Introduction to Probability & Statistics The Central Limit Theorem

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 1 2 3 4 5 6 p(x) 1/6 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

Joint Distributions x 1 2 3 4 5 6 p(x) 1/6 1/6 1/6 1/6 1/6 1/6 X1 X2 X

Joint Distributions x 1 2 3 4 5 6 p(x) 1/6 1/6 1/6 1/6 1/6 1/6 X1 X X2

Distribution of X x 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 p(x) 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36 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

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

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

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

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

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

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 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

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

Example P not breakeven { }

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