Econ 250 Fall 2006 Due November 7 in Class Assignment 2: Probability. Discrete and Continuous Random Variables

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1 Econ 50 Fall 006 Due November 7 in Class Assignment : Probability. Discrete and Continuous Random Variables. A hospital wants to study a new drug for sleep disorders. They know that this disorder happens randomly in the population with probability = :5 (a) What is the minimum sample size to ensure that the expected number of patients with the sleeping disorder is more than 6 (b) With this sample size what is the probability that at least 0% have the sleeping disorder (c) 5 people are randomly given the drug. If the person given the drug has the disease, 90% get better. What is the probability that at least people get better (d) For those receiving the drug, what is the probability of having the disease and not getting better. Answers E[X] = n =) n > 4 say 5 P (X > :5) = P (X 3 j n = 5; = :5) = P ((X j n = 5; = :5) :09 = :9 P (X j n = 5; = :9) P (Y 3 j n = 5; = :) = :008 E(GB j D) = :9 =) E(notGB j D) = : P (D) = :5 P (notgb j D) P (D) = P (notgb \ D) = : :5 = :05. Let X N(3; 4) What is P ( 7 < Y < 4)? Y = 4 3X Y N( 5; 36) E[Y ] = E[4 3X] = = 5 V [Y ] = V [4 3X] = 9 4 = 36 P ( 7 < 7 ( Y < 4) = P 6 5) P ( :33 < Z < :5) = :565 < Y y < 4 ( 5) y 6

2 3. The economics professor knows that it is equally likely that a student writing an exam obtains a grade between 30 and 90. If the performance on the exams are independent, how many exams does the professor need to set before there is at least a 40% chance that the student obtains at least a grade of 80 on one of the exams? Let Y =number of exams over 80.: P (X > 80) = = :7 60 need to take more than one exam P (Y = 0 j n = ; = :83) = :83 :7 0 = :689 So that there is only a 30% chance of at least exam over 80. Need to take a third exam P (Y = 0 j n = 3; = :83) = :83 3 :7 0 = :58 There is a 4% chance of getting greater than an 80% 4. You buy a lottery ticket that costs $ and there is a probability of.00 winning $500. Alternatively, there is a bingo game that costs $0 to play but there is a possibility of winning $000. If you are a risk neutral agent and therefore care only about the expected pay-o, what probability of winning the bingo game is required to be indi erent between the two games of chance? E[Lottery] = : :999 0 = :5 E[Bingo] = ( ) 0 0 = E[Lottery] = :5 = 0:5 000 = : One way to look at the correction factors to the normal distribution in approximating the binomial distribution is that there is insu cient observations to get a good approximation. This question illustrates this point. Let X be the number of successes with the probability of success equal to 0.4 and suppose you want to know P ( X 3) (a) Calculate the exact probability of this event with n = 0 and n = 0 (b) Calculate the normal approximation without continuity corrections with n = 0 and n = 0 (c) Calculate normal approximation with continuity corrections with n = 0 and n = 0 P ( X 3) = P (X 3) P (X = 0) = :38 :006 = :376 exact probability with n = 0 = :06 0 = :06

3 P ( X 3) 4 0 :4 :6 n( ) P ( :94 Z :65) = :9738 :74 = : p 0 :4 :6 P (:5 X 3:5) :5 4 0 :4 :6 n( ) P ( :6 < Z < :3) = :988 :655 = :366 3:5 4 p 0 :4 :6 P ( X 3) 8 0 :4 :6 n( ) P ( 3:95 < Z < ::8) = :993 :9887 = : p 0 :4 :6 P (:5 X 3:5) :5 8 0 :4 :6 n( ) P ( 3:4 < Z < :05) = :9997 :9798 = :09 3:5 8 p 0 :4 :6 As the sample size gets larger the need for continuity corrections lessens as the approximation is getting quite good 9another way to say that is that the integer constraints are no so severe) 3

4 6. Suppose X N(50; 00), calculate the following (a) P (45 < X < 55) (b) P (45 < X 5 < 55) where X 5 refers to the sample mean with 5 observations (c) P (45 < X 0 < 55) P < X < 0 = P ( :5 < Z < :5) = : q 00 5 < X < q A 00 5 = P ( :8 < Z < :8) = : q 00 0 < X < = P ( :58 < Z < :58) = q A 00 0 : 0:885 8 : 7. Suppose X i NID(3; 9) and is independently distributed, calculate 9X P X i > 36 Answer i= P 9 i= P ( X i n p 36 7 > n 3 3 ) P (Z > ) = :5 :343 = : Suppose X N(5; 7); Y is uniformly distributed over the interval (40, 80), and B is a binomial variable with = :3 and n = 0. All variables are assumed to be independent. If 4

5 What is the E[R] and variance V [R]? R = 0 + X + 4Y B: E[R] = 0 + E[X] + 4E[Y ] E[B] = = 48 (0 :3) V [R] = = 74: + 4 (0 : :8) : 9. Suppose you wish to drive across a country that is 65 km wide and you intend to rent a series of cars from Rent-A-Wreck. The distance that the rst car they give you is normally distributed with a mean distance of 500 km and a variance of 500 km. Each subsequent car you rent gets 50% less km on average than the previous one with a 75% reduction in the variance. (a) Try to formulize this problem? (b) What is the probability that the trip can be done using exactly cars? (c) What is the probability that you do the trip with more than 3 cars? (d) If each car costs $00 what is the expected cost of the trip? (e) Approximate the expected length of the farthest trip that can be taken Let X i be the distance travelled by car i Consider rst car X N(500; 500) X N(750; 5) X 3 N(375; 3:5). and so on P (X > 65) = P ( X > P (Z > 50:3) p 500 Consider the distance traveled by srt car and then the second car 5

6 P (X + X > 65) = P X + X ( + ) p > + P (Z > 5) 0 Now the third car: 65 ( ) p (X + X + X 3 > 65) = P X + X + X 3 ( ) 65 ( p > p :5 P (Z > 0) = :5 So n = 3 Farthest trip possible T (geometric series) T = X + X + X 3 + : : : = E[T ] = 500 = 3000 X :5 6