7.5. Sums and Differences of Independent Random Variables http://www.ck12.org
The implications of these can be better understood if you return to example 1.
Example:
The casino has decided to ’triple’ the prizes for the game being played. What are the expected winnings for a person
who plays one game? What is the standard deviation?
Solution:
Recall that the expected value was $2.60 and the standard deviation was $6.46. The simplest way to calculate the
expected value of the tripled prize is 3($2. 60 ), or $7.80, with a standard deviation of 3($6. 46 ), or $19.38. Herec= 0
andd=3. Another method of calculating the expected value would be to create a new table for the tripled prize:
TABLE7.18:
Winnings,x Probability,p
−$2. 00 0. 30
$0. 00 0. 40
$6. 00 0. 20
$20. 00 0. 10New Table
TABLE7.19:
Original Winnings,x New Winnings, 3x Probability,p
−$2. 00 −$6. 00 0. 30
$0. 00 $0. 00 0. 40
$6. 00 $18. 00 0. 20
$20. 00 $60. 00 0. 10The calculations can be done using the formulas or by using the graphing calculator.
Using the graphing calculator:
Notice that the same results are obtained.
This same problem can be changed again in order to introduce the addition and subtraction rules for random
variables. Suppose the casino wants to encourage customers to play more, so begins demanding that customers
play the game in sets of three. What are the expected value (total winnings) and standard deviation now?
Solution:
LetX,YandZrepresent the total winnings on each game played. If this is the case, thenμX+Y+Zis the expected
value of the total winnings when three games are played. The expected value of the total winnings for playing one
game was $2.60 so for three games the expected value is:y