1. Introduction to Probability & Statistics The Central Limit Theorem

2. The Sample Mean    x 1 2 3 4 5 6 p(x) 1/6 1/6 1/6 1/6 1/6 1/6  
Suppose, for our die example, we wish to compute
the mean from the throw of 2 dice: x
p(x) /6 1/6 1/6 1/6 1/6 1/6    xp x ( ) . 3 5
Estimate  by computing the average of two throws:   X   1 2

3. Joint Distributions x
p(x) /6 1/6 1/6 1/6 1/6 1/6 X1 X2 X

4. Joint Distributions x
p(x) /6 1/6 1/6 1/6 1/6 1/6 X1 X X2

5. Distribution of X x
p(x) / /36 3/ /36 5/36 6/ / /36 3/ / /36
Distribution of X
Distribution of X
0.00
0.05
0.10
0.15
0.20 1 2 3 4 5 6
0.00
0.05
0.10
0.15
0.20 1 2 3 4 5 6 7 8 9 10 11

6. Distribution of X n = 2 n = 10 n = 15 0.0 0.1 0.2 0.3 0.4 1.0 1.5 2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
0.00
0.05
0.10
0.15
0.20 1 2 3 4 5 6 7 8 9 10 11
0.0
0.1
0.2
0.3
0.4
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
n = 15

7. Expected Value of X   E X n [ ] .           1 n E X [ ] .2     1 2 n E X [ ] .

8. Expected Value of X       E X n [ ] .           1 n E2     1 2 n E X [ ] .     1 2 n  .    1 n 

9. Variance of X  ( ) . x X n                  n X . 21 ( ) . x X n                  2 1 n X .

10. Variance of X    ( ) . x X n                  n X2 1 ( ) . x X n                  2 1 n X .           1 2 n X  ( ) .        1 2 n ( )    2 n

11. 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
where,

12. 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?

13. Example P not breakeven x { } = < m 500 450 = - > P x { } m 500
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 x { } =
< m
500
450 = -
> P x { } m
500
450

14. Example P not breakeven { } = - > x m 500 450 Recall that
Using the standard normal transformation

15. Example P
not
breakeven { }

16. 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. P Z          50 25 10      P Z 1 58 .
Pr{not breakeven} =
= 0.943