Econ 250 Fall Due at November 16. Assignment 2: Binomial Distribution, Continuous Random Variables and Sampling

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1 Econ 250 Fall 2010 Due at November 16 Assignment 2: Binomial Distribution, Continuous Random Variables and Sampling 1. Suppose a firm wishes to raise funds and there are a large number of independent financial lenders who might lend from 0 to $10 million dollars each. The total amount raised follows a uniform distribution from 0 to $10n million dollars, where n is the number of lenders. If the firm is to have at least 80% chance of raising at least $100 million dollars what is the minimum number of lenders that should be contacted? Solution: This is a uniform distribution question. Let X be the total amount raised. X is distributed uniformly on [0,10n]. The density function is: f(x) = 1 10n The probability of raising more than $100 is at least 80% P (X > 100) = 1 P (X < 100) = n 80% At least 50 lenders should be contracted. n The probability that a person catches a cold during the cold and flu season is 0.4. Assume that 10 people are chosen at random. Solution: This is a binomial question. A success occurs when a person catches a cold during the cold and flu season. The probability of a success is 0.4. Let X be the number of successes in 10 people. (a) What is the probability that exactly 4 people have the flu? P (X = 4) = ( 10 4 )(0.4)4 (0.6) 6 = (b) What is the probability that between 2 and 5 people inclusive have the flu? P (2 X 5) = P (X 5) P (X 1) = = (c) What is the expected number of people with the flu and what is its variance? E[X] = µ = np = = 4 V [X] = σ 2 = np(1 p) = 2.4 1

2 (d) Approximate your answer in (b) using the normal approximation with and without continuity corrections. Without continuity correction: With continuity correction: P (2 X 5) P ( 2 4 X µ 2.4 σ ) = P ( Z ) = F (0.6455) (1 F (1.2910)) = ( ) = P (2 X 5) P ( X µ 2.4 σ = P ( Z ) ) = F (0.9682) (1 F (1.6137)) = ( ) = The following table displays the joint probability distribution of X and Y. Solution: X/Y (a) What is the covariance and correlation between the X and Y? Are these variables independent? Explain. E[X] = µ X = xp (x) = 1.01 E[Y ] = µ Y = yp (y) = 1.02 V [X] = σ X 2 = (X µ X ) 2 P (x) = V [Y ] = σ 2 Y = (Y µ Y ) 2 P (y) = Cov(X, Y ) = E(XY ) µ X µ Y = xyp (x, y) µ X µ Y = X Y Corr = Cov(X, Y ) = = σ X σ Y

3 (b) Calculate the mean and variance of D D = 2 + 4X 2Y E(D) = 2 + 4E(X) 2E(Y ) = 2 + 4(1.01) 2(1.02) = 4 V [D] = 4 2 V [X] V [Y ] 2(4)(2)Cov(X, Y ) = 16(0.4699) + 4(0.7396) 16(0.1698) = 7.76 (c) Write out the joint cumulative distribution. F(X,Y) F(X,0) F(X,1) F(X,2) F(0,Y) F(1,Y) F(2,Y) Suppose we know the number of sales X by any sales person follows a normal distribution with a mean of 61.7 and a standard deviation of 5.2. Solution: X N(61.7, ) (a) What is P (62.5 < X < 64)? P (62.5 < X < 64) = P ( < Z < ) = P ( < Z < ) = F (0.44) F (0.15) = = (b) What is the P (62.5 < X 3 < 64) where X 3 is the average sales from three (independent) sales persons? P (62.5 < X 3 < 64) = P ( 5.2/ < Z < 3 5.2/ ) = P ( < Z < ) 3 = F (0.77) F (0.27) = = (c) What is the value of k such that P (X > k) = 0.63? F ( 0.33) X = Zσ + µ = 0.33(5.2) = (d) What is the value of k such that P (59 < X < k) = 0.54? Thus P (59 < X < k) = P ( < Z < Z k ) = P ( 0.52 < Z < Z k ) 5.2 = F (Z k ) (1 F (0.52)) = 0.54 F (Z k ) = F (0.52) = = Z k 1 k = Z k σ + µ = 1(5.2) =

