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+ 1.5 x IQR + 1.5 x 30.5 b upper boundary = Q3 SOLUTIONS TO TOPIC 5 (STATISTICSAND PROBABILITY) = 137.5 = 183.25 lower boundary = Q1 - 1.5 x IQR NO CALCULATORS = 107 - 1.5 x 30.5 = 61.25 a There are 72 cars in total. 4 + 11 + x + 18 + 12 + 6 + 4 + 2 = .'. x + 57 = There are no data values which are either greater than upper boundary or less than the lower boundary, so are no outliers. 72 72 .'. x = 15 b The modal class is 60 ~ v c v 30 ~ v 40 ~ v 50 ~ v 60 ~ v 70 ~ v 80 ~ v 90 ~ v 100 ~ v < < < < c 70 km h-1. Frequency Cumul. Frequency o o 4 4 15 30 48 60 66 70 72 < 30 < 40 < 50 < 60 < < 11 15 18 12 6 4 2 70 80 90 100 110 • It I 90 I I I I I I I • 100 110 120 130 140 150 160 a From the graph, 100 units were sold for < $200 000. b When N = ~ of 800 = 400, the value = 300 thousand the median = $300 000 c IQR = Q3 - Q1 = $375 000 - $250000 = $125000 cumulative frequency 70 60 /54 ---.----------------------------------50 40 30 20 I 80 d From the graph, 500 units were sold for 300 units were sold for> $330000. .. ----------- P(selling price> '·29..································· '·is..··························· 5 $330000) = ~gg= < $330 000.. ~ a 10 10 20 30 median = 40 60 q,~::::;63 ii Q1::::; 52, .'. 56 km h 70 80 90 100 110 "! QiJ::::; 75 IQR = Q3 - Q1 ::::;75 - 52::::; 23 kmh-1 iii 40% of 72 ::::;29, The line of best fit passes through (120, 22) and (80, so the 40th percentile se 59 km h -1. The line has gradient 2 total mean = -= 11 7 a 7 + 9 + 9 + x + 13 + 13 + 16 7 .'. x = -0.15 Its equation is =11 + 67 = 77 .'. x = 12 = 77 .'. 10 b y = -0.15(100) ii y = -0.15(200) +t C 8 96 = 77 = 19 y- 28 x _ 80 = -0.15 Y - 28 = -0.15x b The total of the scores is 67 + 10 = 77. If Kai scores t goals in the next game, then - 28 - 22 m = --80 - 120 +t .'. t y = + 40 + 40 -0.15x + 12 + 40 = 25 = 10 The first calculation is likely to be more reliable, as it . an interpolated value. 200 grams is outside the domain, however so the second calculation is an extrapolation anc. less reliable. So, she must score 19 goals next time. ,.-A-... 3 ~ ,.-A-... ~ 6 a 22 cm b range = 22 - 4 85 96 98 100 105 106 108 108 112 112 118 120 123 . 125 126 128 133 135 140 144 144 148 148 156 '-v-" Q3 a Q1 = 107 ii median = Q2 = 121.5 iii Q3 = 137.5 iv IQR = 30.5 v range = 156 - 85 = 71 . Mathematics SL - Exam Preparation & Practice Guide (3rd edition) ii 4 cm = 18 cm iii 19 cm iv 17 cm ii IQR = 19 - 11 = 8 cm C 75% of seedlings were taller than 11 cm. d No, the distribution is negatively skewed (skewed to the left). a 7 Die 2 6 5 4 @ 1 b 0 P(sum more than 8) 5 4 @ 2 1 0 1 = P(sum 4 @ 2 1 0 1 2 10 3 2 1 0 1 2 @ 2 1 0 1 2 @ 4 1 0 1 2 @ 4 5 1 b 2 2 4 3 5 6 36 5 18 ii 11 + 8 8 There are * Die 1 a .2 P(T') b 11 + 2 25 19 13 25 25 P(plays at least one sport) 23 13 a 8 19 P(T n E') b a f(x)?: J: 0 on a given interval a:::;; x :::;;b, ~ 3 ii P(T U E') = 1 - P(T' = l. :::;;X2) = r J X2 f(x) dx X1 n E) =1-~xt =1-~=~ a The random variable is discrete. i+i+k+fi=l b = ~x 1 f(x) dx b P(X1:::;; X o /5 = 8 11+8 25 and 6) I H) P(T d 11+8+4 25 9 = {1 + 5, 2 + 4, 3 + 3, 4 + 2, 5 + 1} students. 