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Question 4 (a) Suppose that the length of time t (in days) between sales for an automobile salesperson is modelled as the following distribution: fit)-0.5e-05r, where t2(0 (i) Identify the name of the probability model. Solve for the probability that the salesperson goes more than 5 days without a sale (3 marks) (ii) What is the mean and variance of ? (2 marks) (iii) Apply suitable formulas to find the quantile qo.9s for t. (3 marks) (b) Suppose that a random variable X has probability density function, p.d.f. f(x)-4x3. What is the median of X? (3 marks) (c) The random variables Xi . Xi and Is are independent. Each random variable has a mean value of 5 and the variance is 2. Determine the mean and standard deviation of the function 2X1 +Y2+Xi + 10 (5 marks) (d) Suppose the length of time for one customer to be served at the restaurant can be modelled by the exponential distribution with a mean of 4 minutes. Determine the probability that a customer is served in less than 3 minutes on at least 4 of the next 6 days. (4 marks)

Answer 1

Page 1 cheon I Page 1 C1EGA Suechon (a) (1) t(t) = 0.se -0.56 Donere t7o. the above is an exponential die top buhon with parameter =0.5 Do We need the following P[775] = probability is to hos: tesat losenost

Page 2 Chea Page 2 I $) (1) -"-- I --[e-oso que --[o-ecosys] = -2.5 = 0.082084998 goes Hence, the brobability that the salesperson more than 5 days without a cale = 0.082084948

Page 3 Pages] I Page 3 - Cheap 0400 (11) Mean 04 t = 5(T). tttat to. se at. - tedt. -= 0.5 integral is of Laplace The above transform.

T Page 4 I chega I Page 4 Sraban · the boat a t (n) batt · Et Mean = 0.5 m (lt) 60.5H - MHD 0.5 Elt). mean = 2 =2 days. Now variance of t = 5(T)2 [et] e = ECD2 – 22 = ED2 -4. th tedt. AO Now S(T)² X 5.50 -.stat

T Page 5 T T Page 5 T CHEARI § 69260. 20. 20. Sut - 0.5 . 54² 0.5 lbhé The above integral is of Laplace transform. using the above result we have = 0.5 to (2 ) (0.59 24. = 0.5 X 2 H (2) 0.53 - 2 762) - 2 - Bisſ = 2*4= 8

1 Page 6 CHEES - Date: Page No.: 6 Vanance of Ta 8-4 Se mean a 2 days 34 days Vanana (1). Noo, the CoF of X 12 detined as taloog FHP) = PETZad] - Pfret] tludu

I Page 7 - Cheag I Page : Date : 7 GUD. no 0.54 Ft 0.54 du. -05प --[e-o] --[e0920 ale 56. Ftate-o.so. 1.4(0.95) 0.95 = FO)

I Page 8 TCHBag Page: Date: 8 / 1 aconte ca 0.95% re-0.50. 0.954 re-56. of 0.05 =-e or005=enoso Tabara loq Goth the sides. leq 0.05= -0.50. 32.995232274 -0.55 ..443732274 -0.5 t 5 .991464547 days approx 6 days

T Page Chara 1 Page : Date: 08 t(2) 403. OLULI let in find the CDF ofx. F(X)= P EX ca] Arpat = 4 ] =

l Page : Date : 10. 1 9cb nou + I Page 10 Cheaal contd Medran = M. F(X) = 0.5. or M 0.5=M4 o Ma(0.5 & Ma 0.840896415 Medias = 0.840896 415

T Page 1 Cheaal Page: Date : 1. 1 tx-Xe, to are independent and ECXI) = E(12) 5142) = 5. v Cx1) = roky srix) = 2.- Noo . Mean of 2X1 + x2 +Xe to E2XtX 1Kg 10 = = = 2 ELXD + 6(2)+5763) tyo 2(5)+ 5+5+10 10 +10 ho Mean of 2x1+ Xet Kg Ho. _Nowo Standard dawanon of 244 +X2 +*8710 = V no [2x1+x2tX8 +1] Now var [2X+Xz +43 +0 J.-

T Pagele Page: Date : 2 1 CHEErte - var 2x4+x2+%2+103 a 2 + + = Mar [2x] + vor(K)+ Mur (13) -F connepanas tem caill be mere L since Xus are independent = 22 mar (XV + (X2)+r(x3) = 4 har (x1)+ van (K2) + Var (x3) = (4) (21+ 2+2 = 8 + 4 a 12 SD of 2x+x2 + x3 = 112 - 3.46 4101415 Mean = 30 standard deviation - 2.46 4101615.

Page : Date : 3 1 @ T Page 137 cheap Let X = time in: Minutes Xe Đạp (A) there - suho E(X) = e = a. le Ex) = = 4, f(x) det aro +6) te ² ano ayne 1 PEx 2.3J=$. f(x) da, 3 ale ad CX Va

T Page 14 I tchen Page: 14 Date : @ __ Te 41473 e pa -re 31417 Co Ble-0.75 -0.47 2366 552. 0: 5276 33 447 b. 5 Let y 6 days. y u -Yu a that a clustomer is serrel in 3 minutes for next Brnanizal (nb) Bin (67 0.527633447) We need PEx 7,4] - PT 4] + Plaas focat 10:527633447) (10.2 Poscana | +65 (0.5970334475(1-0.92761344) + 60.52763344736.

t Page 15 LChem [ Pase 15 ] Date : od 30. 25940 528 to 11590228 to 0215774 = 0 3968847