IB Questionbank Mathematics Higher Level 3rd edition11.(a)(i)median = 104 gramsA1Note: Accept 105. (ii)30th percentile = 90 gramsA1 (b)80 49(M1)= 31A1Note: Accept answers 30 to 32.[4] 2.(a)(i)the median height is 1.18A1 (ii)the interquartile range is UQ LQ= 1.22 1.13 = 0.09 (accept answers that round to 0.09)A1A1Note: Award A1 for the quartiles, A1 for final answer. (b)(i)1.00 < h 1.051.05 < h 1.101.10 < h 1.151.15 < h 1.201.20 < h 1.251.25 < h 1.305913241910A1A1Note: Award A1 for entries within 1 of the above values andA1 for a total of 80. (ii)unbiased estimate of the population mean = 1.17A1unbiased estimate of the population varianceuse of or GDC(M1)obtain 0.00470A1 (c)(i)P(h 1.15 m) = (0.3375 or 0.338) (allow (0.325))A1(ii)use of the conditional probability formula P(A | B) = P (A B) / P(B)(M1)obtain (A1)(A1)= A1[13] 3. = 6.91(M1)A1((6.7 6.91)2 + ... + (7.3 6.91)2)(M1)= = 0.0543 (3 sf)A1Note: Award M1A0 for 0.233.[4] 4.(a)(i)a, 2a, 3a, ..., na are n consecutive terms of an APwith first term a and common difference aso their mean is M1A1 = AGN0 (ii)4 + 2 4 + 3 4 + ... + 4n > + 100M1 > 2(n + 1) + 100A12n2 + 2n > 2n + 102attempt to solve(M1)n2 > 51so the minimum value of n that satisfies the condition is 8A1N0Note: Award M1A1(M1)A1 for use of equations if there is aclear conversion to an inequality.(b)(i)M = M1= A1= AGN0 EITHERS = M1A1attempt to simplifyS = = A1= AGN0 ORVar(x) = M1A1attempt to simplifyM1Var(x) = = = = A1AGN0(ii)M = S A1attempt to solveM1n = m, as n > 0A1so, then the set has 2n numbers, x1, ..., xn, y1, ..., ynfrom which the first n are all 0 and the last n are all 1(M1)hence the value of the median is A1N0[17] 5.(a)64A1 (b)(i)90 percentile = 87 minimum mark = 87 (accept 88)(M1)A1(ii)70 percentile = 74 minimum mark = 74 (accept 73)(M1)A1[5] 6.(a)Use of (M1)(A1)A1N3 (b)Either attempting to find the new mean or subtracting 3from their (M1)A1N2[5] 7.(a)Median = 50 (allow 49 or 51)A1 (b)Interquartile range = 60 40 = 20 (allow 59, 61, 39, 41and corresponding difference)A1A1N1Note: Award A1 for correct quartiles, A1 for difference. (c)Time to complete puzzle in secondsNumber of applicants20 < t 301030 < t 35635 < t 40940 < t 451145 < t 501450 < t 602560 < t 8025Notes: Allow 1 on each entry provided total adds up to 100.M1A1[5] 8.(a)Lines on graph(M1)100 students score 40 marks or fewer.A1N2 (b)Identifying 200 and 600A1Lines on graph.(M1)a = 55, b = 75A1A1N1N1[6]9.(a)(i)P(AB) = P(A) + P(B) = 0.7A1 (ii)P(AB) = P(A) + P(B) P(AB)(M1) = P(A) + P(B) P(A)P(B)(M1) = 0.3 + 0.4 0.12 = 0.58A1 (b)P(AB) = P(A) + P(B) P(AB) = 0.3 + 0.4 0.6 = 0.1A1P(AB) = (M1) = = 0.25A1[7] 10.R is rabbit with the diseaseP is rabbit testing positive for the disease (a)P(P) = P(RP) + P(RP)= 0.01 0.99 + 0.99 0.001M1= 0.01089 (= 0.0109)A1Note: Award M1 for a correct tree diagram with correct probability values shown. (b)P(R|P) = M1A1= 10 % (or other valid argument)R1[5]11.