Download The Central Limit Theorem

The Central Limit Theorem
April 4, 2009
The Central Limit Theorem
in terms of
Histograms and Bell Curves
I
The Central Limit Theorem is about histograms looking like
bell curves.
The Central Limit Theorem
in terms of
Histograms and Bell Curves
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The Central Limit Theorem is about histograms looking like
bell curves.
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What types of histograms look like bell curves?
The Central Limit Theorem
in terms of
Histograms and Bell Curves
I
The Central Limit Theorem is about histograms looking like
bell curves.
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What types of histograms look like bell curves?
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Histograms of averages of large samples.
The Central Limit Theorem
in terms of
Histograms and Bell Curves
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The Central Limit Theorem is about histograms looking like
bell curves.
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What types of histograms look like bell curves?
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Histograms of averages of large samples.
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Histograms of sums of large samples.
The Central Limit Theorem in terms of Distributions
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The distribution of sample averages becomes a normal
distribution as the size of the sample becomes large.
The Central Limit Theorem in terms of Distributions
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The distribution of sample averages becomes a normal
distribution as the size of the sample becomes large.
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The distribution of sample sums becomes a normal
distribution as the size of the sample becomes large.
The Central Limit Theorem in terms of Distributions
I
The distribution of sample averages becomes a normal
distribution as the size of the sample becomes large.
I
The distribution of sample sums becomes a normal
distribution as the size of the sample becomes large.
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How large is large?
The Central Limit Theorem in terms of Distributions
I
The distribution of sample averages becomes a normal
distribution as the size of the sample becomes large.
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The distribution of sample sums becomes a normal
distribution as the size of the sample becomes large.
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How large is large? It depends.
The Central Limit Theorem in terms of Distributions
I
The distribution of sample averages becomes a normal
distribution as the size of the sample becomes large.
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The distribution of sample sums becomes a normal
distribution as the size of the sample becomes large.
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How large is large? It depends.
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If the original population is normally distributed, then a
sample of size n = 1 is large enough.
The Central Limit Theorem in terms of Distributions
I
The distribution of sample averages becomes a normal
distribution as the size of the sample becomes large.
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The distribution of sample sums becomes a normal
distribution as the size of the sample becomes large.
I
How large is large? It depends.
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If the original population is normally distributed, then a
sample of size n = 1 is large enough.
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If the original population is nearly normally distributed, then a
sample of size n = 5 might be large enough.
The Central Limit Theorem in terms of Distributions
I
The distribution of sample averages becomes a normal
distribution as the size of the sample becomes large.
I
The distribution of sample sums becomes a normal
distribution as the size of the sample becomes large.
I
How large is large? It depends.
I
If the original population is normally distributed, then a
sample of size n = 1 is large enough.
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If the original population is nearly normally distributed, then a
sample of size n = 5 might be large enough.
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If the original population is not at all normally distributed,
then a sample of size n = 30 might be large enough.
The Central Limit Theorem in terms of Distributions
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No matter what the original distribution is, if we choose a
large enough sample size n, then the distributions of sample
averages and sample sums will be approximately normal.
Where is the Center?
Where is the Center?
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Another way of saying this is, what is µ in the normal
distribution we are approaching?
Where is the Center?
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Another way of saying this is, what is µ in the normal
distribution we are approaching?
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That depends on whether we are talking about averages or
sums.
Where is the Center for Averages?
Let X be the random variable that measures the sample average
and let µX be the mean of the sample average X .
Where is the Center for Averages?
Let X be the random variable that measures the sample average
and let µX be the mean of the sample average X .
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If µX is the population mean, then
µX = µX
Where is the Center for Averages?
Let X be the random variable that measures the sample average
and let µX be the mean of the sample average X .
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If µX is the population mean, then
µX = µX
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That is, the mean of the normal distribution we are
approaching for averages and the mean of the population are
the same.
Where is the Center for Sums?
P
Let
X be the random variable that measures
Pthe sum of the
sample and let µP X be the mean of the sum
X.
Where is the Center for Sums?
P
Let
X be the random variable that measures
Pthe sum of the
sample and let µP X be the mean of the sum
X.
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If µX is the population mean, then
µP X = nµX
Where is the Center for Sums?
P
Let
X be the random variable that measures
Pthe sum of the
sample and let µP X be the mean of the sum
X.
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If µX is the population mean, then
µP X = nµX
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That is, the mean of the normal distribution we are
approaching for sums is the mean of the population multiplied
by the size of the sample n.
How large is the spread?
How large is the spread?
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Another way of saying this is, what is σ in the normal
distribution we are approaching?
How large is the spread?
I
Another way of saying this is, what is σ in the normal
distribution we are approaching?
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That depends on whether we are talking about averages or
sums.
How large is the spread for averages?
Let X be the random variable that measures the sample average
and let σX be the standard deviation of the sample average X .
How large is the spread for averages?
Let X be the random variable that measures the sample average
and let σX be the standard deviation of the sample average X .
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If σX is the standard deviation of the population, then
√
σX = σX / n
How large is the spread for averages?
Let X be the random variable that measures the sample average
and let σX be the standard deviation of the sample average X .
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If σX is the standard deviation of the population, then
√
σX = σX / n
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That is, the standard deviation of the normal distribution we
are approaching for averages is the standard deviation of the
original population divided by the square root of the sample
size n.
How large is the spread for averages?
Let X be the random variable that measures the sample average
and let σX be the standard deviation of the sample average X .
I
If σX is the standard deviation of the population, then
√
σX = σX / n
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That is, the standard deviation of the normal distribution we
are approaching for averages is the standard deviation of the
original population divided by the square root of the sample
size n.
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We call the standard deviation of the averages the standard
error of the mean.
