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Cal Poly A.P. Statistics Workshop – February 16, 2013
Probability 1: Central Limit Theorem
Matt Carlton
Activity A: Ticket, please!
Preface: Understanding that different random samples from the same population yield different
results is central to statistics. The purpose of this activity is to specifically investigate how the
averages of different random samples from the same population differ.
Materials: A roll of numbered tickets with last three digits 001 through 999, 3x5 index cards,
supplies for posters.
At the front of the classroom is a large bowl with numbered tickets in it, as well as a stack of
index cards. Each student does the following:
-
take an index card
stir up the tickets, and then take 10 of them
record the last three digits from all 10 tickets onto the index card
return the tickets to the bowl
Ideally, students should return the tickets quickly, so that other students in the class could
potentially resample them.
Imagine that the tickets in the bowl actually represent the population of all teenagers with Visa
cards. So, each ticket represents a teenager. The last three digits on the ticket represent that
teenager’s Visa card balance; for example, a ticket ending in 374 represents a teenager with a
balance of $374 on her/his Visa card.
Everyone in this classroom now has a random sample from this population.
1. Is everyone’s sample exactly the same?
2. If everyone in the class were to calculate his or her sample average, would all of those
sample averages be the same?
Probability 1
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Cal Poly A.P. Statistics Workshop – February 16, 2013
Our model for the population of teenagers’ Visa card balances has a uniform distribution on the
numbers $1 through $999:
Population Histogram
5
Population Percentage
4
3
2
1
0
0
100
200
300
400
500
600
Visa card balance ($)
700
800
900
1000
This population has a mean of µ = $500 and a standard deviation of σ = $288.39.
Note to teachers: If you only tore off tickets ending in 001 through N, then your population is
uniform on $1 to $N. The mean of this population is µ = (N + 1)/2, and the standard deviation is
σ=
( N 2  1) /12 .
3. If everyone were to calculate his or her sample average, would all of those sample
averages equal $500?
4. Would you be surprised if someone’s sample average were exactly $500? Would you be
surprised if someone’s sample average were close to $500?
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Cal Poly A.P. Statistics Workshop – February 16, 2013
Now, we’re going to explore how averages of different samples, from the same population,
behave. Using your samples, we’re going to explore the behavior of averages based on samples
of size n = 5, then n = 10.
5. Calculate the average of just the first five numbers in your sample. Record your answer:
my sample average =
(n = 5)
6. Make a graph of all the sample averages for every student in the class.
One way to do this is to make a poster, with a number line from 0 to 1000 already in place as the
x-axis. The students create a dot plot, with circular stickers or tiny sticky notes or whatever you
like to use. Ideally, each of these stickers should have the x symbol written on it (or y if that’s
what your book uses).
The x-axis on this poster needs labeling. Students can discuss what they want. An ideal label
would be, Sample average Visa balance (n = 5).
7. Describe the plot of sample averages your class just created (shape, center, spread). How
is this distribution similar to the population distribution, and how is it different?
Students need to recognize that the x-bar distribution is centered around µ = $500, just like the
population is. However, this distribution is less spread out (more concentrated) than the
population distribution. This is a good time to talk about why: for an x-bar to be down near $0
or up near $1000, all numbers in the n = 5 random sample would have to be low (or high). But
an x-bar near µ = $500 can come from various mixtures of high/middle/low numbers, and that’s
much more common.
8. Next, you’re going to repeat this process for all ten numbers in your sample. Make a
prediction: When you plot the averages of samples of size n = 10, what will the pattern be
like? How will it compare to the plot you just made for n = 5?
Probability 1
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Cal Poly A.P. Statistics Workshop – February 16, 2013
9. Calculate the average of all ten numbers in your sample. Record your answer:
my sample average =
(n = 10)
10. Make a graph of all the sample averages (n = 10) for every student in the class.
Make another poster of exactly the same width, with a number line from 0 to 1000 already in
place as the x-axis. An ideal x-axis label would be, Sample average Visa balance (n = 10).
11. Compare and contrast this graph with the previous graph of sample averages, as well as
the population distribution.
12. To understand what happens with even larger samples, we’ll use our TI calculators.
Create a random sample of size n = 40 from the population of integers from 1 to 999, and
store this in L1. randInt(1,999,40)→L1
13. Make graphical and numerical summaries of your sample (the 40 numbers in L1). What
do you notice?
Students need to understand that a single sample, taken randomly, should mimic the original
population. It’s critical that students can distinguish a histogram of one sample from the pictures
of x s from many samples that they have been creating. In particular, they must be disabused of
the idea that the Central Limit Theorem (next page) says the x-bar pattern becomes bell shaped,
not that a sample (or the population) becomes bell-shaped as sample size grows.
