\fPSTAT W 120A: Solutions for Assignment 2.b August 18, 2015 Lesson 7 Poisson Approximation 1. A survey of 3300 people asked them whether they liked apple pie. Suppose that the people were all chosen independently with a 99.98% chance of liking apple pie. (a) The normal approximation is not appropriate in this instance because the expected number of successes is np = 3300(0.9998) = 3299.34 which is very close to n. In particular, n(1 p) = 3300(0.0002) = 0.66 < 9. (b) The probability that all 3300 people will say they like apple pie is P(3300) = p3300 = (0.9998)3300 = 0.5168 sh is ar stu ed d vi y re aC s o ou urc rs e eH w er as o. co m (note that our Poisson approximation is en(1p) = e0.66 = 0.51685 as well.) (c) To use a Poisson approximation in this problem, we have to turn it around from the way that we usually do the approximation. Instead of a small chance of success, there is a large chance of success. I think the easiest thing to do is turn the events around to count the number of failures instead of successes. For this problem, the mean number of failures is = n(1 p) = 3300(0.0002) = 0.66. The event that 3297 of the people will say that they like apple pie is equivalent to having 3 people say they don't like apple pie. 3 e 3! (0.663 0.66 = e 6 = 0.02477 P(3) 2. A jar contains half green marbles and half red marbles. We draw out three marbles without replacement. (a) The probability that all three marbles are red if we started out with 6 of each color in the jar is \u0012 \u0013\u0012 \u0013\u0012 \u0013 6 5 4 P(RRR) = 12 11 10 1 = 0.0909 = 11 Th (b) The probability that all three marbles are red if we started out with 60 of each color in the jar is \u0012 \u0013\u0012 \u0013\u0012 \u0013 60 59 58 P(RRR) = 120 119 118 \u0012 \u0013 1 58 = = 0.1218 4 119 (c) The probability that all three marbles are red if we started out with 600 of each color in the jar is \u0012 \u0013\u0012 \u0013\u0012 \u0013 600 599 598 P(RRR) = 1200 1199 1198 \u0012 \u0013 1 598 = = 0.1247 4 1199 (d) Under independence, the probability would be 1/8 = 0.125. We get closer and closer to that answer as the number of marbles in the jar increases. Thus, a large population makes the experiment look more like a binomial experiment. https://www.coursehero.com/file/11948315/PSTAT-120A-Assignment-2b-Answers/ Lesson 8 Random Variables 3. I roll a fair six-sided and let the random variable X be the number that comes up on the die. (a) P (3) = 1 6 = 0.1667 (b) P{X 4} = P (1) + P (2) + P (3) + P (4) = 0.6667 (c) First, we solve the quadratic inequality, (X 3)2 3 3X 3 3 3 3X 3+ 3 1.268 X 4.732 We use this to argue sh is ar stu ed d vi y re aC s o ou urc rs e eH w er as o. co m P{(X 3)2 3} = P (2) + P (3) + P (4) = 0.5 4. Suppose that X is a random variable with density x P (x) 0 0.00032 1 0.0064 2 0.0512 3 0.2048 4 0.4096 5 0.32768 and that the random variable Y = 9X 2 36X + 4. (a) The values that Y takes are x y 0 4 1 -23 2 -32 3 -23 4 4 5 49 which means that 32 Y 49 or specifically Y {32, 23, 4, 49}. (b) From the table in part (a), we can see that P{Y = 4} = P{X = 0} + P{X = 4} = 0.00032 + 0.4096 = 0.40992 (c) We can also see P{Y = 0} = 0 because none of the values of X lead to this value of Y . Questions Th 5. We perform an experiment that consists of independent trials where the probability of success is p = 0.25. p (a) p Let X be the number of successes in 800 trials. The mean is = 200 and = np(1 p) = 200(0.75) = 12.2474. Using the Normal approximation, \u0012 \u0013 220.5 200 P{X > 220} 1 150 = 1 (1.674) = 1 0.95292 = 0.0471 (b) Let Y be the proportion of trials that come out a success. In other words, Y = X/800. The event {Y > 0.28} is the same as {X > 224} = {X > 224.5}. Thus, P{Y > 0.28} = P{X > 224.5} \u0012 \u0013 224.5 200 1 150 = 1 (2.0004) = 1 0.9773 = 0.0227 https://www.coursehero.com/file/11948315/PSTAT-120A-Assignment-2b-Answers/ (c) Let S bethe number of successes in 8000 trials. The expected