(New page: P(Xi) = .2 (equal chance of picking each case) LET: X1 = 1 X2 = 10 X3 = 100 X4 = 1000 X5 = 10000 REMAINING AVERAGE VALUE IF THE FIRST CASE PICKED IS: X1 - $2777.50 X2 - $2775.25 X3 - $...)
 
 
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P(Xi) = .2 (equal chance of picking each case)
 
P(Xi) = .2 (equal chance of picking each case)
  
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THEREFORE (BY DEFINITION)
 
THEREFORE (BY DEFINITION)
 
E[XI] = .2 * (2777.50 + 2775.25 + 2752.75 + 2527.75 + 277.75) = $2222.20
 
E[XI] = .2 * (2777.50 + 2775.25 + 2752.75 + 2527.75 + 277.75) = $2222.20
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......I THINK

Latest revision as of 17:49, 22 September 2008

P(Xi) = .2 (equal chance of picking each case)

LET: X1 = 1 X2 = 10 X3 = 100 X4 = 1000 X5 = 10000

REMAINING AVERAGE VALUE IF THE FIRST CASE PICKED IS: X1 - $2777.50 X2 - $2775.25 X3 - $2752.75 X4 - $2527.75 X5 - $277.75

THEREFORE (BY DEFINITION) E[XI] = .2 * (2777.50 + 2775.25 + 2752.75 + 2527.75 + 277.75) = $2222.20

......I THINK

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Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

Dr. Paul Garrett