11 +8 25 ( {marked x } iii P(sum P(H) sum :::;;'7) 1 i'B = i. + 4 + 2 = 25 P(more than 3, but less than 8) = P(4:::;; There are 6 outcomes where the difference is 3. As all outcomes are equally possible, the probability of the difference being 3 is ?: 9) 14 .'. k+f2=l . .'. k = fi " ( E(X)=-2(i)+0(i)+3(fi)+5(fi)=1 ,, b=71-a c=44-a a a+b=71 a+c.=44 a + d = 21 d = 21- a a+ b+ c+ d = a) + (44 - a) + (21- a) = .'. 136 - 2a = a= So, b = 53, c = 26, d = Now a + (71- P(at least one of each sex) 100 = ~= ~ 100 100 15 a The scatter diagram suggests there is a very strong positive relationship between age and arrnual income. 18 b The mean point is (27, 20). 3 The line of best fit passes through (27, 20) and (39, 30). The line has gradient m = 30 - 20 39 _ 27 _ 10 -12 3 UL.- 5 "6 ---' Its equation is b p(RnE') ii P(R nE) iii P(R U B) = 11080= = = 0.18 6 5 4 ~ 3 ! 2 ..........•.. _--J _-)j(...... i . ··x There are 36 possible outcomes each shown by a dot point. x = 30, When ~x - ¥ y = ~(30) - ~ = 22.5 So, the annual income is approximately ii When x = 60, d • . 5 6 •• $22500. y = ~(60) - ~ = 47.5 So, the annual income is approximately 11--+-+-:~~*"i-*-1 .-+--'---'---'---'---'---'--....... Di e 1 1 234 = .: y=~x-~ ( h~h~~: =t::::;:~,d __ sum > 9 5 6 - x - 27 y - 20 = 0.53 18\"03l26 = 0.97 15~0 Die 2 a y - 20 -- $47500. The age of 30 is within the given data range but 60 is outside the range. Predicting the income at 30 years is an interpolation and is more reliable than the extrapolation required for 60. Mathematics SL - Exam Preparation & Pradice Guide (3rd edition) 16 Tom Jerry 0.7............H <::::: 0.60.4 __ < 0.3 .........•. M <::::: 0.60.4 __ a o + 4 + 6 + 3a + 8 2:.f a x= H M,f 3a + 18 a+ 15 H ,f M 3a b = 1.65 + 18 = 1.65a + 24.75 = 6.75 a=-- 6.75 1.35 1.35a + (0.3)(0.6) = (0.7)(0.4) = 1.65 6+4+3+a+2 P(only one of them is successful) = 17 - 2:. fx 2 a=5 0.46 b The standard deviation ~ 1.39 {technology} P (at least one is successful) = = = 1 - P(both miss) = 0.88 3 Using technology a 1 - (0.3)(0.4) J.l ~ 1 - 0.12 4 a 90 3.56, b a ~ 0.512 J.l ~ 67.9, a ~ 12.2 + 100 + 93 + 96 + p + 107 + 98 + 98 + 92 = 97 9 . 774+p=873 a p b .§..-11 DB 5 LB LB~!r- .§..-- 99 The ordered data is: 90, 92, 93, 96, 98, 98, 99, 100, 107 .'. the median is 98. a The sum of the frequencies is 30, so 30 drives were chosen. b ft- w = 8 fr equency DB 6 LB 4 w 2 - w~~1 rrb P(2 whites) ii i6 = 12 x-&.= P( different colours) = = 1- [(f2)h~iJ + (~)(fI) = 1- 12;:1l 1 - P(both the same colour) a = ~ a+ b+c+d= a Since a+ b = 17, d Since b + c = 12, BW' 3 + b+ c = 26 230::::;d 235::::; d +c= 26 240::::;d 245::::; d 17 b+9 = 250::::; d 255::::; d 260::::;d 12 b=3 < < < < < < < < Cumul. Frequency 1 235 240 3 1 4 5 7 9 16 3 6 4 19 25 29 1 30 245 250 255 260 265 a+b=17 9 a U +3 = 17 = 14 a e ~ ~ 63.3% The golfer hit a distance of more than 235 metres on 26 out of the 30 drives sampled. a 3 students have both brown hair and blue eyes. = fa = = 12 = 0.25 P(BI but not Br) ii P(Br I Bl) Frequency 230 Distance (m) d=4 4 b h--, 225 230 235 240 245 250 255 260 265 distance (m) 225::::; d c=9 14 v Since the mean, median, and mode are about 'equal, the data appears to be symmetric. 