Probability 1
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Cal Poly A.P. Statistics Workshop – February 16, 2013
14. Record your sample average.
my sample average =
(n = 40)
15. Make a graph of all the sample averages (n = 40) for every student in the class. Describe
that distribution of sample averages, and compare it to the previous two.
What you’ve just observed are three important patterns in statistical behavior. Consider sampling repeatedly from a
particular population. Write
µ = the population mean
σ = the population standard deviation
n = the common size of every sample
Also write X for a sample average, which varies from sample to sample.
Fact #1: The center (mean) of the distribution of all sample averages is µ, and that’s true no matter what the sample
size n was. In short-hand,
µX =µ
Fact #2: The spread (standard deviation) of the distribution of all sample averages is related to both the population
standard deviation, σ, and the common sample size, n. Specifically,
X 

n
Fact #3: As the sample size n increases, the shape of the distribution of all sample means becomes more and more
Normal, and that’s true no matter what the population shape was. In practical terms, we say that when n is “large,”
the distribution of X is approximately Normal
This last fact is known in statistics as the Central Limit Theorem.
Probability 1
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Cal Poly A.P. Statistics Workshop – February 16, 2013
16. Use Fact #1 and Fact #2 to find the theoretical mean and standard deviation of all three of
the sample mean graphs (n = 5, 10, and 40). Label the graphs.
17. Suppose I take a simple random sample of 40 teenagers with Visa cards and record the
balance for each person. Approximately, what is the probability that my sample average
will be with $50 of the population mean, µ = $500?
Probability 1
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Cal Poly A.P. Statistics Workshop – February 16, 2013
Activity B: Let’s make guacamole
Oscar works at an avocado orchard. His job is to randomly select avocados from those that have
been picked at the orchard and put them into “four-packs” for sale at the local farmer’s market.
For every four-pack, Oscar is required to write the average weight of the four avocados on the
bottom of the container.
Weights of individual avocados follow a Normal distribution, with a mean of 7.5 ounces and a
standard deviation of 0.9 ounces.
We can think of each four-pack as an SRS of size n = 4 from the population of all avocados at
this orchard. Our goal is to investigate how the average weights of four-packs vary. We’ll begin
with a simulation.
1. To simulate 100 four-packs of avocados, create four columns of simulated avocado
weights, each with 100 entries.
randNorm(7.5,0.9,100)→L1
randNorm(7.5,0.9,100)→L2
randNorm(7.5,0.9,100)→L3
randNorm(7.5,0.9,100)→L4
2. Each row of four numbers represents the weights of four individual avocados that will be
combined to make a four-pack. To find the average weight of each of your four-packs,
compute the sample average of these four rows and store them.
(L1+L2+L3+L4)/4→L5
3. Using a graph as well as summary statistics, describe the shape/center/spread of the 100
sample averages you have created.
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Cal Poly A.P. Statistics Workshop – February 16, 2013
Now let’s use our knowledge of how sample averages behave.
4. What is the theoretical mean of the distribution of average weights of all four-packs of
avocados? How close is this to the mean of your 100 simulated values?
5. What is the theoretical standard deviation of the distribution of average weights of all
four-packs of avocados? How close is this to the standard deviation of your 100
simulated values?
6. Would it be appropriate to apply the Central Limit Theorem to this example? Why or
why not?
There’s one more fact to know about how sample averages behave.
Fact #4: If the population from which samples are being taken is itself Normal, then the distribution of sample
averages will also be Normal, and that’s true no matter the sample size. That is,
X is always Normal if the population is Normal
So, the distribution of X is (at least approximately) Normal in two situations:
-
When n is large, no matter the population shape (Fact #3/Central Limit Theorem/approximate)
When the population is Normal, no matter n (Fact #4/exact)
7. What is the theoretical shape of the distribution of average weights of all four-packs of
avocados? Why? How close is this to the shape of your 100 simulated values?
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Cal Poly A.P. Statistics Workshop – February 16, 2013
Draw two overlaid curves: one showing the population distribution of individual avocado
weights, and one showing the distribution of average weights for avocado four-packs.
8. What is the probability that a randomly selected four-pack of avocados has an average
weight of less than 7.2 ounces?
Probability 1
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Cal Poly A.P. Statistics Workshop – February 16, 2013
Practice Problem A
Wait times at the security checkpoint at a small, municipal airport have a symmetric distribution
with a mean of 10 minutes and a standard deviation of 3 minutes.
