P_PracticeExam6_05-1..

More documents

Recommendations

Info

$KSU Math Circle Summer Camp Application Form.pdf$

SECTION 10 for !" < x < +" and y > 0. Therefore, Answer E. Pr Y > X 2 ( ) = +# +# $ $ x 2 "# = 1 ! 28. ( 1 ) , X ( 2) , , X ( 400) +# 1 ! e" x2 e " y dydx = 1 " x2 e ! $ % sY x 2 ( )dx = 1 " x2 e ! $ e "2 x2 $ e dx = 2 % "# 1 2 % 1 ( + ' 2 * $ e dx 1 "# % 2! ! ### 2 " ### $ +# ASM Study Manual for Course P/1 Actuarial Examination. © Copyright 2004-2007 by Krzysztof Ostaszewski - 282 - +# "# " 1 2 & ) ( x + % ( 2 1 + INTEGRAL OF PDF OF N 0, 1 & 2 & ) ) ( ' ( 2* + + ' * +# "# " x2 dx = = 2 % 1 2 %1 = 2 2 . X … are order statistics from a continuous probability distribution with a finite mean, median m and variance ! 2 . Let ! be the cumulative distribution function of the Pr X ! m ( ) standard normal distribution. Which of the following is the best approximation of ( 220) using the Central Limit Theorem? & 20 # A. ! ( 0.05" ) B. 0.0049 C. 0.0532 D. 0.0256 E. ' $ ! % ( " Solution. Let F X be the cumulative distribution function of the distribution under consideration. We have: 400 ( ) = " Pr Exactly j random sample elements are ! m Pr X( 220) ! m j =220 Pr( N ! 220) = Pr N > 219.5 Continuity correction ( ) = 400 # 400& " $ % j ' ( ( FX ( m) ) j =220 j 400) j ( 1) FX ( m) ) = 400 400 # 400& # 1& " $ % j ' ( $ % j =220 2' ( . The last expression is the probability that a binomial random variable with parameters n = 400, p = 0.5 has at least 220 successes. The mean of that binomial distribution is 200 and the variance of it is 100. Denote that binomial variable by N. We have: ( N " 200 ) = Pr > 10 219.5 " 200 # & $ % 10 ' ( ) Answer D. 29. N is a Poisson random variable such that Pr( N ! 1) = 2 " Pr( N = 2). ) 1 " * ( 1.95) ) 0.0256. =
Find the variance of N. A. 0.512 B. 1.121 C. 1.618 D. 3.250 E. 5.000 Solution. Let ! be the mean and the variance of N. We have Pr( N ! 1) = " 0 # e $" + 0! "1 # e $" = 2 # Pr N = 2 1! Therefore, 1+ ! = ! 2 , so that ! 2 " ! " 1 = 0, and 1 ± 1 + 4 ! = . 2 Because the parameter ! must be positive, 1 + 5 ! = " 1.618. 2 This is both the mean and the variance of N. Answer C. ( ) = 2 # " 2 # e $" PRACTICE EXAMINATION NO. 6 30. There are two bowls with play chips. The chips in the first bowl are numbered 1, 2, 3, …, 10, while the chips in the second bowl are numbered 6, 7, 8, …, 25. One chip is chosen randomly from each bowl, and the numbers on the two chips so obtained are compared. What is the probability that the two numbers are equal? 1 1 1 1 1 A. B. C. D. E. 2 5 10 40 50 Solution. There are 10 ! 20 = 200 pairs of chips that can be picked, and of these there are 5 pairs of identical numbers, so that the probability desired is 5 1 = 200 40 . Answer D. ASM Study Manual for Course P/1 Actuarial Examination. © Copyright 2004-2007 by Krzysztof Ostaszewski - 283 - 2! .
Page 1 and 2: PRACTICE EXAMINATION NUMBER 6 1. An
Page 3 and 4: ! 1 $ A. # & " # ! % & B. ! 1 $ # &
Page 5 and 6: PRACTICE EXAMINATION NO. 6 17. The
Page 7 and 8: PRACTICE EXAMINATION NO. 6 27. X is
Page 9 and 10: We have: 40 " ( ) !2 1 = C 10 + x 0
Page 11 and 12: with density function ! 2x, for 0 <
Page 13 and 14: PRACTICE EXAMINATION NO. 6 ! 4 1.5
Page 15 and 16: PRACTICE EXAMINATION NO. 6 Let W =
Page 17 and 18: "# PRACTICE EXAMINATION NO. 6 The d
Page 19 and 20: 20 ! z 10 PRACTICE EXAMINATION NO.
Page 21 and 22: Similarly, Pr X( 7) < m PRACTICE EX
Page 23 and 24: This density at 1 ! 1$ , " # 2 2% &
Page 25 and 26: The survival function for the Weibu
Page 27 and 28: PRACTICE EXAMINATION NO. 6 21. Let
Page 29 and 30: PRACTICE EXAMINATION NO. 6 Let z0.6
Page 31: 26. For a Poisson random variable N

P_PracticeExam6_05-1..

You also want an ePaper? Increase the reach of your titles

Delete template?

Save as template?