Homework 5 Solutios p329 # 12 No. To estimate the chace you eed the expected value ad stadard error. To do get the expected value you eed the average of the box ad to get the stadard error you eed the stadard deviatio of the box. You ca do it without the actual umbers as log as you have the average ad the stadard deviatio. Sice the umbers are bouded betwee -10 ad 10 you ca assume that with 1000 draws the probability histogram for the sum will be approximately ormal. To use the ormal distributio you eed to kow the expected value ad the SD of the the ormal distributio (also called the SE). The expected value of the sum of 1000 draws is just 1000(AVG of the box). The SD for the sum is 1000(SD of the box). So you do have eough iformatio. p352 # 11 I this questio the box that the car is i is fixed, sice the car has already bee hidde. Suppose the car is i box B. Let X i be a radom variable that couts whether the i-th coi flip is heads. The we are lookig at the sum B+X 1 +X 2 + +X 100. So if the sum is 65 the we have 65 = B +X 1 + +X 100 so we ca solve for B to get B = 65 (X 1 +X 2 + +X 100 ). Now we are lookig for the probability histogram for this sum. We ca compute the expected value: 1
E(65 (X 1 + + X 100 )) = 65 E(X 1 + + X 100 ) = 65 100E(X 1 ) = 65 50 = 15 Sice this sum is goig to follow a ormal distributio (by the Cetral Limit Theorem) the Porsche would be most likely to be i the boxes closest to the expected value. So you should look i 10,11,12,13,14,15,16,17,18,19 ad 20. Here the expected value for B is computed just like i the last questio, which comes out to be 45. So you would look i boxes 40, 41, 42, 43, 44, 45, 46, 47, 48, 49 ad 50. Part (c) I geeral if the sum is N you would look i boxes N 55 to N 45. Part (d) To figure out the chaces you eed the SD. Remember, that the SD is ot chaged whe we add a costat to each elemet i a list. So we ca compute the SD as follows: SD(65 (X 1 + + X 100 )) = 65 SD(X 1 + + X 100 ) = 100SD(X 1 ) = 100 1 2 = 5 The chace the would be the area uder the ormal curve from N 55.5 to N 44.5. I SU these poits become 1.1 ad 1.1. So the area uder the curve, ad thus the chace to get the Porsche, is 72.87%. p371 # 1 Number of heads Percet of heads Number of tosses Expected value SE Expected value SE 100 50 5 50% 5% 2,500 1,250 25 50% 1% 10,000 5,000 50 50% 0.5% 1,000,000 500,000 500 50% 0.05% 2
p372 # 5 We are lookig at the sum of 50 radom guests. Let X i deote the weight of the i-th guest. The we are tryig to estimate the sum X 1 + + X 50. We ca compute the expected value ad the SD as usual: E(X 1 + + X 50 ) = 50E(X 1 ) = 50(150) = 7500 SD(X 1 + + X 50 ) = 50SD(X 1 ) = 50(35) 247.5 Now 4 tos is 8000 pouds, so we are iterested i the chace that this sum will be over 8000. I SU 8000 is 8000 7500 2, so we are lookig for the area to the right of 2 i the 247.5 ormal distributio. This is about 2.5%. Souds like the tower of terror :P p383 # 3 The expected value i 100 draws is 1 red marble. (E(X 1 + +X 100 ) = 100E(X 1 ) = 100 1 The SE is: SE = SD(X 1 + + X 100 ) = 1 99 100 100 100 1 You ca t draw fewer tha 0 red marbles, so the chace is 0. Part (c) The area i the ormal approximatio that correspods to drawig less tha 0 red marbles is the area left of 0.5. I SU this is 0.5 1 = 1.5. This is 6.68%. (The book gets its 1 aswer by usig 0 istead of 0.5, but here it is importat to pay attetio to the edpoits, because it chages the aswer so much sice the SD ad expected value are so small.) Part (d) No. There have t bee eough draws to make the ormal curve look like the probability histogram. There are two factors that affect whether the ormal curve approximates the probability histogram; how ubalaced the box is ad how may draws have happeed. If the box is really ubalaced like this oe you eed more draws for the probability histogram to be ormal. I this case with 10000 draws it would be much closer. 100 ). 3
p383 # 4 False. That is how you would regularly compute the cofidece iterval, but as is show i the last problem, the ormal curve does ot approximate the probability histogram well here. So we ca t compute the 95% cofidece iterval i this way :( p391 # 2 Here you just compute the percet i the sample, which is also 47%. It is likely to be off by (0.47)(0.53) 1 SE, which is 2.2%. 500 The previous sample is a simple radom sample, so a 95% cofidece iterval ca be computed as usual. We computed the SE i the previous step, so a 95% cofidece iterval is 47% ± 4.4%. p392 # 4 I the households iterviewed 379 had oe or more cars. This is 75.8%. The SE is (0.758)(0.242) = 1.9%. 500 p393 # 13 I this box P(1) = 0.25, P(2) = 0.5 ad P(5) = 0.25. So the probability histogram is (ii). The probability histogram for the sum of 100 draws should look like a ormal distributio accordig to the Cetral Limit Theorem, so it is (iii). (i) is irrelevat. Worksheet 1 We kow their wiig proportio is curretly 0.375. So the SE for this after 16 games is (0.375)(0.625) 16 = 0.121. The 95% cofidece iterval is just the observed proportio ± 2SE. This is 0.375 ± 0.242. 4