(a)A1Note: Award A1 for intercepts of 0 and 2 and a concave down curvein the given domainNote: Award A0 if the cubic graph is extended outside the domain [0, 2]. (b)(M1)Note: The correct limits and = 1 must be seen but may be seen later.A1 = 1M1(A1)k = A1[6]12.(a)METHOD 1P(3 defective in first 8) = M1A1A1Note: Award M1 for multiplication of probabilities with decreasingdenominators.Award A1 for multiplication of correct eight probabilities.Award A1 for multiplying by .= A1METHOD 2P(3 defective DVD players from 8) = M1A1Note: Award M1 for an expression of this form containing three combinations.= M1= A1 (b)P(9th selected is 4th defective player3 defective in first 8) = (A1)P(9th selected is 4th defective player) = M1= A1[7] 13.(a)M1k = A1Note: Only FT on positive values of k. (b)(i)E(X) = M1= = (= 0.111)A1 (ii)median given by a such that P (X < a) = 0.5M1 = 2(A1)(a + 2)3 0 = 6a = 2 (= 0.183)A1[7] 14.(a)P(X < 30) = 0.4P(X < 55) = 0.9or relevant sketch(M1)given Z = P(Z < z) = 0.4 = 0.253...(A1)P(Z < z) = 0.9 = 1.28...(A1) = 30 + (0.253...) = 55 (1.28...) M1 = 16.3, = 34.1A1Note: Accept 16 and 34.Note: Working with 830 and 855 will only gain the two M marks. (b)X ~ N(34.12..., 16.28...2)late to school X > 60P(X > 60) = 0.056...(A1)number of students late = 0.0560... 1200(M1) = 67 (to nearest integer)A1Note: Accept 62 for use of 34 and 16.(c)P(X > 60 | X > 30) = M1 = 0.0935 (accept anything between 0.093 and 0.094)A1Note: If 34 and 16 are used 0.0870 is obtained. This should be accepted. (d)let L be the random variable of the number of students wholeave school in a 30 minute intervalsince 24 30 = 720A1L ~ Po(720)P(L 700) = 1 P(L 699)(M1) = 0.777A1Note: Award M1A0 for P (L > 700) = 1 P(L 700) (this leads to 0.765). (e)(i)Y ~ B(200, 0.7767)(M1)E(Y) = 200 0.7767 = 155A1Note: On FT, use of 0.765 will lead to 153. (ii)P(Y > 150) = 1 P(Y 150)(M1)= 0.797A1Note: Accept 0.799 from using rounded answer.Note: On FT, use of 0.765 will lead to 0.666.[17] 15.(a)P(x < 1.4) = 0.691 (accept 0.692)A1 (b)METHOD 1Y ~ B(6, 0.3085...)(M1)P(Y 4) = 1 P(Y 3)(M1)= 0.0775 (accept 0.0778 if 3 s.f. approximation from (a) used)A1 METHOD 2X ~ B(6, 0.6914...)(M1)P(X 2)(M1)= 0.0775 (accept 0.0778 if 3 s.f. approximation from (a) used)A1(c)P(x < 1 | x < 1.4) = M1 = = 0.0966 (accept 0.0967)A1[6] 16.(a)P(x = 0) = 0.607A1 (b)EITHERUsing X ~ Po(3)(M1)ORUsing (0.6065...)6(M1)THENP(X = 0) = 0.0498A1 (c)X ~ Po(0.5t)(M1)P(x 1) = 1 P(x = 0)(M1)P(x = 0) < 0.01A1e0.5t < 0.01A10.5t < ln (0.01)(M1)t > 9.21 monthstherefore 10 monthsA1N4Note: Full marks can be awarded for answers obtained directly from GDCif a systematic method is used and clearly shown. (d)(i)P(1 or 2 accidents) = 0.37908A1E(B) = 1000 0.60653... + 500 0.37908...M1A1 = $796 (accept $797 or $796.07)A1 (ii)P(2000) = P(1000, 1000, 0) + P(1000, 0, 1000)+ P(0, 1000, 1000) + P(1000, 500, 500)+ P(500, 1000, 500) + P(500, 500, 1000)(M1)(A1)Note: Award M1 for noting that 2000 can be written both as2 1000 + 1 0 and 2 500 + 1 1000.