F09 - Further Special Discrete Probability Distributions - Assignment (Solutions)

National Junior College Mathematics Department 2017 2016 – 2017 / H2 FM / Further Special Discrete Probability Distributions Page 1 of 4 National Junior College 2016 – 2017 H2 Further Mathematics Topic: Further Special Discrete Probability Distributions Assignment Solutions Qn Suggested Solution MS 1(a)  It is equally likely to snow on any one of the days in the winter season.  Snowing occurs independently across all days in the winter season. B1 B1 1(b)(i) Let X denote the number of days from 1 December for it to first snow. Then X ~ Geo(0.2). Hence P(X = 20) = (0.2)(1– 0.2)19 = 0.0028823 B1 1(b)(ii) P(X < 5) = 1 – (1– 0.2)4 = 0.5904 B1 1(b)(iii) P(X > 20 | X > 5) = P(X > 15) = (1 – 0.2)15 = 0.035184 M1, A1 1(b)(iv) Let n be the least number of days from 1 December for the probability that the first snow falls on or before the nth day is at least 0.95.   1 (1 0.2) 0.95 0.8 0.05 ln 0.8 ln 0.05 ln 0.8 ln 0.05 ln 0.05 ln 0.8 13.4 n n n n n n         Therefore, least value of n = 14. M1 M1 A1 2(a)  Particles are emitted from the material at a constant mean rate.  Particles are emitted from the material independently of one another across the hour. B1 B1 2(b)(i) Let X denote the number of particles that are emitted in a randomly chosen 10 hour period. Then X ~ Po(2.8). Hence P(X > 2) = 1 – P(X ≤ 2) = 0.53055 M1 A1 2(b)(ii) Let Y be the number of periods with more than 2 particles emitted each, out of three randomly chosen 10 hour periods. Then Y ~ B(3, 0.53055). Hence P(Y = 2) = 0.39643 M1 A1 2(b) (iii) Let W denote the number of particles that are emitted in a randomly chosen m minute period. Then 7~ Po 0.7 ~ Po .150 1500 m mW W           For the probability that no particles are emitted in a m minute period to be at least 0.99,       7 1500 7 1500 07 1500 7 1500 P 0 0.99 e 0.990! e 0.99 7 ln 0.991500 ln 0.99 2.1536 m m m W m m           Therefore, the longest time period is 2.15 minutes. M1 M1 M1 A1 www.KiasuExamPaper.com 818

National Junior College Mathematics Department 2017 2016 – 2017 / H2 FM / Further Special Discrete Probability Distributions Page 2 of 4 Qn Suggested Solution MS 2(b)(iv) Let U and V denote the number of particles emitted in the first n hours and the last 20 – n hours of a randomly chosen 20 hour period. Then 7~ Po ,25 nU      140 7~ Po 25 nV      and 28~ Po .5U V                        7 140 7 25 25 28 5 7 140 7 25 25 28 5 4 2 6 47 1 25 P 4 | 6 0.324135 P 4 6 0.324135P 6 P 4 2 0.324135P 6 P 4 P 2 0.324135P 6 e 7 e 140 7 4! 25 2! 25 0.324135 e 28 6! 5 e e 6! 4!2!e n n n n n U U V U U V U V U V U V U V U V n n                                                            240 7 25 628 5 4 27 140 7 25 25 4 228 28 5 5 4 2 4 2 2 2 3 2 0.324135 6! 0.3241354!2! 15 1 0.32413520 20 1 0.02160920 20 1 0.14720 20 20 1176 20 1176 0 12.644, 14, or 6.6436 n n n n n n n n n n n n n n                                                   (rej.) Therefore, there is exactly one integer solution for n, n = 14. M1 M1 M1 M1 M1 A1 www.KiasuExamPaper.com 819

National Junior College Mathematics Department 2017 2016 – 2017 / H2 FM / Further Special Discrete Probability Distributions Page 3 of 4 Qn Suggested Solution MS 3 Since f is the probability distribution function for a geometric distribution,   1 f ( ) 1 , for 1, 2,3, x x p p x     , where p is a positive constant such that 0 < p < 1. Note that f(1) = p. Hence, 2 1 2 f (1)g( ) f (2 )1 f (1) 2 (1 )1 x x x p p pp            Since 0 < p < 1, 1 < 2 – p < 2 and 0 < 1 – p < 1, therefore g(x) > 0 for all .x             2 1 1 1 2 1 1 2 2 2g( ) (1 )1 2 (1 )1 2 (1 ) 1 1 (1 ) 2 1 1 2 2 1 11 2 x x x x x px p pp p p pp p p p p p p p p p p p p p p p p p                                          Therefore, g is a probability distribution function.                     1 2 1 1 2 2 1 1 1 2 1 22 22 22 E g( ) 2 (1 )1 2 (1 )1 2 (1 ) 2 1 1 2 2 2 1 22 x x x x x x x X x x px p pp p p x pp p p x p p p p p p p p p p p pp p                                                M1 M1 B1 M1 A1 M1 M1 A1 www.KiasuExamPaper.com 820

National Junior College Mathematics Department 2017 2016 – 2017 / H2 FM / Further Special Discrete Probability Distributions Page 4 of 4 Qn Suggested Solution MS                      1 2 1 1 2 2 1 1 1 2 2 32 32 3 23 2 E 1 ( 1)g( ) 2( 1) (1 )1 2 ( 1)(1 )1 2 ( 1) (1 ) 2 2 1 1 2 2 2 2 2 2 2 2 x x x x x x x X X x x x px x p pp p p x x pp p p x x p p p p p p p p p p p p p p                                                                   2 22 E ETheref 1 E 2 1 2 ore, 2 X X X X p pp p       and                         22 2 2 22 22 2 22 2 2 22 and Var E E 2 1 1 2 22 1 1 22 21 2 2 1 2 2 X X X p p p pp p p pp p p p p p p p p p p p                         M1 M1 A1 M1 M1 A1 www.KiasuExamPaper.com 821