Mean and Standard Deviation

Definition

The mean of a probability distribution is given by

Since each x is multiplied by its probability, x values with small probabilities contribute less to the mean while x values with larger probabilities contribute more to the mean. In this way we get an average value for x.

Definition

The expected value of a discrete random is denoted by E and is given by

Note that the expected value and the mean are the same.

Definition

The variance for a probability distribution is given by

The square root of this gives the standard deviation.

Definition

The standard deviation for a probability distribution is given by

Example 1

Find the mean (expected value), variance, and standard deviation for the probability distribution given below.

Solution .

The details of the calculations are shown in the work that follows.

 x      prob(x)       x*prob(x)    ((x - mean)^2)*prob(x)  

---------------------------------------------------------------

0     .3             0                 .50700                    

1.    .25            .25               .22500E-1                 

2.    .3             .6                .14700                    

3.    .15            .45               .433500                   

SUM   1.00           1.30              1.110000                  

                     mean              variance                  

                                       1.053565375               

                                       standard deviation        

Example 2

Find the mean (expected value), variance, and standard deviation for the probability distribution given below.

Solution .

Here are the details.

 x      prob(x)       x*prob(x)    ((x - mean)^2)*prob(x)  

---------------------------------------------------------------

0     .25            0                 .250000                   

1.    .5             .5                0                         

2.    .25            .50               .250000                   

SUM   1.00           1.00              .500000                   

                     mean              variance                  

                                       .7071067812               

                                       standard deviation        

Example 3

Suppose you have five marbles in an urn and that each marble has a unique integer on it from 1 through 5. In this experiment a marble is randomly selected and the number on it is recorded and the marble is placed back in the urn. Find the mean (expected value), variance, and standard deviation for this probability distribution.

Solution .

In this problem x takes on the values of 1, 2, 3, 4, and 5. Since each marble has the same chance of being selected ( we are replacing the marbles), P(x) = 1/5 = .2 for each x. The calculations for the mean, variance, and standard deviation are given below.

 x      prob(x)       x*prob(x)    ((x - mean)^2)*prob(x)  

---------------------------------------------------------------

1.    .2             .2                .800                      

2.    .2             .4                .200                      

3.    .2             .6                0                         

4.    .2             .8                .200                      

5.    .2             1.0               .800                      

SUM   1.0            3.0               2.000                     

                     mean              variance                  

                                       1.414213562               

                                       standard deviation