= 3(0.6065...)2(0.01437...) + 3(0.3790...)2(0.6065...)M1A1= 0.277 (accept 0.278)A1[18]17.P (six in first throw) = (A1)P (six in third throw) = (M1)(A1)P (six in fifth throw) = P(A obtains first six) = (M1)recognizing that the common ratio is (A1)P(A obtains first six) = (by summing the infinite GP)M1= A1[7] 18.EITHERlet yi = xi 12M1A1x = y = 3A1M1A1= 10(9 + 4) = 130A1ORM1A1A1x = 3, (M1) = 10(9 + 100)A1 = 1090 2400 + 1440 = 130A1[6] 19.(a)M1A1M1A1ea + 1 = 1 A1Note: Accept e0 instead of 1.ea = ea = a = A1a = ln 2AG(b)M1A1eMa + 1 = eMa = A1Ma = ln 2M = = 2A1 (c)P(1 < X < 3) = M1A1= e3a + eaA1P(X < 3|X > 1) = M1A1= A1= A1= (e3a + ea)= A1= A1Note: Award full marks for P (X < 3X > 1) = P (X < 2) = or quotingproperties of exponential distribution.[20]20.(a)the total area under the graph of the pdf is unity(A1)area = = A1= c = 6A1 (b)E(X) = 6(M1)= A1Note: Allow an answer obtained by a symmetry argument.[5] 21.recognition of X ~ B(M1)P(X = 3) = A1P(X = 2) = A1A1[4] 22.weight of glass = XX ~ N(160, 2)P(X < 160 + 14) = P(X < 174) = 0.75(M1)(A1)Note: P(X < 160 14) = P(X < 146) = 0.25 can also be used.P = 0.75(M1) = 0.6745...(M1)(A1) = 20.8A1[6] 23.(a)number of patients in 30 minute period = XX ~ Po(3)(A1)P(X = 0) = 0.0498(M1)A1 (b)number of patients in working period = YY ~ Po(12)(A1)P (X < 10) = P (X 9) = 0.242(M1)A1 (c)number of working period with less than 10 patients = WW ~ B(6, 0.2424...)(M1)(A1)P(W 3) = 0.966(M1)A1Note: Accept exact answers in parts (a) to (c). (d)number of patients in t minute interval = XX ~ Po(T)P(X 2) = 0.95P(X = 0) + P(X = 1) = 0.05(M1)(A1)eT(1 + T) = 0.05(M1)T = 4.74(A1)t = 47.4 minutesA1[15] 24.(a)p + q = 0.44A12.5p + 3.5q = 1.25(M1)A1p = 0.29, q = 0.15A1 (b)use of Var(X) = E(X2) E(X)2(M1)Var(X) = 2.10A1[6] 25.(a)required to solve P = 0.8(M1) = 0.842... (or equivalent)(M1) = 7.13 (days)A1N1 (b)P (survival after 21 days) = 0.337(M1)A1[5]26.(a)required to solve e + e = 0.123M1A1solving to obtain = 3.63A2N2Note: Award A2 if an additional negative solution is seen but A0 if onlya negative solution is seen. (b)P(0 < X < 9)= P(X 8) P(X = 0) (or equivalent)(M1)= 0.961A1[6] 27.(a)(i)mean = 6(A1)P(5 customers before 10:00) = = 0.161(M1)A1 (ii)P(2 in 30 mins5 in 60 mins)= (M1)(A1)= A1= (accept 0.312 or 0.313)A1 (b)(i)P(T > t) = P 0 or 1 arrivals in [0, t] = R1 (ii)the distribution function is given byF(t) = 1 M1A1the probability density is given, for t > 0, byf(t) = F(t)(M1)= A1= A1(iii)E(T) = M1= M1A1= M1A1= = 20A1Note: Accept a method based on adding two exponential variables.