How large is the spread for sums?
P
Let
X be the random variable that measures the sum of P
the
sample and let σP X be the standard deviation of the sum
X.
How large is the spread for sums?
P
Let
X be the random variable that measures the sum of P
the
sample and let σP X be the standard deviation of the sum
X.
I
If σX is the standard deviation of the population, then
√
σP X = σX n
How large is the spread for sums?
P
Let
X be the random variable that measures the sum of P
the
sample and let σP X be the standard deviation of the sum
X.
I
If σX is the standard deviation of the population, then
√
σP X = σX n
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That is, the standard deviation of the normal distribution we
are approaching for sums is the standard deviation of the
original population multiplied by the square root of the sample
size n.
The Central Limit Theorem for Averages
Suppose X is a random variable with mean µX and standard
deviation σX .
If we draw a sample of size n, then as n increases the distribution
of the random variable X , which finds the sample mean, becomes
increasingly normal and
σX
X ∼ N µX , √
n
The Central Limit Theorem for Sums
Suppose X is a random variable with mean µX and standard
deviation σX .
If we draw a sample of P
size n, then as n increases the distribution
of the random variable
X , which finds the sample sum, becomes
increasingly normal and
X
√ X ∼ N nµX , σX n
Example 1
Suppose students at Cortland attend 1.2 football games per year
on average and with a standard deviation of 0.95.
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If we ask 25 students how many games they attended last
year, what is the probability that the average number of
games attended is less than 1?
Example 1
Suppose students at Cortland attend 1.2 football games per year
on average and with a standard deviation of 0.95.
I
If we ask 25 students how many games they attended last
year, what is the probability that the average number of
games attended is less than 1?
sol’n First, we find the standard deviation for X
√
σX = 0.95/ 25 = 0.19
Example 1
Suppose students at Cortland attend 1.2 football games per year
on average and with a standard deviation of 0.95.
I
If we ask 25 students how many games they attended last
year, what is the probability that the average number of
games attended is less than 1?
sol’n First, we find the standard deviation for X
√
σX = 0.95/ 25 = 0.19
Then, we use our calculator to find
P(X < 1) = normalcdf(−100, 1, 1.2, 0.19) = .15
note: -100 could be any number a, so long as P(X < a) = 0.
Example 1a
Example 1a
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If we ask 100 students how many games they attended last
year, what is the probability that the average number of
games attended is less than 1?
Example 1a
If we ask 100 students how many games they attended last
year, what is the probability that the average number of
games attended is less than 1?
√
sol’n σX = 0.95/ 100 = 0.095
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P(X < 1) = normalcdf(−100, 1, 1.2, 0.095) = 0.018
note: -100 could be any number a, so long as P(X < a) = 0.
Example 2
Suppose the average time waiting in line at a certain grocery store
is six minutes and that the standard deviation is 3. Suppose we ask
30 people how long they waited in line.
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What is the probability that the the average wait time for
these people is between 5 and 7 minutes.
Example 2
Suppose the average time waiting in line at a certain grocery store
is six minutes and that the standard deviation is 3. Suppose we ask
30 people how long they waited in line.
What is the probability that the the average wait time for
these people is between 5 and 7 minutes.
√
sol’n The standard error of the mean is 3/ 30 = .548.
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Example 2
Suppose the average time waiting in line at a certain grocery store
is six minutes and that the standard deviation is 3. Suppose we ask
30 people how long they waited in line.
What is the probability that the the average wait time for
these people is between 5 and 7 minutes.
√
sol’n The standard error of the mean is 3/ 30 = .548.
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P(5 < X < 7) = normalcdf(5, 7, 6, .548)
= .932
Example 2
Suppose the average time waiting in line at a certain grocery store
is six minutes and that the standard deviation is 3. Suppose we ask
30 people how long they waited in line.
What is the probability that the the average wait time for
these people is between 5 and 7 minutes.
√
sol’n The standard error of the mean is 3/ 30 = .548.
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P(5 < X < 7) = normalcdf(5, 7, 6, .548)
= .932
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Find the probability that a person waits between 1 and 10
minutes.
Example 2
Suppose the average time waiting in line at a certain grocery store
is six minutes and that the standard deviation is 3. Suppose we ask
30 people how long they waited in line.
What is the probability that the the average wait time for
these people is between 5 and 7 minutes.
√
sol’n The standard error of the mean is 3/ 30 = .548.
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P(5 < X < 7) = normalcdf(5, 7, 6, .548)
= .932
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Find the probability that a person waits between 1 and 10
minutes.
We have no way of knowing. The CLT does not apply.
Example 3
According to cortland.org, the average yearly precipitation in
Cortland is 39.73 inches. Let us suppose that the standard
deviation is 1.1 inches.
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Find the probability that it rains more than 1000 inches in
Cortland over the next two decades.
Example 3
According to cortland.org, the average yearly precipitation in
Cortland is 39.73 inches. Let us suppose that the standard
deviation is 1.1 inches.
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Find the probability that it rains more than 1000 inches in
Cortland over the next two decades.
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The average over 20 years should be 20(39.73) = 794.6.
The√standard deviation of the sums should be
1.1 20 = 4.919.
The distribution of the sums should be N(794.6, 4.919).
Example 3
According to cortland.org, the average yearly precipitation in
Cortland is 39.73 inches. Let us suppose that the standard
deviation is 1.1 inches.
I
Find the probability that it rains more than 1000 inches in
Cortland over the next two decades.
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The average over 20 years should be 20(39.73) = 794.6.
The√standard deviation of the sums should be
1.1 20 = 4.919.
The distribution of the sums should be N(794.6, 4.919).
P( rainfall > 1000) = normalcdf(1000, 10000, 794.6, 4.919)
=0