number of successes is = 2000 and = 1500 = 38.72983. Using the Normal approximation, \u0012 \u0013 2020.5 2000 P{S > 2020} 1 1500 = 1 (0.52931) = 1 0.7017 = 0.2983 (d) Let T be the proportion of trials that come out a success. In other words, T = S/8000. The event {T > 0.28} is the same as {S > 2240} = {S > 2240.5}. sh is ar stu ed d vi y re aC s o ou urc rs e eH w er as o. co m P{T > 0.28} = P{S > 2240.5} \u0013 \u0012 2240.5 2000 =1 1500 = 1 (6.21) 0 If you put a large number into the Normal CDF calculator, you will get out 1 though the true probability is actually slightly less than that. As a result, it is not impossible for T > 0.28 only very unlikely. My computation puts the answer at nearly 4.3 1010 . (e) Notice that while S is less likely to be within 20 of the mean than X, it is much more likely to be within 3% of the proportion. The answer then depends on what you mean by closer. The way it is written I think the statement is incorrect. The chance of S being close to its expectation is smaller (this can be seen in that the standard deviation goes up from 12.2 to 38.7.) 6. Suppose that 47% of registered voters approve of the job President Obama is doing, and 47% of registered voters disapprove of the job he is doing. If we choose 2014 voters independently, and let X = the number that approve. Let Y = the number that disapprove. p (a) For X, = 2014(0.47) = 946.58 and = 2014(0.47)(0.53) = 22.3984. The event {X > 1012} is equivalent to {X > 1012.5} so \u0012 \u0013 1012.5 946.58 P{X > 1012} 1 22.3984 = 1 (2.943) = 1 0.9984 = 0.0016 (b) The event {972 < Y < 1052} is equivalent to {972.5 < Y < 1051.5}. \u0012 \u0013 \u0012 \u0013 1051.5 946.58 972.5 946.58 P{972 < Y < 1052} 22.3984 22.3984 Th = (4.684) (1.157) = 1 0.8764 = 0.1236 Note that (4.684) is actually a little less than 1, but it is close enough for our calculation here to treat it as essentially 1. (c) The events A = {X > 1012} and B = {Y > 1012} are not independent because only one of them could happen. It is possible for more than half of the people to approve or more than half to disapprove, but those two can't happen at the same time in one sample. Thus, P(A | B) = 0 which is not P(A). 7. I have a box with 6 marbles in it: 4 are red and 2 are green. I draw marbles out of the box, one at a time, without replacement. Let X be a random variable equal to the number of red marbles that I draw before I draw the first green marble. Let Y be a random variable equal to the number of green marbles that I draw out of the first three marbles drawn. https://www.coursehero.com/file/11948315/PSTAT-120A-Assignment-2b-Answers/ (a) The X random variable can be any of {0, 1, 2, 3, 4}. If we draw a green marble first then we have drawn 0 red marbles and X = 0. The most red marbles we can draw is 4 because after that there are only green marbles in the box. The Y random variable has range {0, 1, 2, 3}. (b) We want the probability mass function PX (x) for each value that x can take. sh is ar stu ed d vi y re aC s o ou urc rs e eH w er as o. co m 2 P (0) = P{first marble is green} = = 0.3333 6 \u0012 \u0013 4 4 2 = = 0.2667 P (1) = P{Red then Green} = 6 5 15 \u0012 \u0013\u0012 \u0013 4 3 2 1 P (2) = P{RRG} = = = 0.2000 6 5 4 5 \u0012 \u0013\u0012 \u0013\u0012 \u0013 4 3 2 2 2 P (3) = P{RRRG} = = = 0.1333 6 5 4 3 15 \u0012 \u0013\u0012 \u0013\u0012 \u0013 4 3 2 1 1 P (4) = P{RRRR} = = = 0.0667 6 5 4 3 15 We can write this pmf as a table x P (x) 0 0.3333 1 0.2667 2 0.2 3 0.1333 4 0.0667 (c) We have already seen that P{X 1} = PX (0) + PX (1) = 9 = 0.6 15 For Y , we consider draws without replacement, \u0012 \u0013\u0012 \u0013 4 3 2 P{Y = 0} = P{RRR} = = 0.2 6 5 4 \u0012 \u0013\u0012 \u0013\u0012 \u0013 4 3 2 = 0.6 P{Y = 1} = P{GRR, RGR, RRG} = 3 6 5 4 Th Therefore, P{Y 1} = 0.2 + 0.6 = 0.8, and it is more likely than {X 1}. https://www.coursehero.com/file/11948315/PSTAT-120A-Assignment-2b-Answers/ Powered by TCPDF (www.tcpdf.org)