Controller Posted February 28, 2007 Share Posted February 28, 2007 Here is the extra credit problem: Suppose that you are a car salesperson. Whenever you approach a potential new customer, you have a 10% chance of selling him or her a new car. Let the random variable N be the number of customers you talk to until you first sell a car (for example, if you do not sell a car to your first two prospects and then you do sell one to the third, N = 3). What is the expected value E[N]? What is the variance V[N]? Note: I thought the probability P(N=K) = 1/10^k, but the prof. tells me that is wrong. Any out there who can help??? Quote Link to comment Share on other sites More sharing options...
Gonkis Posted February 28, 2007 Share Posted February 28, 2007 It's been 15 years since I had prob/stats in college, but isn't the expected value the average of the possible outcomes? Chances of selling a car to the first person is .1(1) = .1. Since it's a 10% chance, numbers say you'll sell one by the time N=10. (100% chance you'll sell a car if you see 10 people) Add up E(N) = .1(N) where N=1 to 10 .1+.2+.3+.4+.5+.6+.7+.8+.9+1 = 5.5 Expected value is 5.5 people. I'm probably wrong. I don't remember variance, but I think it's the square of the standard deviation? Quote Link to comment Share on other sites More sharing options...
wiegie Posted February 28, 2007 Share Posted February 28, 2007 It's been 15 years since I had prob/stats in college, but isn't the expected value the average of the possible outcomes? Chances of selling a car to the first person is .1(1) = .1. Since it's a 10% chance, numbers say you'll sell one by the time N=10. (100% chance you'll sell a car if you see 10 people) This is not correct. If you see 10 people, there is a 34.867844% chance that you won't have sold any of them a car. I can't remember off the top of my head how to answer the original questions, but perhaps it will come to me later on this afternoon (no promises though). Quote Link to comment Share on other sites More sharing options...
Bill Swerski Posted February 28, 2007 Share Posted February 28, 2007 I don't remember variance, but I think it's the square of the standard deviation? I believe that's correct. And, IIRC, skewness is (SD)^3 and kurtosis is (SD)^4. Quote Link to comment Share on other sites More sharing options...
Hat Trick Posted February 28, 2007 Share Posted February 28, 2007 This is not correct. If you see 10 people, there is a 34.867844% chance that you won't have sold any of them a car. I can't remember off the top of my head how to answer the original questions, but perhaps it will come to me later on this afternoon (no promises though). And here I was about to come in here and say Wiegie will be along shortly to answer this Quote Link to comment Share on other sites More sharing options...
Yukon Cornelius Posted February 28, 2007 Share Posted February 28, 2007 17 Quote Link to comment Share on other sites More sharing options...
spain Posted February 28, 2007 Share Posted February 28, 2007 And here I was about to come in here and say Wiegie will be along shortly to answer this Wiegie is NOT a real economist... Quote Link to comment Share on other sites More sharing options...
Ursa Majoris Posted February 28, 2007 Share Posted February 28, 2007 kurtosis I had that but fortunately my doctor put me on a regimen of pills and it went away. Quote Link to comment Share on other sites More sharing options...
Yukon Cornelius Posted February 28, 2007 Share Posted February 28, 2007 2 Quote Link to comment Share on other sites More sharing options...
cliaz Posted February 28, 2007 Share Posted February 28, 2007 once you ge above 1+1 = 3 i'm lost Quote Link to comment Share on other sites More sharing options...
Controller Posted February 28, 2007 Author Share Posted February 28, 2007 Expected value is defined at the summation of k from 1 to infinity of: k times (the probability of N = k) Variance is defined as the summation of k from 1 to infinity of: [k minus (expected value of n)] squared times the probability of N = k) If that helps. Quote Link to comment Share on other sites More sharing options...
Cameltosis Posted February 28, 2007 Share Posted February 28, 2007 I was told there would be no math. Quote Link to comment Share on other sites More sharing options...
Tford Posted February 28, 2007 Share Posted February 28, 2007 Last time, I promise, I guess my first post was right. The probability of selling a car is P(success)=1-0.9^K The probability of N=K is P(N=K) = 1 - 0.9^K - P(success at K-1) since the sample stops when a success is reached thus the probability of success at N=1 is 0.1 and the probability of success at N=2 is 0.19, then P(N=2)=0.19-0.1=0.09 This is easy to accomplish with an Excel spreadsheet. Multiply K by the probability of N=K and sum your answers. You should get 9.96. Quote Link to comment Share on other sites More sharing options...
muck Posted February 28, 2007 Share Posted February 28, 2007 Not sure if this helps... The probabilty of NOT selling a car to any given person is 90% ... so, with any random group of 10 individuals, you WON'T sell a single car 34.868% of the time (i.e., 0.9^10) ... or, said another way ... you WILL sell at least one car 65.132% (1 - 34.868%) of the time. Quote Link to comment Share on other sites More sharing options...
