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Errata for ASM Exam C/4 Study Manual (Sixteenth Edition Second Printing) Sorted by Page 1 Errata and updates for ASM Exam C/Exam 4 Manual (Sixteenth Edition Second Printing) sorted by page Note the change
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Errata for ASM Exam C/4 Study Manual (Sixteenth Edition Second Printing) Sorted by Page 1 Errata and updates for ASM Exam C/Exam 4 Manual (Sixteenth Edition Second Printing) sorted by page Note the change in wording in Practice Exam 1:1 (page 1261), and the changed answer choices for Practice Exam 1:8 (page 1359). [3/25/214] On page 11, in the third displayed line of Section 1.4, change e t x to e t X. [6/12/214] On page 23, in the solution to exercise 1.15, the second displayed line should be.16 = exp z 2 z 2 2 (ln 5 ln 2) = exp ln 2/5 = z 2 /2 [1/31/214] On page 53, two lines above Continuity correction, change call to called. [1/18/214] On page 93, in the solution to exercise 5.9, on the last line, change 2 to 1. [6/27/214] On page 99, on the first line of the answer to Example 6F, the first sentence is inaccurate; E[(X 5) + ] is not calculated previously. However, it is easy to calculate: E[(X 5) + ] =.2(1 5) +.1(2 5) = 25. [2/22/214] On page 99, 2 lines after the answer to Example 6H, change θ d to θ = d. [3/27/214] On page 13, in exercise 6.4, in the second bullet, change r % to 1r %. [3/27/214] On page 112, in the solution to exercise 6.2, on the second displayed line, place a parenthesis around 2, 2, + d 2. [2/5/214] On page 165, in exercise 9.13, although the exercise can be solved mechanically, the values in the table are impossible if losses are nonnegative. If F (1) =.3, then E[X 1] (1.3)(1) = 7. [4/1/214] On page 171, in the solution to exercise 9.4, on the first line, add is after it. [6/2/214] On page 176, in the solution to exercise 9.19, on the last displayed line, there should be a.8 on the left hand side, so that the part preceding the first equals sign is.8(e[x 5] E[X 1]) [9/1/214] On page 26, in the solution to exercise 11.26, on the third line, add 1 plus before the second mean :... so its mean is 1 plus the mean of an unshifted.... [4/1/214] On page 237, in the warning box on the second line, change variance distribution to variance formula. [8/1/214] On page 274, in the solution to exercise 15.29, on the third line from the end, change to [7/3/214] On page 332, in the solution to exercise 19.1, on the last line, change 5 to 4. [6/17/214] On page 332, in the solution to exercise 19.3, on the last line, change X 3 to S 3. [4/18/214] On page 339, in exercise 2.11, delete the first line with s. [6/23/214] On page 361, in exercise 21.9, on the sixth line, replace MSE β (ψ)/ MSE β (φ) with MSE ψ (β)/ MSE φ (β). [6/23/214] On page 365, in the solution to exercise 21.9, on the first displayed line, replace MSE β (ψ) with MSE ψ (β). On the second displayed line, replace M S E β (φ) with MSE φ (β). [7/29/214] On page 382, in exercise 23.11, on the last line, delete the hat on top of v (θ ). 2 Errata for ASM Exam C/4 Study Manual (Sixteenth Edition Second Printing) Sorted by Page [6/18/214] On page 386, in the solution to exercise 23.11, on the second line, change θ n is binomial to θ n is a binomial proportion. [7/29/214] On page 386, in the solution to exercise 23.13, on the fifth line, change q = 1 7 to q = 6 7. [6/19/214] On page 43, in exercise 24.29, on the fourth line, change states to times. [4/23/214] On page 475, in the solution to exercise 27.9, on the third displayed line, change F n (17) = F n (16) to F n (17) F n (16). [6/11/214] On page 499, in the solution to exercise 28.5, on the second line, replace toal with total. [2/2/214] On page 499, in the solution to exercise 28.8, replace the bottom 3 lines of the page with Total exposure in months is 24(36) + 41(34) (29) (25) = The estimate of 3 q 25, by formula (28.2), is 3q 25 = nd j = 3(1) = 3/(3215/12) = e j 3215/12 The standard deviation, by formula (28.4), is Var( q j (1 q j ) (.11198)( ) 3 ˆq 25 ) = = = e j /n (3215/12)/3 [6/11/214] On page 5, in the solution to exercise 28.9, on the fifth line, replace (33-5,38-5,x ) with (33.5,38-5,s ). [4/17/214] On pages 51, in the solution to exercise 28.13, on the first line of the page, change e 19/12 to e 12/19. On the third line of the page, change the final answer to [8/24/214] On page 51, in the solution to exercise 28.15, on the third line from the end, change e 17/12 to e 12/17. On the second line from the end, change 12/17 to 12/22. Change.5456 to On the last line, change.391 to.392. [4/19/214] On page 53, in the