course Mth 152 assignment #011011. `query 11
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20:11:51 `questionNumber 110000 **** Query 12.6.6 rnd # table to simulate 50 one-and-one foul shooting opportunities if 70% prob of success; 2 shots Give the results of your tally. How does your empirical probability compare with the theoretical probability?
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RESPONSE --> For zero I got 11/49=.2245 for one I got 10/49=.2041 for two I got 28/49=.5714 Susan has a 70% foul shot record that means she has 30% failures, so misses on 1st shot would be .3, then for one hit and one miss it would be .7*.3=.21, and two hits would be .7*.7=.49. I'm not sure where to go from here, but it took a long time to do this one.
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20:12:00 `questionNumber 110000 ** In 1-and-1 shooting you only get a second shot if you make the first. So there are 3 possibilities: miss the first, don't get another shot make the first, get another shot and make it make the first, get another shot and miss it prob of 0 = prob of miss on first shot = .3 prob of 1 = prob of hit on first and miss on 2d = .3 * .7 = .21 prob of 2 = prob ot hit * prob of hit = .49. 'Hits' happen with 70% or .7 probability, misses with probability 30% or 3. The theoretical probability of 2 misses is probability of miss * probability of miss = .30 * .30 = .09. The theoretical probability of 2 miss and 1 hit is probability of miss * probability of hit + probability of miss *hit probability of miss = .30 * .70 + .70 * .30 = .21 + .21 = .42. The theoretical probability of 2 hits is probability of hit * probability of hit = .70 * .70 = .49. Note that these probabilities add up to .09 + .42 + .49 = 1, as they must since these three events cover all possibilities. To use the table, randomly pick a starting point. Let numbers 1-7 correspond to making the free throw, with 8, 9 and 0 corresponding to misses. Go down the list, or across the list in an order you decided before looking at the list. Read two digits from the list and see if they correspond to two 'hits', two 'misses' or a 'hit' and a 'miss'. Record your result as 'hit-miss', 'miss-hit', 'miss-miss' and 'hit-hit'. Read two more digits and record your result as 'hit-miss', 'miss-hit', 'miss-miss' and 'hit-hit'. Continue until you have the required number of results. Tally how many times you got 0 'hits', 1 'hit', 2 'hits' etc.. Any outcome that starts with a 'miss' corresponds to zero point. 'Hit-miss' corresopnds to 1 point and 'hit-hit' corresponds to 2 points. Determine the percent of time you got each number of points, and compare to the theoretical probabilities .09, .42 and .49. *&*& **
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RESPONSE --> ok
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20:19:11 `questionNumber 110000 Query 12.6.12 rnd walk start N then right, left or straight with prob 1/2, 1/6, 1/3; 1 st 2 columns of table
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RESPONSE --> I set up proportions and got 1/2 to be 3/6; 1/6 is, of course 1/6; 1/3 is 2/6. After this I had no idea what to do next!
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20:19:22 `questionNumber 110000 ** Your probabilities are given as 1/2, 1/6 and 1/3. These can all be expressed in terms of the common denominator 6: 1/2 = 3/6, 1/6 = 1/6, 1/3 = 2/6. So a move to the right has 3 chances out of 6, a move to the left has 1 chance out of 6 and a move straight has 2 chances out of 6. You can simulate this by letting the three digits 1, 2, 3 stand for a move to the right, the single digit 4 for a move to the left and the two digits 5, 6 for a straight move. The remaining digits 0, 7, 8, 9 don't stand for anything, and if you land on one of these numbers you just move to the next number. So according to your the first two columns of you table, how many times do you move to the right, how many to the left, how many straight and where do you end up? **
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RESPONSE --> ok
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20:20:14 `questionNumber 110000 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> This was the hardest section for me as you can see by my finishing it so late.
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