Answers to Section 25 of Marcel B. Finan's "A Probability Course for the Actuaries"
Dr. Finan's study guide is an excellent resource for those preparing to take Actuarial Exam 1/P on probability.
Problems that require numerical answers are answered here, but it is still the responsibility of the student to provide his or her own work for these problems. These answers are meant to enable students to independently verify the correctness of their reasoning by checking to see if the end result they obtained is correct. Questions from the study guide that require proofs or diagrams are not addressed here, as the end result of those questions is known in advance, and it is the responsibility of the student to provide the procedure for getting there.
Section 25
Answer 25.1a. about 0.2389
Answer 25.1b. about 0.1423
Answer 25.1c. about 0.6188
Answer 25.1d. about 88.12
Answer 25.2a. 0.7157
Answer 25.2b. about 0.8926
Answer 25.2c. about 0.0238
Answer 25.3a. 0.5
Answer 25.3b. about 0.9876
Answer 25.4a. about 0.4772
Answer 25.4b. about 0.004
Answer 25.5. about 0.0228= 57/2500
Answer 25.6a. about 0.9452
Answer 25.6b. about 0.8185
Answer 25.7a. about 0.2578
Answer 25.7b. about 0.832
Answer 25.8. about 0.223584
Answer 25.9. The standard deviation of X is 75.
Answer 25.10a. P(Y= 1060)=e-2/[30√(2π)]
Answer 25.10b. x= 1098
Answer 25.11a. P(X > 5.5)= about 0.4721
Answer 25.11b. P(4 < X < 6.5) = about 0.1389
Answer 25.11c. P(X < 8) = about 0.6664
Answer 25.11d. P(|X − 7| ≥ 4) = about 0.5808
Answer 25.12. σ =about 0.01287
Answer 25.13. about 0.8664
Answer 25.14a. about 0.1736
Answer 25.14b. E(X) = about 1070.4
Answer 25.15. 0.1151
Answer 25.16a. 0.1056
Answer 25.16b. 362.84
Answer 25.17. 1:18 p.m.
Answer 25.18. about 0.1788
Answer 25.19a. about 0.4129
Answer 25.19b. about 0.2993
Answer 25.19c. about 26.81 milliamperes
See a list of all the answer keys to Dr. Finan's study guide for Exam 1/P here.
Published by G. Stolyarov II
G. Stolyarov II is a science fiction novelist, independent essayist, poet, amateur mathematician, composer, author, and actuary. View profile
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4 Comments
Post a CommentChris S, thank you for showing your work. After examining it, I think that you are correct, and I have revised the answer keys accordingly.
Here are my calculations:
25.14 (a) [Q'n is, paraphrasing: exam scores are normally distr w/ mean 1000, st dev 160, what should the mean be set to, w/o changing the st dev, such that 1400 is the 98th percentile]. For 1400 to be 98th percentile, 1400 must be, from the table on p 248, about 2.06 st devs above the mean. 2.06 * 160 = 329.6. 1400 - 329.6 = 1070.4.
25.19 [Q'n: mean 12, st dev 9] (a) p x > 14 = 1-phi((14-12)/9) = 1 - phi(2/9) = ~0.4129
(b) p 9 < x < 16 = phi((16-12)/9) - phi((9-12)/9) = 0.67 - (1 - .6293) = 0.2993
(c) 0.95 =~ phi(1.65). So 1.65 = (x - 12)/ 9. x = 26.85 milliamperes.
Thanks!
Chris S, thank you for your comments. I would ask that you present your calculations for these problems here (multiple comment posts may be required, as there are length limitations to each post). This will enable me to check your work against the official answers. Dr. Finan's answer key on these problems agrees with mine, but it is possible that you are correct; I just need to see how you arrived at your answers.
The answer to question 25.14 appears to be incorrect -- should be ~1070 I think.
Also the answers to all three of 25.19 were apparently calculated with variance 9 rather than st dev 9 as per the question.