solution to exercise 28.26, on the fourth line, change.5(35) to.5(7). [8/24/214] On page 51, in the solution to exercise 28.14, on the third line from the end, change e 13/12 to e 12/13. [3/14/214] On page 541, in the solution to exercise 3.32, the answer key should be (B) instead of (C). [6/24/214] On page 579, in exercise 32.15, in answer choice (B), the exponent 3 in the denominator should be outside the parenthesis, so that the denominator is f (1;θ ) 3. [5/8/214] On page 626, in exercise 33.36, the correct solution, starting with the second line, is ˆα = n ln(1 + xi ) If the sample mean x i /n = x, then at least one of the x i s is greater than or equal to x. The other x i s are nonnegative. So n i =1 ln(1 + x i ) ln(1 + x ). As x, ln(1 + x ). Therefore ˆα. (A) [8/4/214] On page 642, 2 lines from the end of the page, change e 825/1, to e 1,/825. [5/5/214] On page 643, on the fifth displayed line of the answer to Example 34E, change the minus sign after m a sign. [1/7/214] On page 653, in exercise 34.2II, change Crarner to Cràmer. to a plus [4/23/214] On page 661, in exercise 34.32, on the last line, change Var to VaR. Errata for ASM Exam C/4 Study Manual (Sixteenth Edition Second Printing) Sorted by Page 3 [6/26/214] On page 664, in the solution to exercise 34.7, delete negative. [4/23/214] On page 669, in the solution to exercise 34.32, change the last four lines to The variance of the estimate is g θ = ( ln.5)1/τ = ( ln.5) 1/2 = g θ ( ln.5)1/τ 5( ln.5)1/2 = ln( ln.5) = ln( ln.5) τ τ 2 4 ( )(62,5) + (2, )(.8) = 191,741 = [6/27/214] On page 671, the solution to exercise is incorrect. The correct solution is The length of the 2-sided confidence interval for the proportion is 2 times 1.96 (the 97.5 th percentile of a standard normal distribution) times the standard deviation of the estimate for the proportion. We will use the delta method. In the previous exercise, we estimated ˆθ = 125. Let g (θ ) be Pr(X 25 θ ). Then g (θ ) = S(25 θ ) = e 25/θ By the delta method Var g ( ˆθ ) Var( ˆθ ) g ( ˆθ ) 2 Since ˆθ is the sample mean, its variance is the variance of the distribution divided by the size of the sample. The variance of an exponential with mean θ is θ 2, so the variance of ˆθ is Var( ˆθ ) = θ 2 where as usual we approximate θ with its estimate. 1 g (θ ) = 25 θ 2 e 25/θ g ( ˆθ ) = e 25/125 = e 2 Var g ( ˆθ ) e 2 =.2e 2 2 The standard deviation of g ( ˆθ ) is.2e 2, and the length of the confidence interval is 2(1.96)(.2e 2 ) =.161. (E) [6/3/214] On page 753, in exercise 39.1, at the end of the last bullet, change good to fit to goodness of fit. [6/3/214] On page 763, in exercise 39.23, after the table, add A Pareto distribution is fit to the data. [9/27/214] On page 77, in the solution to exercise 39.6, on the sixth line, change 441 to 447. On the last line, change 446 to 447 in two places, and change the final answer to 258. [12/23/214] On page 771, replace the solution to exercise with the following: The value of the chi-square statistic does not depend on the number of parameters; it depends only on the number of actual and expected observations in each group. So (B) is false. The Kolmogorov-Smirnov statistic 4 Errata for ASM Exam C/4 Study Manual (Sixteenth Edition Second Printing) Sorted by Page depends on the data points themselves, so (C) is false. The Kolmogorov-Smirnov statistic does not depend on the number of parameters of the distribution, so (D) is false. However (A) is true, since grouping data differently will lead to different statistics. [6/19/214] On page 778, in the paragraph beginning Thus if an exponential model, on the fourth line, delete one that. [4/23/214] On page 78, on the second-to-last line of the answer to Example 4C, replace 2( ) with 2( ). [5/19/214] On page 799, in the solution to exercise 41.5, on the first displayed line, change F ( x ) to F (x ). [6/9/214] On page 81, in the solution to exercise 41.13, in the table, change Transformed gamma to Generalized Pareto. [1/12/214] On page 852, in the solution to exercise 44.18, replace the line that is three lines from the bottom with y 2 p = (.12 ) = [8/21/214] On page 89, in the paragraph beginning with 1., on the first line, change x to θ. [5/28/214] On page 971, exercise is a duplicate of exercise [8/3/214] On page 145, in the solution to exercise 53.17, replace the 9 lines starting with We want to calculate with We want to calculate Pr(1, and p .1) Pr(p .1 1, ) = Pr(1, ) Pr(1, and p .1) = Pr(1, and p .1) + Pr(1, and p.1) We will obtain the numerator by integrating, from to.1, Pr(1, ) = p (1 p) over the density function of p. We will obtain the second term of