Elementary Statistics and Inference. Elementary Statistics and Inference. 17 Expected Value and Standard Error. 22S:025 or 7P:025.

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1 Elementary Statistics and Inference S:05 or 7P:05 Lecture Elementary Statistics and Inference S:05 or 7P:05 Chapter 7 A. The Expected Value In a chance process (probability experiment) the outcomes of the chance process vary, but cluster around the expected value or the average. Example: Toss a coin 00 times. P( H ) = on each toss you would expect about 50% of the outcomes to be heads this is the expected value. Suppose in 00 tosses you obtained 57 heads, the difference between 57 and 50 is called chance error (+7). 3

2 Example: Suppose you selected 00 draws from the box, with replacement: 5 You would expect about ¾ of the 00 draws to be a, and about ¼ of the draws to be a 5. The expected sum would be (5 x 5) + (75 x ) = = The formula for the Expected Sum is: E( = (average of the number in the box) (number of draws) E ( sum ) = 00 = 00 4 E( = n (avg of box) = n X Note: see description on page 89 5 Example (page 89): Example : Suppose you are going to Las Vegas to play Keno. Your favorite bet is a dollar on a single number. When you win, they give you the dollar back and two dollars more. When you lose, they keep the dollar. There is a chance in 4 to win. About how much should you expect to win (or lose) in 00 plays, if you make this bet on each play? $.00 -$.00 -$.00 -$.00 P( W ) = 4 n=00 plays 6

3 average of the box = + ( 3) = $.5 4 E(= 00 avg of box = 00 (-$ $.5) = $ You would expect to lose $5 in 00 plays. 7 Exercise Set A (p. 90) #,, 3, 4, 5. Find the expected value for the sum of 00 draws at random with replacement from the box (a) 0 6 (b) (c) (d) Someone is going to play roulette 00 times, betting a dollar on the number 7 each time. Find the expected value for the net gain. (See pp ) 9 3

4 6. A game is fair if the expected value for the net gain equals 0: on the average, players neither win nor lose. A generous casino would offer a bit more than $ in winnings if a player staked $ on red-and-black in roulette and won. How much should they ypay to make it a fair game? (Hint: Let X stand for what they should pay. The box has 8 tickets X and 0 tickets -$. Write down the formula for the expected value in terms of X and set it equal to 0.) 0 8 X 0 Expected gain = = X 0 = 0 8X = 0 X = $. B. The Standard Error of the Sum Given the box: Avg of box = 5 / 5 = 3 4

5 In 5 draws with replacement, Expected Sum = E(=n avg of box E(=5 3=75 The actual sum will be Sum=expected value + chance error The chance error is a function of the standard deviation of the box. The chance error, called Standard Error (SE) is: SE = n (SD of box) See Note p. 9 3 In the box above, the SD of box =. (0 3) + ( 3) + (3 3) + (4 3) S = S = = = S = SE = 5 () = 0 + (6 3) 4 For the box example n = 5 draws with replacement The E( = n avg box = 5(3) = 75 SE( = n SD of box = 5() = 0 5 5

6 In 5 draws with replacement, we would expect the sum of the draws to be 75 give or take 0. Note: The sum of draws is likely to be around the expected value, give or take the standard error. Note: Observed values in a chance process are rarely more than or 3 standard errors away from the Expected Value. 6 Exercise Set B (p. 93) #,,

7 9 C. The Normal Curve For a large number of draws from a box, with replacement, the sum of the draws is approximately normally distributed. 0 Example: Suppose 5 draws with replacement are made from the box, with tickets as shown: X = avg of box = 3, S = SD of box = E( = n avg of box = 5 (3) = 75 SE ( = n SD of box = 5() = 0 Now the sums are approximately normally distributed with mean = 75, and S = 0. Find probability (chance) that the sum for any 5 draws will be between 50 and 00. 7

8 98.76% SE(=0 50 E(= sum sum mean Z = SE Using Normal Curve table we find 98.76% of the scores between ±.50 Standard Deviations from the mean. Example. In a month, there are 0,000 independent plays on a roulette wheel in a certain casino. To keep things simple, suppose the gamblers only stake $ on red at each play. Estimate the chance that the house will win more than $50 from these plays. (Red-or-black pays even money, and the house has 0 chances in 38 to win. Solution: What is probability that casino will gain $50.00 or more from 0,000 plays of roulette. 3 0 $ $ avg of box = expected gain on one play = = $ (.05) + 8(.05) S = 38 0(.95) + 8(.05) S = 38 S =

9 SD of box = $.998 $.00 E( = n avg of box E( = 0,000(.05) = $ SE( = n SD of box = 0,000() = 00 5 Find Probability (Chance) that sum is greater than $50. SE(=00 50 E(=500 0 X Z Z = = = Area between ±.50 = 98.76% Area less than -.50 = =.6% So about 99.3% for casino to win more than $50 in 0,000 roulette plays. Exercise Set C (pp ) #,, 3, 4, 5 7 9

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