[19]28.(A1)PA1P(L) = A1P(R|L) = (M1)= A1[5] 29.X ~ N(, 2)P(X 5) = 0.670 = 0.4399...M1A1P(X > 7) = 0.124 = 1.155...A1solve simultaneously + 0.4399 = 5 and + 1.1552 = 7M1 = 3.77 (3 sf)A1N3the expected weight loss is 3.77 kgNote: Award A0 for = 3.78 (answer obtained due to early rounding).[5] 30.(a)P(X = 1) + P(X = 3) = P(X = 0) + P(X = 2)mem + (M1)(A1)m3 3m2 + 6m 6 = 0(M1)m = 1.5961 (4 decimal places)A1 (b)m = 1.5961 = 0.582(M1)A1[6] 31.+123123423453456(a)let T be Tims score(i)P(T = 6) = (= 0.111 to 3 s.f.)A1 (ii)P(T 3) = 1 P(T 2) = 1 (= 0.889 to 3 s.f.)(M1)A1 (b)let B be Bills score(i)P(T = 6 and B = 6) = (= 0.012 to 3 s.f.)(M1)A1 (ii)P(B = T) = P(2)P(2) + P(3)P(3) + ... + P(6)P(6)= M1= (= 0.235 to 3 s.f.)A1(c)(i)EITHERP(X 2) = M1A1because P(X 2) = P((a, b, c, d)a, b, c, d = 1, 2)R1or equivalentP(X 2) = AGORthere are sixteen possible permutations, which areCombinationsNumber1111111124112261222422221M1A1Note: This information may be presented in a variety of forms.P(X 2) = A1= AG (ii)x123P(X = x)A1A1(iii)E(X) = (M1) = = (= 2.79 to 3 sf)A1E(X2) = = = (= 7.98 to 3 sf)A1Var(X) = E(X2) (E(X))2(M1) = 0.191 (to 3 s.f.)A1Note: Award M1A0 for answers obtained using rounded values(e.g. Var(X) = 0.196). (d)CombinationsNumber33116322112P(total is 8 M1A1since P(X = 3) = P(total is 8(X = 3)) = M1= (= 0.277)A1[21]32.(a)M1A1Note: Award M1 for LHS and A1 for setting = 1 at any stage.A1A1k = AG (b)E(X) = (M1)= 1A1Note: Accept answers that indicate use of symmetry.[6] 33.(a)H ~ N(166.5, 52)P(H 170) = 0.242...(M1)(A1)0.242... 63 = 15.2A1so, approximately 15 students (b)correct mean: 161.5 (cm)A1variance remains the same, i.e. 25 (cm2)A2[6] 34.(a)X ~ Po(0.6)P(X 1) = 1 P(X = 0)M1= 0.451A1N1 (b)Y ~ Po(2.4)(M1)P(Y = 3) = 0.209A1(c)Z ~ Po(0.6n)(M1)P(Z 3) = 1 P(Z 2) > 0.8(M1)Note: Only one of these M1 marks may be implied.n 7.132.. . (hours)so, Mr Lee needs to fish for at least 8 complete hoursA1N2Note: Accept a shown trial and error method that leads to a correct solution.[7] 35.(a)A1A1Note: Award A1 for a diagram with two intersecting regions and atleast the value of the intersection. (b)A1 (c)A1[4]36.(a)using the law of total probabilities:(M1)0.1p + 0.3(1 p) = 0.22A10.1p + 0.3 0.3p = 0.220.2p = 0.08p = = 0.4p = 40% (accept 0.4)A1 (b)required probability = M1= (0.182)A1[5] 37.(a)X ~ N(998, 2.52)M1P(X > 1000) = 0.212AG (b)X ~ B(5, 0.2119...)evidence of binomial(M1)P(X = 3) = (0.2119...)3 (0.7881...)2 = 0.0591 (accept 0.0592)(M1)A1 (c)P(X 1) = 1 P(X = 0)(M1)1 (0.7881...)n > 0.99(0.7881...)n < 0.01A1Note: Award A1 for line 2 or line 3 or equivalent.n > 19.3(A1)minimum number of bottles required is 20A1N2 (d) = 1.1998 (accept 1.2)M1A1 = 0.3999 (accept 0.4)M1A1 = 999 (ml), = 2.50 (ml)A1A1 (e)(i)M1A112m2 4m3 m4 = 0(A1)m = 6, 0, 2 m = 2A1N2 (ii)P(X > 2) = 1 P(X 2)(M1)= 1 P(X = 0) P(X = 1) P(X = 2)= 1 e2 2e2 = 0.323A1[20] 38.