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10_W6_S32_chapter_16_corrected_Fall_11

10_W6_S32_chapter_16_corrected_Fall_11 - Week Seven...

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Unformatted text preview: Week Seven Sessions Two (Chapters 16) Professor Esfandiari Main concepts in Chapter 16 Radom variables: Random variables are based on the outcome of a random event. We Will use Capital X to denote the random variable and the lower—case x to denote a particular value that a random variable can take. Discrete random variables can take on specific values and we can list all of the possible outcomes. Example of a discrete random variable: If you take a multiple choice test with three questions in a language you have never heard of and choose the correct answers at random, the following probability model will result for this discrete random variable. This model provides you with all of the X values that the random variable can assume as well as the probabilities assigned to each of those values. Remember these are independent events. x = Number Score Probability of correct you P (X = X) (X - M)"2 * P(X) answers receive picked at in the random exam _-_—- WWWW 0.2373 1 X = 5 [3*(1/4A1)*(3/4’\2)] 1*27/64 = 27/64 (1-0.75)"2* 27/64 = =27/64 = 0.421875 RWW WRW WWR 2 X = 10 [3*(1/4A2)(*3/4A1)]= 2*9/64= 18/64 (2—0.75)"2*9/64 = 9/64 =0.28125 0.2197 RRW WRR RWR 3 X = 15 3*1/64 = 3/64 = (3-0.75)’\2*1/64 = 0.046875 0.0791 Sum ofall p.= Ex P(x) = (#2 =E(x - p)"2* P(x) = probabilities = 48/64 = 0.75 0.5625 (within rounding 27/64 + error) 27/64+9/64+1/64 = 64/64 = 1 Mean, Variance, and SD of a discrete random variable: M = E(X) = EX P(X) The mean or the expected value of a discrete random variable or Mue is equal to the expected value of x which is equal to the sum of X multiplied by p(x) o"2=VAR\/(X)=2( x - mAz * P(x) Variance of the discrete random variable is equal to x minus [1. (mue) squared multiplied by p (x) and added. 0 = SD (X) = Square Root ofoAZ 0r VOAZ Also: uor E(X) = N * P and 0A2 or Var(X) = n * p* (l-p) For the example given on page one: uor E(X) = N * P = 3*1/4 = 3A = 0.75 0A2 or Var(X) = 3 * 1/4* % = 9/16 = 0.5625 The effect of linear transformation on mean, variance, and SD Transformation Add a constant Constant gets added Variance does not change to X to the mean Var (X + c) = Var QC) E X + c = E X + c Subtract a Constant gets Variance does not change constant from subtracted from the X Var (X — c) = Var (X) Standard deviation does not change SD X + c = SD X Standard deviation does not change SD (X - c) = SD (X) Standard deviation is multiplied by the absolute value of C SD (X * c) = SD (X)* lo! Standard deviation gets divided by the absolute value of C SD (X /c) = SD (X)/ ]c| Multiply X by a constant Variance is multiplied by the square of C Var (X * c) = Var (X)*c/\2 Mean gets multiplied by the constant B (X*c) = E (X)*c Divide X by a constant Variance gets divided by the square of C Var (X /c) = Var (X)/c/‘2 Mean gets divided by the constant B (X/c) = E (X)/c Continuous Random Variables Continuous random variables can take on any value . Normal model is an example of a continuous random variable. As we saw you could use the Z scores to estimate the area of any region of interest. Z=(X- (0/0 The reason that we use parameters as measures of center (u) and spread (O) is that we assume that the normal model holds true in the population for certain attributes like height, weight, measures of mood, etc. The mean and variance of the sum of two random variables E (X + Y) = E(X) + E(Y) The mean of the sum of two random variables is the sum of the means E (X — Y) = E(X) — E(Y) The mean of the sum of difference of random variables is the difference of the means If two random variables are independent (iftwo quantitative random variables are independent, then their coefficient of correlation is equal to zero) Var (X + Y) = Var (X) + Var (Y) Var (X - Y) = Var (X) - Var (Y) Additional exercise to be done in class: Given the follovving data sets, without any calculation, estimate the value of variance and standard dev1ation Mean, variance, and SD have been rounded. 6. You bet! You roll a die. If it comes. up a 6, you win $100. If not, you get to roll again. If you get a 6 the second time, you win $50. If not, you lose. a) Create a probability model for the amount you Win at this game. b) Find the expected amount you'll Win. c) How much would you be willing to pay to play this game? . Batteries. In a group of IQ_§§geries, choose 2 batteries at random. A a) Create a probability model for th ' batteries you get. 6 nUmber Of gOOd b) What’s the expected number of good ones c) What’s the standard deviation? 3 are dead. You you get? i 40. Pets. The American Veterinary Association claims that , the annual cost of medical care for dogs averages $100, with a standard deviation of $30, and for ca $120, with a standard deviation of $35. a) What’s the expected difference in the cost of medical care for dogs and cats? b) What’s the standard deviation of that difference? c) If the difference in costs can be described by a Normal model, what's the probability that medical expenses are higher for someone’s dog than for i her cat? ts averages Solution to problems selected from Chapter 16 6. You bet! a) $100 1/6 * 5/6)=5/36 5/6 * 5/6=25/36 13) u or E (Amount won) = 100*1/6 + 50*5/36 + 0*25/36 = 23.61 c) Answers may vary. In the long run, the expected payoff of this game is $23.61 per play. Any amount less than $23.61 would be a reasonable amount to pay in order to play. Your decision should depend on how long you intend to play. If you are only going to play a few times, you should risk less. 25. Batteries 3) -__- 100d _ood =6l90 =42/90 b) u or E (Number of goods) = 0*6/90 + 1*42/90 + 2*42/90 = 1.4 0) 0’2 or Var (number good goods) = (0-1.4)A2*0 + (1-1.4)A2*1 + (2—1.4)’\2*2 = 0.3733 (1) o or SD( Number good) = square root of 0.3733 = 0.61 batteries. 40) Pets a) u = E(dogs — cats) = E(dogs) - E(cats) = 100 — 120 = (—$20) b) b) o = SD (dogs — cats) = Var(dogs)'+ Var(cats) = square root of 302+ 35“ 2e $46.10 Z = X - a/a Z = (0 — (-20)/46.10 Z = 0.4338 (You are interested in the area above Z = 0.4338) The expected cost of the dog is greater than that of the cat when the difference in cost is positive (greater than 0). According to the Normal model, the probability of this occurring is about 0.332 ...
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