| 
                    
                       
                     Lesson 4.2 Expected Value
                    Law of Large numbers
                       
                    The mean of random variable X is μ. If we do an
                      experiment over and over again, the average gets
                      closer and closer to μ. This is known as the Law
                      of Large Numbers. 
                    Example: If we toss a fair coin 20,000 times and
                      let X = the number of heads, then the mean of X,
                      μ, is very close to 1/2. 
                    
                    Expected Value
                    The expected value is known as "the long-term"
                      average or mean, μ. This means that over the long
                      term of doing an experiment over and over, you
                      would expect this average every time you perform
                      the experiment. 
                    To find the expected value or mean, μ, simply
                      multiply each value of the random variable by its
                      probability and add the products. 
                    Example: In Section 4.1, a PDF table was created.
                      It is shown below.  
                    X = the number of patients out of two cured 
                    P(a cure for one patient) = 5/6 and  
                    P(no cure for one patient) = 1/6. 
                    
                      
                        
                          
                            | 
                               X 
                             | 
                            
                               P(X) 
                             | 
                           
                          
                            | 
                               0 
                             | 
                            
                               P(X=0) = 1/36 
                             | 
                           
                          
                            | 
                               1 
                             | 
                            
                               P(X=1) = 10/36 
                             | 
                           
                          
                            | 
                               2 
                             | 
                            
                               P(X=2) = 25/36 
                             | 
                           
                        
                       
                    
                    To find the expected value, add another column
                      and label it X*P(X). Calculate each product
                      (X*P(X)) and then add (sum) the products for the
                      expected value. 
                    
                       
                        
                          
                            | 
                               X 
                             | 
                            
                               P(X) 
                             | 
                            
                               X*P(X) 
                             | 
                           
                          
                            | 
                               0 
                             | 
                            
                               P(X=0) = 1/36 
                             | 
                            
                               (0)(1/36) = 0 
                             | 
                           
                          
                            | 
                               1 
                             | 
                            
                               P(X=1) = 10/36 
                             | 
                            
                               (1)(10/36) = 10/36 
                             | 
                           
                          
                            | 
                               2 
                             | 
                            
                               P(X=2) = 25/36 
                             | 
                            
                               (2)(25/36) = 50/36 
                             | 
                           
                        
                       
                    
                    Add the last column: 0 + 10/36 + 50/36 = 60/36 =
                      1.67. The expected value is one and two-thirds
                      patients cured. That means that over the
                      long-term, we would expect 1 2/3 patients out of 2
                      patients to be cured. Another way to look at it is
                      as follows. If we surveyed 6 patients who had
                      taken the drug, we would expect 5 to be cured. 
                    
                    Expected Value Problem
                     Example 
                    The following game problem is an example of an
                        expected value problem. Close the window
                      when you are finished viewing the example. You
                      will return here. 
                    Please continue to the next section of this
                      lesson.  
                      
                    
                      
                    Up » 4.1
                        Discrete Probability »
                      4.2 Expected Value » 4.3 Binomial Probability »
                        4.4 Poisson Probability 
                    Lesson 1
                      | Lesson 2 | Lesson 3 | Lesson 4 | Lesson 5 | Lesson 6 | Lesson 7 | Lesson 8 | Lesson 9 | Lesson 10 | Lesson 11 | Lesson 12  
                   |