the denominator by integrating the same integrand from.1 to.2. Replace the left side on the next line with Pr(1, and p .1. Replace the left side three lines after that line with Pr(1, and p.1). Replace the final line of the solution with Pr(p .1 1, ) = 37/12 37/ = = (B) [5/6/214] On page 179, in the solution to exercise 55.16, on the second line, change.2(.5) +.8(3) to.8(.5) +.2(3). [6/26/214] On page 1154, in the solution to exercise 59.16, on the second-to-last line, change 17,46,223 to 17,46,223. [6/17/214] On page 116, on the first displayed line, change the integrand to u du [6/13/214] On page 1169, exercise 6.19 is a duplicate of exercise 6.2. [7/9/214] 3 On page 1177, in the solution to exercise 6.17, on the last line, change to.125. [1/14/214] On page 1186, two lines from the bottom, change to [5/26/214] On page 1192, in exercise 61.2, on the displayed line, e x /3 should be e x /3. Errata for ASM Exam C/4 Study Manual (Sixteenth Edition Second Printing) Sorted by Page 5 [6/23/214] On page 1196, in exercise 61.21, on the first line, change X (l ) to X (1). [12/3/213] On page 1198, in the solution to exercise 61.13, on the last displayed line, change to [6/16/214] On page 12, in the solution to exercise 61.22, on the last line, change 2e to (2e ). [6/17/214] On page 1225, in exercise 63.16, two lines after the enumerated list, change longer times to shorter times. [6/2/214] On page 124, 2 lines above Example 64F, change ˆσ 2 to s 2. [12/4/213] On page 1261, in question 1, on the last line, change counts to costs. [1/27/214] On page 1359, in question 8, replace the answer choices with (A) 43.8 (B) 44.1 (C) 44.4 (D) 44.7 (E) 45. [6/5/214] On page 1395, in the solution to question 16, on the second line, in the exponent on (1 e 1/θ ), change 94 to 84. [9/18/214] On page 1398, in the solution to question 26, on the first line, change 1 to 16 in two places. [1/7/214] On page 1399, in the solution to question 29, on the seventh line, change to d q (x q )f (x ) dx.5. [3/22/214] On page 1416, in the solution to question 5, change the two Age 45 s in the heading of the table to Age 35. Change the three subscripts 45 in the two lines after the table, and the subscript 4 in the displayed formula, to 35. [2/8/215] On page 1444, in the solution to question 15, on the second to last line, change #3 to #2. [12/9/214] On page 1458, in the solution to question 21, on the second line, change.25 (x i 2.75) 2 to.25(1 2.75) 2 +.5(3 2.75) (4 2.75) 2. [5/19/214] On page 1476, the solution to question 8 has typos and is unclear. Here is a clearer solution: The function transforming X to Y is y = g (x ) = x 1 = 1/x. Notice that x = g 1 (y ) = 1/y. The derivative of g 1 (y ) is dg 1 dy = 1 y 2 By the formula for transforming densities of random variables, f Y (y ) = f X (1/y ) 1 = (1/(1y ))4 e 1/1y y 2 (1/y )Γ (4) 1 e 1/1y = 1 4 Γ (4)y 5 1 Not surprisingly, inverting X results in an inverse gamma. If you check the tables, you will see that the new parameters are α = 4 and θ = 1/1. The mode of an inverse gamma according to the tables is θ /(α + 1) = 1/(1 5) = 1/5. (B) If you did not recognize the density as an inverse gamma, you could still find the mode by differentiating it and set the derivative equal to. It is easier to differentiate the log: ln f Y = 4 ln 1y 1 + ln y lnγ (4) 2 ln y 1y d ln f Y dy y 2 = 4 1y + 1 1y 2 1 y = 5 y + 1 1y 2 = 6 Errata for ASM Exam C/4 Study Manual (Sixteenth Edition Second Printing) Sorted by Page 5y = y = 1 5 [1/29/214] On page 1485, in the solution to question 3, on the last line, interchange and.5. [1/27/214] On page 1489, in the solution to question 5, on the second displayed line, change ( ) to ( ) 2. [1/27/214] On page 151, the solution to question 8 is incorrect. The correct solution is From the tabular values with d = 2. We also have E[X ] = E[X 2] + Pr(X 2)e (2) = (2) = 199 E[X ] = E[X 5] + Pr(X 5)e (5) Let s get a lower bound for Pr(X 5) by using the values of E[X 5]. 2 Plugging this into (*), we get 199 = E[X 5] + 18 Pr(X 5) (*) E[X 2] = 2 Pr(X 2) + = 1 19 = 2(.9) + 2 E[X 5] = 5 Pr(X 5) + = 5 Pr(X 5) Pr(X 5) Pr(2 X 5) = 5 Pr(X 5) Pr(X 5) = 3 Pr(X 5) + 19 (**) Pr(X 5) Pr(X 5) = 21 Pr(X 5) Pr(X 5) 18 Pr(X 5) 6 7 Then plugging into (**), 6 E[X 5] = Errata for ASM Exam C/4 Study Manual (Sixteenth Edition Second Printing) Sorted by Page 7 To prove that this lower bound is attained, here is a discrete random variable satisfying the question s assumptions with E[X 5] = : x Pr(X = x ) [6/17/214] On page 1522, in the solution to question 3, on the last line of the page, change to
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