes)
Imagine that you are about to start a blog and wish to drive as much traffic to it
as possible. You have $10,000 set aside for marketing. You know that a friend
of yours recently started a blog and was able to attract 100,000 visitors after
spending only $5,000 on marketing. She did this by having a lottery for people
who post comments on the site.
1. Your friend set up an account with a large search engine company to place ads
for her blog on other websites. She agreed to pay $0.02 per visitor that the ads
attracted to her site, up to 100,000 visitors. How much of her marketing budget
did she set aside for the search engine?
Answer: $0.02 x 100,000 = $2,000
In order to make her ads stand out, she wanted to make a lottery that sounded
as attractive as possible. She came up with the following possible ads:
a) “Post a comment on Sara’s blog and be automatically entered in a drawing for
a Gizmophone2000.”
b) “Post a comment on Sara’s blog and be automatically entered in a drawing
for one of 10 Gizmophone2000 Nanos.”
c) “Post a comment on Sara’s blog and be automatically entered in a drawing for
a Gizmophone2000 Nano, or one of 1,000 other prizes.
2. If Sara’s search engine company requires her ad to state the odds of winning
a prize, and she will not administer the lottery until she has 100,000 entries,
which of the above three ads should she pick as the most likely to attract
someone’s mouseclick? Justify your answer.
Answer: Ad “a” has odds of 1 in 100,000 of winning. Ad “b” has odds of 1 in
10,000. Ad “c” has odds better than 1 in 100. The answer may vary, but usually
a potential visitor would rather have a better chance of winning something of
lesser value as opposed to a lesser chance of winning something of greater
value. This generality suggests that “c” is the best choice, but there are other
acceptable answers, as long as they are well-justified.
Unit 7 | 10
UNIT 7
ACTIVITY
Making Sense of Randomness
FACILITATOR GUIDE
3
3. Find the expected value of each version of the promotion.
Answer: The expected value is the sum of the products of the payoff and the
probability of getting that payoff. All three versions have the same expected
value of $0.03. The values of each prize can be deduced from the amount spent
and the number of prizes.
4. Suppose that Sara goes with choice “c”; the actual breakdown of prizes is
as follows:
1 × Gizmophone2000 Nano
2 × $100 gift certificates to iTones.com
200 × $5 gift certificates to iTones.com
798 × Sara’s blog keychain flashlights (valued at $1.87 each)
What are the odds of winning something worth $100 or more?
3
Answer: 100,000 = 1 in 33,333
5. Sara estimates that if she draws 100,000 visitors to her blog during this
promotion, at least 10% will continue to visit the site once a week. Each of
these visits (both during the promotion and during the month) counts as a
“hit.” She estimates that out of every 100 hits, 50 visitors will click on the ads
of her sponsors, for which she is paid $0.03 each. She further estimates that
one visitor out of 100 will buy a Sara’s blog t-shirt, for which she nets $5. How
much net income can she expect as a direct result of her promotion in her first
6 months if she starts the lottery promotion at the beginning of the first month?
(We are concerned only with the direct effects of the promotion, so ignore any
new visitors and website costs.)
Answer: Six months is about 24 weeks. In the first week, Sara gets 100,000 hits,
and each subsequent week, she gets 10,000 hits. This totals 330,000 hits in 6
months. The money she makes on ads will be $0.03 × 0.5 × 330,000 = $4,950.
The money she makes on t-shirts will be $5 × 0.01 × 330,000 = $16,500. She
spent $5,000 dollars on the promotion, so her net is $16,450.
Unit 7 | 11
UNIT 7
ACTIVITY
Making Sense of Randomness
FACILITATOR GUIDE
3
B
(15 minutes)
As a group, design a similar promotion for your own blog. Remember that you
have $10,000 for marketing; suppose that your goal is to attract 500,000 visits
as a result of the promotion. Feel free to choose how much you spend on search
engine ads, what prizes you will offer, and how you will structure your prize
scheme. Include the expected value of a typical entry. Your promotion should
consist of a prize structure, the associated probabilities, the expected value of
an entry, and the text for an attractive web ad.
Give the groups about 10 minutes to design their promotions and then have each
group share what they came up with. If time allows, have a discussion about
the various assumptions that went into this activity. How reasonable are the
values given?
Unit 7 | 12
UNIT 7
CONCLUSION
DISCUSSION
Making Sense of Randomness
FACILITATOR GUIDE
(30 minutes)
HOW TO RELATE TOPICS IN THIS UNIT TO STATE OR NATIONAL STANDARDS
Facilitator’s note:
Have copies of national, state, or district mathematics standards available.
Mathematics Illuminated gives an overview of what students can expect when
they leave the study of secondary mathematics and continue on into college.
While the specific topics may not be applicable to state or national standards
as a whole, there are many connections that can be made to the ideas that
your tstudents wrestle with in both middle school and high school math.
For example, in Unit 12, In Sync, the relationship between slope and calculus
is discussed.
Please take some time with your group to brainstorm how ideas from Unit 7,
Making Sense of Randomness could be related and brought into your classroom.
Questions to consider:
Which parts of this unit seem accessible to my students with no “frontloading?”
Which parts would be interesting, but might require some amount of
preparation?
Which parts seem as if they would be overwhelming or intimidating to students?
How does the material in this unit compare to state or national standards?
Are there any overlaps?
How might certain ideas from this unit be modified to be relevant to your
curriculum?
WATCH VIDEO FOR NEXT CLASS (30 minutes)
Please use the last 30 minutes of class to watch the video for the next unit: Unit
8, Geometries Beyond Euclid. Workshop participants are expected to read the
accompanying text for Unit 8, Geometries Beyond Euclid before the next session.
Unit 7 | 13
UNIT 7
Making Sense of Randomness
FACILITATOR GUIDE
NOTES
Unit 7 | 14