4 5. Sales at a local electrical wholesaler consist of both over-the-counter sales as well as deliveries. During the course of a month, over-the-counter sales have a mean of $96,780 with a standard deviation of $12,520. Over the same time period, deliveries average $229,620 with a standard deviation of $234,100. Assume that over-the-counter sales and deliveries have a correlation of.2. Solution: Let S C denote over-the-counter sales and S D denote deliveries. S C N(96780, ) and S D N(229620, ) Corr(S C, S D ) = 0.2 (a) What is the mean, variance and distribution of all sales S? All sales S = S C + S D E[S] = µ S = E[S C ] + E[S D ] = = V [S] = σ s 2 = V [S C ] + V [S D ] + 2(Corr)( V (S C ))( V (S D )) = (0.2)(12520)(234100) = Now we can see that S N(326400, ) (b) What is the P (222, 900 < S < 240, 400)? 222, P (222, 900 < S < 240, 400) = P ( < Z < = P ( < Z < ) = F (0.4369) F (0.363) = Suppose X is uniform distribution over the interval 0 to 150. (a) Find the mean and variance. µ X = a + b = = σ 2 (b a)2 = = = , ) (b) Find the value that leaves.05 in the lower tail and also the value that leaves.05 in the upper tail = X L = 150 X U 150 X L = 7.5 X U =

5 (c) Suppose that you do not know that the variable is uniform but are given the mean and variance from (a). Calculate the same magnitudes for (b). F (Z L = 1.645) = 0.05 X L = ( 1.645) = F (Z U = 1.645) = 0.05 X U = (1.645) = (d) Draw the two distributions to explain these results. 7. Two classes of statistics have grades that are normally and independently distributed with C 1 N(75, 12) and C 2 N(80, 22). (a) What is the expected difference and its variance? C = C 1 C 2 The expected difference and its variance: E[C] = µ C = E[C 1 ] E[C 2 ] = = 5 V [C] = σ C 2 = V [C 1 ] + V [C 2 ] = = 34 (b) What is the probability that the difference from picking 1 student from each class is between -1 and 1? 1 ( 5) P ( 1 < C < 1) = P ( 34 < Z < 1 ( 5) 34 ) = P ( < Z < ) = F (1.03) F (0.69) = (c) To give the class the same mean as the second class, the professor adds 5 to all grades. Explain why this does not leave the two classes equivalent. Which class would you prefer to be in? 8. A car company says their car gets a mean of 45km per liter with a standard deviation of 6. Suppose we assume a normal distribution. Solution: (a) Suppose some sales representative claims you will get at least 47 80% of the time, what can you tell him? P (X 47) = P (Z ) = P (Z 1 ) = 1 F (0.333) = (b) If we have 4 cars and take the average, calculate the P (44 < X 4 < 46). P (44 < X 4 < 46) = P ( 6/ < Z < 4 6/ 4 ) = P ( < Z < ) = F (0.33) (1 F (0.33)) = 2(0.6293) 1 =

6 9. Suppose you wish to drive across a country that is 2625 km wide and you intend to rent a series of cars from Rent-A-Wreck. The distance that the first car they give you is normally distributed with a mean distance of 1500km and a variance of 500km. 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 2 cars? (c) What is the probability that you do the trip with more than 3 cars? (d) If each car costs $100 what is the expected cost of the trip? (e) Approximate the expected length of the farthest trip that can be taken. Let Xi be the distance travelled by car i... and so on Consider the first car X1 N(1500, 500) X2 N(750, 125) X3 N(375, 31.5) P (X 1 > 2625) = P ( X 1 µ 1 σ 1 > ) P (Z > 50.3) 0 Consider the distance traveled by the first car and then the second car P (X 1 + X 2 > 2625) = P ( X 1+X 2 (µ 1 +µ 2 ) σ 2 1 +σ 2 2 Now the third car: P (X 1 + X 2 + X 3 > 2625) = P ( X 1+X 2 +X 3 (µ 1 +µ 2 +µ 3 ) σ 1 2 +σ 2 2 +σ 3 2 So n = 3 Expected cost of the trip is $300. Farthest trip possible T (geometric series) > 2625 ( ) ) P (Z > 15) 0 > 2625 ( ) ) P (Z > 0) = 0.5 T = X 1 + X 2 + X = E[T ] = = X 1 6

Chapter 5. Continuous Random Variables and Probability Distributions. 5.1 Continuous Random Variables

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