Download The Central Limit Theorem

The Central Limit Theorem
You may be wondering why Normal distributions seem to have a special place
among the probability distributions we have studied. Indeed, the Empirical Rule that we
looked at several weeks ago came directly from a normal distribution. The reason that
the normal distribution occurs so frequently in the real world is because of the Central
Limit Theorem. We begin by stating a “business student” version of this theorem.
Let X be computed by taking a simple random sample of size n from a
population with mean µ and variance σ2. Then for “large n,” X will have an
approximate Normal distribution.
As is always the case when sampling from an infinite population or from a finite
population of size N where N >> n,
E[ X ]  
and
Var ( X )  
2
n
Since the mean and variance uniquely specifies the normal distribution in question, you
have everything you need to know about the distribution of X . In particular, you can
compute probabilities about X .
For a good normal approximation, “large n” means either
(1) n > 1 when sampling from a population which itself has a normal distribution,
or
(2) n > 30 when sampling from populations with arbitrary distributions.
Now, what does the Central Limit Theorem say in general? Whenever you are
adding random variables (typically assumed to have an identical distribution) together,
like
S  X1  X 2  X 3    X n
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the sum S will have an approximate normal distribution (when properly normed) as long
as there are not terms in the sum that dominate the sum. In essence, this says that the
sum obtained by adding a bunch of numbers of approximately the same size will have an
approximate normal distribution.
How does this apply to X ?
Example 1
Incomes in a community are normally distributed with µ = $30,000 and σ =
$4,000.
Q1: If we select an income at random, what is the probability that the income exceeds
$32,000?
Q2: Now suppose we take a random sample of size 4. What is the probability that the
average income in the sample exceeds $32, 000?
Let X = the average income in the sample of size 4.
What is the distribution of X ?
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We have
X 
X 
What is the probability we wish to compute?
In order to standardize X , we need to subtract its mean and divide by its standard
deviation:
Z
X  X
X

X 

n
Thus


 X   32,000  30,000 
P{ X  32,000}  P 



4000


n
4


2000 

 P Z 

2000 

 P{Z  1}
 0.1587
Example 2
A company which produces sealing lids considers its production process to be
properly adjusted if the average diameter of a lid is 4 inches. The standard deviation of
the diameters is 0.012 inches. Someone has suggested that the machine is in need of
adjustment, so the foreman has taken a sample of 100 lids and has found that
x  4.003 inches
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Everyone seems to believe that this is really close to 4 inches and production should
continue. Should the production facility be shut down to make the adjustment?
Assume for a moment that µ = 4 inches; that is, there is no need to make an adjustment.
The sample mean we saw was a little larger than this. If, in fact, µ = 4 inches, what is the
probability of seeing a sample mean as large as we saw?


 X   4.003  4.000 
P{ X  4.003}  P 


0.012


n
100 

.003 

 P Z 

.0012 

 P{Z  2.5}
 0.0082
What do you think?
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