(a) Consider a random sample of 60 airline passengers. Find the approximate probability that
their average wait time at the security checkpoint is less than 9 minutes.
(b) Consider a single randomly selected passenger. What is the probability this passenger’s
wait time is more than 10 minutes?
(c) Consider a random sample of six airline passengers. Compute the probability that exactly
two of these six passengers will wait more than 10 minutes.
(d) Consider again a random sample of six airline passengers. Compute the probability that at
most two of these six passengers will wait more than 10 minutes.
Probability 1
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Cal Poly A.P. Statistics Workshop – February 16, 2013
Practice Problem B
The heights of Cal Poly women follow a Normal distribution, with a mean of 64 inches and a
standard deviation of 2.5 inches.
(a) Find the probability that a randomly selected Cal Poly woman is between 62 inches and
65 inches tall.
(b) Consider a random sample of 10 Cal Poly women. Find the probability that their average
height is between 62 inches and 65 inches.
(c) Consider again a random sample of 10 Cal Poly women. Find the probability that all 10
of them have heights between 62 inches and 65 inches.
(d) Consider again a random sample of 10 Cal Poly women. Find the probability that exactly
four of them have heights between 62 inches and 65 inches.
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Cal Poly A.P. Statistics Workshop – February 16, 2013
Central Limit Theorem questions are among the lowest-scoring free response questions on the
A.P. Statistics exam. Here are some recent examples, with scores out of 4:
2010 FR #2
2009 FR #2
2007 FR #3
2006 FR #3
mean = 0.78, sd = 1.02
mean = 0.84, sd = 1.01
mean = 1.16, sd = 0.94
mean = 0.64, sd = 0.96
Teaching/Exam tips:
-
-
-
-
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When you make graphs of sampling distributions, label each item with the correct
statistical symbol (such as x ; the same goes for p̂ ).
For C.L.T. activities, make sure all graphs have the same x-scale, so the decreased
variability with increased n is evident.
Test students, both verbally in class and on quizzes/exams, on the right and wrong
interpretations of the Central Limit Theorem. Activities and visuals help, but students still
are willing to believe some incorrect statements:
o “When n is large, samples are more Normal.”
o “When n is large, the population is Normal.”
Don’t just tell your students the “right answer”; go through these mis-interpretations
carefully and make sure they understand why each is wrong.
The term “sampling distribution” causes havoc, especially when students first learn of the
concept. Start with a soft term like “pattern,” as in “as n increases, the pattern of x-bar
values becomes more Normal.”
At the same time, notation-speak can be helpful: “x-bar” and “mu” are easier for
students’ ears to distinguish than “population mean” and “sample mean.” Of course, they
have to get the vocabulary distinction down, too, eventually!
Students need to understand the conceptual difference between σ and σ/ n . While σ
indicates how much an individual value typically differs from µ, σ/ n measures how far
a sample average (an x ) typically differs from µ.
It may be worth pointing out that σ is really just a special case of σ/ n ; specifically, that
when n = 1 our “sample” is just a single individual, and the formula σ/ n reduces to σ.
Have students read C.L.T.-style questions and label the information provided (µ, σ, n).
Allow students to use their calculators, but do not allow them to use “TI-speak” in their
written responses.
Don’t let students get away with calling something Normal if it’s really only
approximately Normal. They need to include the word “approximately.”
Probability 1
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Cal Poly A.P. Statistics Workshop – February 16, 2013
Solutions
Activity A: Ticket, Please!
No, everyone’s sample should be different, since the samples were taken independently.
No, since these are all different samples, and there is natural sampling variation.
No, for the same reason as 2.
It would be surprising for someone to get exactly $500, since it’s rare for a sample mean
to exactly equal the population mean. On the other hand, we’d naturally expect some
sample means to be close to the population mean.
5. answers will vary
6. see notes
7. The pattern of x values (sample means) in the graph is roughly symmetric and moundshaped, with a mean/center of around $500, and a standard deviation that’s somewhat
less than the population standard deviation.
8. Hopefully, students will predict a symmetric shape, centered at $500, with even less
spread than the previous graph
9. answers will vary
10. see notes
11. The pattern of x values (sample means) with n = 10 is centered around $500, just like the
population and the pattern of sample means for n = 5. This new graph has even less
spread than the previous two, and is somewhat more bell-shaped.