(a)P(X 2) = 0.4(M1)P(X = 0) + P(X = 1) + P(X = 2) = 0.4(A1)A1mean, = 3.11A1 (b)using a GDC(M1)mode = 3A1[6]39.P MG = (M1)= M1A1A1= = A1[5] 40.M1A1A1Note: Award M1 for the integral equal to 0.5 A1A1 for the correct limits. = 0.5M1A1A1Note: Award M1 for correct integration A1A1 for correct substitutions.(A1)M = 264A1[8] 41.the waiting time, X ~ N (18, 42)(a)P(X > 25) = 0.0401(M1)A1 (b)P X < 20X > 15= (A1)Note: Only one of the above A1 marks can be implied.= = 0.601(M1)A1[6]42.(a)Ying:Number of heads0123P(M1)A1Mario:Number of heads012P(M1)A1P(Ying wins) = = (M1)A1P(Mario wins) = = (M1)A1P(draw) = 1 = A1Yings winnings:X5102Pexpected winnings = M1A1= 0A1(b)P(Ying wins on 1st round) = (A1)P (Ying wins on 2nd round) = (M1)(A1)P (Ying wins on 3rd round) = etc.(A1)P(Ying wins) = (M1)= M1A1= (= 0.727)A1[20] 43.(a)Any consideration of (M1)0A1N2(b)METHOD 1Let the upper and lower quartiles be a and -aM1? A1? ? A1? ? = a = A1Since the function is symmetrical about t = 0,interquartile range is R1 METHOD 2M1A1? A1? ? a = A1The interquartile range is R1[7] 44.(a)Using (M1)4c + 6c + 6c + 4c = 1(20c = 1)A1A1N1 (b)Using E(X) = (M1)= (1 ? 0.2) + (2 ? 0.3) + (3 ? 0.3) + (4 ? 0.2)(A1)= 2.5A1N1Notes: Only one of the first two marks can be implied.Award M1A1A1 if the x values are averaged only ifsymmetry is explicitly mentioned.[6] 45.EITHERUsing P(A | B) = (M1)0.6P(B) = P(A ? B)A1Using P(A ? B) = P(A) + P(B) - P(A ? B) to obtain0.8 = 0.6 + P(B) - P(A ? B)A1Substituting 0.6P(B) = P(A ? B) into above equationM1 ORAs P(A | B) = P(A) then A and B are independent eventsM1R1Using P(A ? B) = P(A) + P(B) - P(A) ? P(B)A1to obtain 0.8 = 0.6 + P(B) - 0.6 ? P(B)A1 THEN0.8 = 0.6 + 0.4P(B)A1P(B) = 0.5A1N1[6] 46.(a)P(X ? 84) = P(Z ? -1.62...) = 0.0524(M1)A1N2Note: Accept 0.0526.(b)P(Z ? z) = 0.01 ? z = - 2.326...(M1)P(X ? x) = P(Z ? z) = 0.01 ? z = -2.326...x = 81.4(accept 81)A1N2 (c)P(X ? 84) = 0.12 ? z = -1.1749...(M1)mean is 88.3(accept 88)A1N2[6] 47.METHOD 1(M1) Let P(I) be the probability of flying IS Air, P(U) be the probabilityflying UN Air and P(L) be the probability of luggage lost.P(I | L) = (M1)= A1A1A1= A1METHOD 2Expected number of suitcases lost by UN Air is 0.18 ? 70 = 12.6M1A1Expected number of suitcases lost by IS Air is 0.23 ? 65 = 14.95A1P(I | L) = M1A1= 0.543A1[6] 48.(a)mean for 30 days: 30 ? 0.2 = 6.(A1)(M1)A1N3 (b)P(X > 3) = 1 - P(X ? 3) = 1 - e-6(1 + 6 + 18 + 36) = 0.849(M1)A1N2 (c)EITHERmean for five days: 5 ? 0.2 = 1(A1)P(X = 0) = e-1 (= 0.368)A1N2ORmean for one day: 0.2(A1)P(X = 0) = (e-0.2 )5 = e-1 (= 0.368)A1N2 (d)Required probability = e-0.2 ? e-0.2 ? (1 - e-0.2)M1A1= 0.122A1N3 (e)Expected cost is 1850 ? 6 = 11100 eurosA1 (f)On any one day P(X = 0) = e- 0.2Therefore, M1A1N2[13]49.Attempting to find the mode graphically or by using f ?