Download Central Limit Theorem

Introduction to
Probability & Statistics
The Central Limit Theorem
The Sample Mean
Suppose, for our die example, we wish to compute
the mean from the throw of 2 dice:
x
p(x)
1
2
3
4
5
6
1/ 1/ 1/ 1/ 1/ 1/
6
6
6
6
6
6
   xp( x)  3.5
Estimate  by computing the average of two throws:
X1  X2 1
X
 X1  X2 
2
2
Joint Distributions
x
1
p(x)
1/
X
X2
1
1/6
2
1/6
3
1/6
4
1/6
5
1/6
6
1/6
2
6
3
4
5
6
1/ 1/ 1/ 1/ 1/
6
6
6
6
6
X1
1
1/6
2
1/6
3
1/6
4
1/6
5
1/6
6
1/6
Joint Distributions
x
1
p(x)
1/
X
X2
1
1/6
2
1/6
3
1/6
4
1/6
5
1/6
6
1/6
2
6
3
4
5
6
1/ 1/ 1/ 1/ 1/
6
6
6
6
6
X1
1
1
1/6
2
1.5
1/36
1.5
1/6
1/36
2
1/36
2
1/36
3
1/36
3.5
1/36
3.5
4
1/36
4
1/36
1/36
4.5
1/36
1/36
1/36
1/36
5.5
1/36
5.5
1/36
1/36
5
5
5
1/36
1/36
1/36
1/36
1/36
4.5
4.5
4.5
1/36
1/36
1/36
1/6
4
4
4
6
3.5
1/36
1/36
1/36
1/6
3.5
3.5
3.5
5
3
1/36
1/36
1/36
1/6
3
3
3
4
2.5
1/36
1/36
1/36
1/6
2.5
2.5
2.5
3
2
1/36
6
1/36
1/36
Distribution of X
x
1
p(x)
1.5
1/
36
2/
36
2
2.5
3
3.5
4
3/
4/
5
6
36
36 /36 /36
5/
36
Distribution of X
0.20
0.15
0.15
0.10
0.10
0.05
0.05
0.00
0.00
2
3
4
5
5.5
6
4/
3
2/
1/
36 /36
36
36
Distribution of X
0.20
1
4.5
5
6
1
2
3
4
5
6
7
8
9
10
11
Distribution of X
n=2
n = 10
0.20
0.4
0.15
0.3
0.10
0.2
0.05
0.1
0.00
n = 15
0.4
0.3
0.2
0.1
6.0
5.5
5.0
4.5
4.0
3.5
3.0
2.5
2.0
1.5
1.0
0.0
6.0
11
5.5
10
5.0
9
4.5
8
4.0
7
3.5
6
3.0
5
2.5
4
2.0
3
1.5
2
1.0
0.0
1
Expected Value of X
 X1  X2  ...  Xn 
E [ X]  E 

n

1
 E [ X 1 ]  E [ X 2 ]  ...  E [ X n ]
n
Expected Value of X
 X1  X2  ...  Xn 
E [ X]  E 

n

1
 E [ X 1 ]  E [ X 2 ]  ...  E [ X n ]
n
1
  1   2  ...   n 
n
1
 n   
n
Variance of X
 X1  X2  ...  Xn 
 ( x)   



n
2
2
21
21
1 X 


X2   ...    Xn 
 
1  
n 
n 
n 
2
Variance of X
 X1  X2  ...  Xn 
 ( x)   



n
2
2
21
21
1 X 


X2   ...    Xn 
 
1  
n 
n 
n 
2
2
2
2
2
 1
    ( X1 )   ( X2 )  ...   ( Xn )
 n
2
 1 ( n 2 )   2
  
n
 n
Distribution of x
Recall that x is a function of random variables,
so it also is a random variable with its own
distribution. By the central limit theorem, we
know that
x  N ( , x )
where,
x 
x
n
Example
Suppose that breakeven analysis indicates we
must have average daily revenues of $500. A
random sample of 10 days yields an average
of only $450 dollars. What is the probability
we will not breakeven this year?
Example
Suppose that breakeven analysis indicates we must
have average daily revenues of $500. A random
sample of 10 days yields an average of only $450
dollars. What is the probability we will not
breakeven this year?
P {not breakeven}  P { < 500 x  450}
 P { -  > - 500 x  450 }
Example
P {not breakeven}  P { -  > - 500 x  450 }
Recall that
x  N ( , x )
Using the standard normal transformation
Z
x -

n
Example


 x -  450 - 500 
P {not breakeven}  P 






n
10



-50

 P Z 


10






Example
In order to solve this problem, we need to know the
true but unknown standard deviation . Let us
assume we have enough past data that a reasonable
estimate is s = 25.


-50 

Pr{not breakeven} = P Z 
P
25

10 
Z  - 1.58
= 0.943