12. randInt(1,999,40)→L1
13. Each sample of 40 values should look like the original population. In particular, the
histogram should be relatively uniform, the average should be near $500, and the
standard deviation should be near the population SD of $288.39. see notes for more info
14. The pattern of x-bar values (sample means) with n = 40 is symmetric and bell-shaped,
perhaps approximately Normal, with a mean around $500 and a much smaller standard
deviation than any of the previous graphs.
15. For every sample size, the theoretical mean of the x distribution is µ = $500. The
theoretical standard deviations of the three x distributions are:
1.
2.
3.
4.
a. for n = 5: $288.39/ 5 = $128.97
b. for n = 10: $288.39/ 10 = $91.20
c. for n = 40: $288/39/ 40 = $45.60
16. With n = 40, the goal is to approximate P($450 ≤ X ≤ $550). We know that X has a
mean of $500 and a standard deviation of $45.60. Also, according to the Central Limit
Theorem, X has an approximately Normal distribution. Therefore,
P($450 ≤ X ≤ $550) ≈ P(–1.10 ≤ Z ≤ 1.10) ≈ .7287
Probability 1
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Cal Poly A.P. Statistics Workshop – February 16, 2013
Activity B: Let’s make guacamole
1. randNorm(7.5,0.9,100)→L1, then L2 and L3 and L4
2. (L1+L2+L3+L4)/4 → L5
3. answers will vary; a histogram should appear symmetric and bell-shaped, with a mean of
roughly 7.5 and a standard deviation of roughly 0.45
4. The theoretical mean of this X distribution is µ = 7.5 ounces (which should be close to
most students’ average L5 value)
5. The theoretical standard deviation of this X distribution is σ/ n = 0.9/ 4 = 0.45 ounces
(which should be close to most students’ L5 standard deviation)
6. No, we cannot apply the Central Limit Theorem to this example. The sample size is just n
= 4, and the Central Limit Theorem only applies to X distributions when n is large.
7. Since the population of individual avocado weights is Normal, the X distribution is
Normal no matter the sample size (even as small as n = 4). (each student’s histogram
should be approximately Normal)
8. Students should hand-draw a picture like this:
Original population
X-bar distribution (n=4)
4
5
6
7
8
9
10
weight/sample average weight (ounces)
11
9. The goal is to find P( X ≤ 7.2). Since the population of avocado weights is Normal, X is
also Normally distributed, but with a mean of 7.5 ounces and a standard deviation of
0.9/ 4 = 0.45 ounces. So,
P( X ≤ 7.2) = P(Z ≤ –0.67) ≈ .2525
Probability 1
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Cal Poly A.P. Statistics Workshop – February 16, 2013
Practice Problem A
(a) The goal is to find P( X ≤ 9 minutes). According to the Central Limit Theorem, X has an
approximately Normal distribution (because n = 60 is large), with a mean equal to µ = 10
minutes and a standard deviation equal to σ/ n = 3/ 60 = .387 minutes. So,
P( X ≤ 9) ≈ P(Z ≤ –2.19) ≈ .0143
(b) Since the distribution of individual wait times is symmetric, its mean and median are the
same, and so half the wait times are below 10 minutes (and half are above). So, the
probability that a passenger’s wait time is more than 10 minutes is 1/2, or .5.
(c) Let Y equal the number of passengers, out of 6, that wait more than 10 minutes. Then Y is
Binomial, with n = 6 and p = .5 from part (b). So,
P(Y = 2) = C(6,2) (.5)2 (1 – .5)4 = .234575
(d) Using the same variable Y defined in part (c),
P(Y ≤ 2) = P(Y = 0, 1, 2) = .015625 + .093750 + .234575 = .34375
Practice Problem B
(a) Since heights are Normally distributed with µ = 64 and σ = 2.5,
P(62 ≤ X ≤ 65) = P(–0.8 ≤ Z ≤ 0.4) = .4436
(b) Let X equal the average height of a random sample of 10 Cal Poly women. Since the
population of individual heights is Normal, X is also Normal (even with n = 10), but with
a mean of µ = 64 inches and a standard deviation of σ/ n = 2.5/ 10 = 0.79 inches. So,
P(62 ≤ X ≤ 65) = P(–2.53 ≤ Z ≤ 1.266) = .8915
(c) Since there is a .4436 probability that a single woman’s height is between 62 and 65
inches, and random sampling implies the 10 women’s heights are independent, the
probability is (.4436)…(.4436) = (.4436)10 = .0003.
(d) Let Y equal the number of women among these 10 with heights between 62 and 65
inches. Then Y is Binomial, with n = 10 and p = .4436 from part (a). So,
P(Y = 4) = C(10,4) (.4436)4 (1 – .4436)6 = .2413
Probability 1
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