ECON HW #3 - Copy

ECON 227 HW#3 (40 points total) Please write your answers on separate papers and submit them through file upload. (12 points) The following table provides a probability distribution for the random variable   x . (a) Compute   E ( x ), the expected value of   x . (b) Compute   σ 2 , the variance of   x   (to 1 decimal). (c) Compute σ, the standard deviation of   x   (to 2 decimals). x f(x) x*f(x) x^2*f(x) 3 0.25 3*0.25=0.75 3^2*0.25=2.25 6 0.5 6*0.5=3 6^2*0.5=18 9 0.25 9*0.25=2.25 9^2*0.25=20.25 sum(f(x))=1 sum(x*f(x))=6 sum(x^2*f(x))=40.5 ∑f(x)=1 ∑x×f(x)=6 ∑x2×f(x)=40.5 A) The expected value of x is, E(x)=∑x×f(x) E(x)=6 The expected value of x is E(x)=6 1. (8 points) Let y be a random variable distributed as shown in the accompanying table. y 0 1 2 3 xxx p ( y ) 0.4 0.3 0.2 0.1 (a) Find the mean of y or, E ( y ) . (b) Find the standard deviation of y , or sd ( y ) . 1

ECON 227 HW#3 (40 points total) E[Y] = (0.4*1) + (0.3*2) + (0.2*3) + (0.2*4) = 2 E[1=Y ] = (0.4/1) + (0.3/2) + (0.2/3) + (0.2/4) = 0.64 V [Y ] = E[Y 2] - E[Y ]^2 = [(0.4 * 1^2) + (0.3 * 2^2) + ( 0.2 * 3^2) + (0.1 * 4^2)] - [ [ 2 ^ 2 ] = [0.4+1.2+1.8+1.6] - [4] = 1 2). (12 points) For borrowers with good credit scores, the mean debt for revolving and installment accounts is $15,015 ( BusinessWeek , March 20, 2006). Assume the standard deviation is $3540 and that debt amounts are normally distributed. (a) What is the probability that the debt for a randomly selected borrower with good credit is more than $18,000 (to 4 decimals)? (b) What is the probability that the debt for a randomly selected borrower with good credit is less than $10,000 (to 4 decimals)? (c) What is the probability that the debt for a randomly selected borrower with good credit is between $12,000, and $18,000 (to 4 decimals)? Given   = 15015,   = 3730 For normal distribution, P(X < x ) = p( z < x -   /  ) a) p( X > 18000) = p( z > 18000 - 15015 / 3730) = p( z > 0.8003) = 1 - p( z < 0.8003) = 1 - 0.78823 = 0.2118 b) p( X < 10000) = P( z < 10000- 15015 / 3730) = p( z < -1.3445) = 1 - p( z < 1.3445) = 1 - 0.91061 = 0.0894 c) p( 12000 < X < 18000) = p( X < 18000) - p( X < 12000) = p( z < 18000 - 15015 / 3730) - p( z < 12000 - 15015 / 3730) = p( z < 0.8003) - p ( z < - -0.8083) = p( z < 0.8003) - ( 1 -p( z < 0.8083) ) 2

ECON 227 HW#3 (40 points total) 4