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Category > Math Posted 21 Apr 2017 My Price 8.00

mean for a random variable

Threads Main View

To find the mean for a random variable we multiply each possibility by its probability and then we add them together.  The symbol for the mean of a random variable is E(X), pronounced "E of x" the short way and "Expected Value" the long way.

 

Like this:

 

X = {1, 2, 3, 4, 5, 6}

 

P(1) = 0.400

P(2) = 0.200

P(3) = 0.100

P(4) = 0.150

P(5) = 0.050

p(6) = 0.100

 

E(X) = ∑  [x * P(x)]

 

E(X) = 1*0.400 + 2*0.200 + 3*0.100 + 4*0.150 + 5*0.050 + 6*0.100

      =  0.400    + 0.400     + 0.300    + 0.600     + 0.250    + 0.600

      =  0.800                   + 0.900                    + 0.850

      =  1.700                                                 + 0.850

      =  2.550

 

Thus the mean is 2.550.  Notice that this isn't even one of the values in the set { }.  That's okay.  We still use it and say that on average we can expect to have 2.550 or whatever these things are we're discussing.

 

Please invent a random variable (be sure it follows the rule that the probabilities must add to 1).  Or find a random variable in the text.

 

Then find E(X) showing all the steps.Threads Main View

To find the mean for a random variable we multiply each possibility by its probability and then we add them together.  The symbol for the mean of a random variable is E(X), pronounced "E of x" the short way and "Expected Value" the long way.

 

Like this:

 

X = {1, 2, 3, 4, 5, 6}

 

P(1) = 0.400

P(2) = 0.200

P(3) = 0.100

P(4) = 0.150

P(5) = 0.050

p(6) = 0.100

 

E(X) = ∑  [x * P(x)]

 

E(X) = 1*0.400 + 2*0.200 + 3*0.100 + 4*0.150 + 5*0.050 + 6*0.100

      =  0.400    + 0.400     + 0.300    + 0.600     + 0.250    + 0.600

      =  0.800                   + 0.900                    + 0.850

      =  1.700                                                 + 0.850

      =  2.550

 

Thus the mean is 2.550.  Notice that this isn't even one of the values in the set { }.  That's okay.  We still use it and say that on average we can expect to have 2.550 or whatever these things are we're discussing.

 

Please invent a random variable (be sure it follows the rule that the probabilities must add to 1).  Or find a random variable in the text.

 

Then find E(X) showing all the steps.

Answers

(11)
Status NEW Posted 21 Apr 2017 01:04 AM My Price 8.00

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