(x) = 12x(2 - 3x)(M1)Mode = A1Use of (M1)A1M1A1N4[6] 50.(a)X ~ B(n, 0.4)(A1)Using P(X = x) = (M1)P(X = 2) = A1N3 (b)P(X = 2) = 0.121A1Using an appropriate method (including trial and error) to solvetheir equation.(M1)n = 10A1N2Note: Do not award the last A1 if any other solution is givenin their final answer.[6] 51.(a)(i)P(4.8 < X < 7.5) = P(-0.8 < Z < 1)(M1)= 0.629A1N2Note: Accept P(4.8 ? X ? 7.5) = P(-0.8 ? Z ? 1). (ii)Stating P(X < d) = 0.15 or sketching an appropriately labelleddiagram.A1(M1)(A1)d = (-1.0364...)(1.5) + 6(M1)= 4.45 (km)A1N4 (b)Stating both P(X > 8) = 0.1 and P(X < 2) = 0.05 or sketching anappropriately labelled diagram.R1Setting up two equations in m and s(M1)8 = m + (1.281...)s and 2 = m - (1.644...)sA1Attempting to solve for m and s (including by graphical means)(M1)s = 2.05 (km) and m = 5.37 (km)A1A1N4Note: Accept m = 5.36, 5.38. (c)(i)Use of the Poisson distribution in an inequality.M1P(T ? 3) = 1 - P(T ? 2)(A1)= 0.679...A1Required probability is (0.679...)2 = 0.461M1A1N3Note: Allow FT for their value of P(T ? 3). (ii)t ~ Po(17.5)A1(M1)= 0.0849A1N2[21] 52.(a)X2 is a geometric random variableA1with A1Therefore E(X2) = A1 (b)X3 is a geometric random variable with A1Therefore E(X3) = A1(c)E(X4) = E(X5) = E(X6) = A1A1A1E(X1) = 1(or X1 = 1)A1Expected number of tosses = M1= 14.7AG[10] 53.Let X denote the number of imperfect glasses in the sample(M1)For recognising binomial or proportion or PoissonA1(X ~ B(200, p) where p-value is the probability of a glass beingimperfect)Let H0: p-value = 0.02 and H1: p-value > 0.02A1A1EITHERp-value = 0.0493A2Using the binomial distribution p-value = 0.0493 > 0.01 we accept H0R1ORp-value = 0.0511A2Using the Poisson approximation to the binomial distribution sincep-value = 0.0511 > 0.01 we accept H0R1ORp-value = 0.0217A2Using the one proportion z-test sincep-value = 0.0217 > 0.01 we accept H0R1Note: Use of critical values is acceptable.[7] 54.(a)(M1)= A1Note: Do not accept 2.73.(b)H0: the data can be modelled by a Poisson distributionA1H1: the data cannot be modelled by a Poisson distributionA1Numberof calls012345? 6Observedfrequency91222101188Expectedfrequency14.28919.46917.68412.0476.5664.701A3Note: Award A2 for one error, A1 for two errors, A0 for threeor more errors.Combining last two columns(M1)Note: Allow FT from not combining the last two columns and / orgetting 2.98 for the last expected frequency.EITHER(M1)(A1)= 8.804(accept 8.8)A1 = 6 - 2 = 4, A1A1Hence 8.804 is not significant since 8.804 < 9.488 and weaccept H0R1ORp-value = 0.0662(accept 0.066) which is not significant sinceA50.0662 > 0.05 and we accept H0R1N0[14] 55.