cre8tiff Posted February 28, 2007 Share Posted February 28, 2007 (edited) Here is the extra credit problem: Suppose that you are a car salesperson. Whenever you approach a potential new customer, you have a 10% chance of selling him or her a new car. Let the random variable N be the number of customers you talk to until you first sell a car (for example, if you do not sell a car to your first two prospects and then you do sell one to the third, N = 3). What is the expected value E[N]? What is the variance V[N]? Note: I thought the probability P(N=K) = 1/10^k, but the prof. tells me that is wrong. Any out there who can help??? With your personality, I'd say the probability of you selling anything is 0. Edited February 28, 2007 by cre8tiff Quote Link to comment Share on other sites More sharing options...
Grits and Shins Posted February 28, 2007 Share Posted February 28, 2007 All other things being equal I don't do probablities. Quote Link to comment Share on other sites More sharing options...
wiegie Posted February 28, 2007 Share Posted February 28, 2007 Not sure if this helps... The probabilty of NOT selling a car to any given person is 90% ... so, with any random group of 10 individuals, you WON'T sell a single car 34.868% of the time (i.e., 0.9^10) ... or, said another way ... you WILL sell at least one car 65.132% (1 - 34.868%) of the time. muck cheated off of me This is not correct. If you see 10 people, there is a 34.867844% chance that you won't have sold any of them a car. I can't remember off the top of my head how to answer the original questions, but perhaps it will come to me later on this afternoon (no promises though). :oldrazz: Quote Link to comment Share on other sites More sharing options...
TimC Posted February 28, 2007 Share Posted February 28, 2007 Pssst, get the Vette. Quote Link to comment Share on other sites More sharing options...
Ursa Majoris Posted February 28, 2007 Share Posted February 28, 2007 Pssst, get the Prius. Fixed Quote Link to comment Share on other sites More sharing options...
wiegie Posted February 28, 2007 Share Posted February 28, 2007 I'm not 100% certain about this, but I think what we are dealing with here is a geometric distribution. The expected value of such distribution = (1-p) / p = (1-.1)/.1 = .9/.1 = 9 The variance of the distribution = (1-p)/p^2 = .9/(.1)^2 = .9/.01 = 90 (where p = the probability of selling the car) I could be wrong, so don't bet the house on it. Quote Link to comment Share on other sites More sharing options...
TimC Posted February 28, 2007 Share Posted February 28, 2007 Listen to me, I am the smarterest. Everything, and I mean everything, in life is 50%. The customer will either buy the car or he won't. You will either wake up this morning or you won't. Quote Link to comment Share on other sites More sharing options...
Tford Posted February 28, 2007 Share Posted February 28, 2007 (edited) I'm not 100% certain about this, but I think what we are dealing with here is a geometric distribution. The expected value of such distribution = (1-p) / p = (1-.1)/.1 = .9/.1 = 9 The variance of the distribution = (1-p)/p^2 = .9/(.1)^2 = .9/.01 = 90 (where p = the probability of selling the car) I could be wrong, so don't bet the house on it. If you assume a geometric distribution, the expected value is not (1-p)/p but it is in fact E(N) = 1/p leaving you with an expected value of 10, very close to my crude methodology answer of 9.96. The variance is the same however. (1-p)/p is the expected value of consecutive failures, not the success expected value http://en.wikipedia.org/wiki/Geometric_distribution Edited February 28, 2007 by Tford Quote Link to comment Share on other sites More sharing options...
wiegie Posted February 28, 2007 Share Posted February 28, 2007 If you assume a geometric distribution, the expected value is not (1-p)/p but it is in fact E(N) = 1/p leaving you with an expected value of 10, very close to my crude methodology answer of 9.96. The variance is the same however. (1-p)/p is the expected value of consecutive failures, not the success expected value http://en.wikipedia.org/wiki/Geometric_distribution I have seen different formulas given for the expected value (the one I gave and the one you gave). I'm not sure exactly which one is correct, but I used the version of the formula given in my old graduate Probability & Statistics textbook (DeGroot). Quote Link to comment Share on other sites More sharing options...
Tford Posted February 28, 2007 Share Posted February 28, 2007 I used Probability and Statistics for Engineers by Soong. A wiki reference helped too I guess. Quote Link to comment Share on other sites More sharing options...
Azazello1313 Posted February 28, 2007 Share Posted February 28, 2007 Listen to me, I am the smarterest. Everything, and I mean everything, in life is 50%. The customer will either buy the car or he won't. You will either wake up this morning or you won't. otherwise known as clubfoothead's parachute theory of statistics. Quote Link to comment Share on other sites More sharing options...
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