The 99.7% cofidece iterval is the observed proportio ±3SE, which is 0.375 ± 0.363. Part (c) Fially a 50% cofidece iterval is the observed proportio ±0.68SE, which is 0.375±0.082. Worksheet 2 This questio is a tricky oe. It is easiest if we thik about it like questio 4 from last week. Let X i be a radom variable that couts whether or ot the Red Sox wo the i-th game ad similarly let Y i cout wis for the Yakees. Now let Z i = Y i X i. We are iterested i how may games we eed to play for a 95% cofidece iterval for the variable Z 1+ +Z to ot iclude 0. This meas that we wat 0 = E( Z 1+ +Z ) 2SD( Z 1+ +Z ). The first term ca be rewritte: E( Z 1 + + Z ) = E( (Y 1 X 1 ) + + (Y X ) ) = E( Y 1 + + Y ) E( X 1 + + X ) = 0.500 0.375 = 0.125 Ufortuately the SD is a foul beast. First we compute SD(Z 1 ). Z 1 has outcomes -1, 0 ad 1. -1 happes whe the Yakees lose ad the Red Sox wi, which happes with probability (0.5)(0.375) = 0.1875. 0 happes whe they both wi or both lose, which happes with probability (0.5)(0.375)+(0.5)(0.625) = 0.5. Fially 1 happes whe the Yakees wi ad the Sox lose, which happes with probability (0.5)(0.625) = 0.3125. So ow we ca compute the expected value for Z 1 : E(Z 1 ) = 1(0.1875) + 0(0.5) + 1(0.3125) = 0.125 Note: there is a much easier way to compute this. E(Z 1 ) = E(Y 1 X 1 ) = E(Y 1 ) E(X 1 ) = 0.5 0.375 = 0.125. Now we ca compute the SD... ugh :( Z 1 has outcomes -1, 0 ad 1, so (Z 1 E(Z 1 )) 2 has outcomes ( 1.125) 2 = 1.26563, ( 0.125) 2 = 0.015625 ad 0.875 2 = 0.765625. So SD(Z 1 ) = E((Z 1 E(Z 1 )) 2 ) = 1.26563(0.1875) 0.015625(0.5) + 0.765625(0.3125) = 0.484375 = 0.695971. Ok, ow we ca solve the problem. We are iterested i the differeces of their wiig percetages. That is give by the value Z 1+ +Z. Now we compute the expected value ad 5
SD of this: E( Z 1 + + Z ) = E(Z 1 +... Z ) = E(Z 1) +... E(Z ) = E(Z 1) = E(Z 1 ) = 0.125 SD( Z 1 + + Z ) = SD(Z 1 + + Z ) SD(Z1 ) = = SD(Z 1) = 0.695971 So after games the 95% cofidece iterval for the differece of the wiig percetages is 0.125 ± 2 0.695971. We are iterested i whe the left edpoit of this iterval is 0, i.e. 0 = 0.125 1.39. We solve this for ad get = 124. Whew! Worksheet 3 If they keep wiig at a.375 rate the their 95% cofidece iterval would be.375 ± (0.375)(0.625) 2. So we wat to solve for whe the right edpoit for this iterval is 0.5. This happes whe = 60. Worksheet 4 For the Red Sox to wi 90 games they would eed to wi 84 out of the remaiig 146 which is a proportio of 0.575. So if their true wiig proportio is this or less I will lose. Right ow we have a sample of 16 games. For this sample the observed wiig proportio is 0.375 with a SE of 0.121. I SU 0.5 becomes 0.575 0.375 = 1.65. So we are lookig for the area to 0.121 the left of 1.65 o the ormal distributio, which is about 95%. Thus we ca say that the Red Sox s true wiig percetage is less tha 0.575 with 95% cofidece. For the bet this meas that the expected value of the bet is 1000(0.05) 2(0.95) = 48.1. So I should take it! 6
Worksheet 5 After 80 games the Red Sox are still wiig at a rate of 0.375. Now our SE is chaged (0.375)(0.625) though; it is = 0.05. To get a 99.99% cofidece iterval we eed to take 3.9 80 SE s, which gives us the iterval 0.375 ± 0.195. Worksheet 6 First let s compute our cofidece that the Red Sox wiig percetage is greater tha 0.500. This is the area to the right of 0.500 i SU, which is 0.500 0.375 = 2.5. The area is the 0.62%. 0.05 (This meas that i oly 0.62% of samples is the true proportio 2.5 SEs greater tha the sample proportio.) Now we compute the same thig for the Yakees. First their wiig percetage is 46 (0.575)(0.425) 0.575. Now, their SE is = 0.06. So i SU 0.5 is 0.500 0.575 = 1.25. So, i 80 0.06 the Yakees case we are 10.6% cofidet that their true wiig percetage is less tha 0.500. These two evets are idepedet, so to fid our cofidece that both would occur at the same time we ca just multiply our cofidece i each evet. So we are 100(0.0062)(0.106) = 0.07% cofidet that the Yakees wiig percetage is less tha 50% ad the Red Sox s is greater tha 50%. 80 = 7