(a)A1M1A1(b)(i)H0: Data can be modelled by a normal distributionH1: Data cannot be modelled by a normal distributionA1(ii)The expected frequencies areIntervalx < 1.601.60 ? x < 1.651.65 ? x < 1.701.70 ? x < 1.751.75 ? x < 1.80x ?1.80Exp Freq8.0430.1966.3175.6044.7516.10A1A1A1A1A1A1M1A1Degrees of freedom = 3A1Critical value = 6.251 or p-value = 0.35A1The data can be modelled by a normal distribution.R1[15] 56.(a)P(Z = n) = M1A1= M1A1= A1This shows that Z is Poisson distributed with mean (l + m).R1 (b)The result is (trivially) true for n = 1.A1Assuming it to be true for n = k, i.e. M1Consider M1A1which, using (a) is Po(km + m) i.e. Po([k + 1]m)A1Hence proved by induction since true for n = k ? true forn = k + 1 and we have shown true for n = 1.R1[12] 57.(a)1, 2, 3, 4A1(b)P(Y = 1) = A1P(Y = 2) = A1P(Y = 3) = A1P(Y = 4) = A1 (c)E(Y) = 1 M1= 2A1[7] 58.(a) (= 0.463)M1A1 (b)E(X) = = 2.31M1A1 (c)M1(A1)m = 2.41A1[7] 59.(a)X ~ Po(3.2)P(X = 4) = = 0.178A1 (b)(i)Var(Y) = E(Y2) E2(Y)(M1)m = 5.5 m2A1m = 1.90 (or m = 2.90 which is invalid)A1(ii)Y ~ Po(1.90)P(Y = 3) = (M1)= 0.171A1 (c)Required probability = 0.171 0.178 = 0.0304 (accept 0.0305)(M1)A1[8] 60.(a)X ~ N(231, 1.52)P(X < 228) = 0.0228(M1)A1Note: Accept 0.0227. (b)(i)X ~ N(, 1.52)P(X < 228) = 0.002 = 2.878...M1A1 = 232 gramsA1N3 (ii)X ~ N(231, 2) = 2.878...M1A1 = 1.04 gramsA1N3 (c)X ~ B(100, 0.002)(M1)P(X 1) = 0.982...(A1)P(X 2) = 1 P(X 1) = 0.0174A1[11] 61.X ~ Po ()P(X = 10) = 2P(X = 9)(M1)A1A1 = = 10 2 = 20A1E(X) = 20A1[5] (c)X ~ B(100, 0.002)(M1)P(X 1) = 0.982...(A1)P(X 2) = 1 P(X 1) = 0.0174A1[11] 62.P(A B) = P(A) + P(B) P(A B)M1= A1M1A1P(A/B) = M1A1[6] 63.(a)Using P(X = x) = 1(M1) k 1 + k 2 + k 3 + k 4 + k 5 = 15k = 1M1A1k = AGN0 (b)Using E(X) = xP(X = x)(M1)= 0 A1= A1N2[6]64.(a)Using (M1)A1A1A1k = 1k = A1 (b)E(X) = M1Let u = 4 x2(M1)A1When x = 0, u = 4A1When x = 1, u = 3A1E(X) = M1= A1= AG(c)The median m satisfiesM1A1A1A1m = 2 sinA1We need to determine whether Consider the graph of y = sin xM1Since the graph of y = sin x or 0 x is concave downwards and it follows by inspection that R1hence m = R1[20] 65.(a)P(RR) = (M1)= A1N2(b)P(RR) = A1Forming equation 12 15 = 2(4 + n) (3 + n)(M1)12 + 7n + n2 = 90A1 n2 + 7n 78 = 0A1n = 6AGN0 (c)EITHERP(A) = A1P(RR) = P(A RR) + P(B RR)(M1)= = A1N2ORA1P(RR) = M1= A1N2(d)P(1or 6) = P(A)M1P(ARR) = (M1)= M1= A1N2[13] 66.P(X > 90) = 0.15 and P(X < 40) = 0.12(M1)Finding standardized values 1.036, 1.175A1A1Setting up the equations 1.036 = (M1) = 66.6, = 22.6A1A1N2N2[6] 67.(a)Total number of ways of selecting 4 from 30 = (M1)Number of ways of choosing 2B 2G = (M1)P(2B or 2G) = = 0.368A1N2 (b)Number of ways of choosing 4B = , choosing 4G = A1P(4B or 4G) = (M1)= 0.130A1N2[6]68.(a)P(3 X 5) = P(X 5) P(X 2)(M1)= 0.547A1N2 (b)P(X 3) = 1 P(X 2)(M1)= 0.762A1N2 (c)P(3 X 5X 3) = (M1)= 0.718A1N2[6] 69.(a)A2 (b)Mode = 2A1 (c)Using E (X) = (M1)Mean = A1= (A1)= (1.51)A1N2(d)The median m satisfies M1A1(A1) m4 + 2m2 12 = 0m2 = = 2.60555...(A1)m = 1.61A1N3[12]