30 b + c = 12 d U --, a+b=17 c b - c An estimate of the mean is 245. An estimate of the median is 243. The modal class is 240::::; d < 245. + (12)(-&')] B(ID' 18 o - So, if the golfer hit 100 drives, we would expect 100 x ~ ~ 87 drives to travel more than 235 metres. 0.3 6 CALCULATORS a Using technology, x= 7.52 marks b The mode = 8 marks c There are 25 data values, so the median is the 25i1 .'. Mathematics = 13th data value. 46 the median is 8 marks. SL - Exam Preparation & Practice Guide (3rd edition) El 66 70 x a P(X > peA) =:r;- Let 60) ::::;0.309 0.63 = x Over a 100 year period we would expect more than 60 mm of rain to occur 31 times. b P(50 < X < 60) ::::;0.669 Over a 100 year period we would expect between 50 mm and 60 mm of rain to occur 67 times. 7 The tree diagram uses: F for female L for left-handed M for male R for right-handed 0.13 + 0.36 - 0.36x 0.27 = 0.64x .'. x = peA) ::::;0.422 = 12 P(AUE) peA U .'. H peA n E) L = 1- f2 = H + P(B) - peA n E) = 0.46 + ~- peA n E) = 0.46 + ~- H 1 - P((AUE)') E) = peA) peA nE) ::::; 0.258 3VF~R 13 2nd draw a ~M~L 1st draw R 0.76 <15 <: + (0.62)(0.24) ::::;0.198 n L) = (0.38)(0.13) ::::;0.249 b P(F I L) = P(F P(L) P R P 7/ a P(L) = (0.38)(0.13) 2."- s ......•. 10~R<: 7..-P ..:l.. 0.1982 9 ......•. R 8 a r::::;0.898 There is a strong correlation between the age of contestants and the time taken to complete the task. b peat least one red ticket) = 1 - P(no red tickets) = 1- P(PP) =l-fox~ 0.930x - 1.56 by::::; ii We expect that an increase of one year in age will add 0.93 minutes to the time to complete the task. = 1- ~ 9 ay::::; 1.09x - 0.781 b 0.883 r::::; 48 {using technology} 96 ii c There is a strong positive correlation between the times =~+~ cd Subject D is an o~t!ier. e y::::; 1.094x - 0.7S<i3 42 96 y = 48 gives 48::::;1.094x - 0.7813 iii =~+~ 63 96 a r::::;0.995 14 b There appears to be a very strong, positive correlation between monthly rainfall and crop yield. c The least squares regression line is y::::; 0.606x d + 9.4 If x =0, then y::::; 0.606(0) + 9.4. 20 y::::; 0.606(12) + 9.4 o 11 So, the expected yield is approximately data range; we expect it to be unreliable, especially as it predicts such a large yield when there is no rain at all. In contrast, the estimate in ii is an interpolation, and we would expect the estimate to be reliable. = peA) + P(E) - peA n E) peA U E) = peA) so peA + P(E) n E) 12 /: time (minutes) ~ ~ 13 14 15 16 17 16.7 tonnes. e The estimate in i is an extrapolation as it is outside the Hence, ------ 1~ 9.4 tonnes. ::::;16.7 But A and E are independent, ,( I: 15 So, the expected yield is approximately 11 peA U E) 7 16 Finishing times 35 .-----:-:---;:------------cumulative frequency 30 ~-25 ~--------------/7~~--------- ::::;9.4 ii If x = 12, then or RP) =fox~+fox~ x::::; 44.6 So, we expect that an adrenaline injection would reduce the mouse's time to 44.6 s, a reduction of about 3.4 seconds. 10 7 15 P(purple ticket second) = P(PP 48.7813::::; 1.094x .'. P(one ticket of each colour) = P(PR or RP) =fox~+fox~ taken to complete the obstacle course of mice without adrenalin~ and the time'S of mice with adrenaline. Substituting in 8 15 a median re 15th score se 13.8 (or 13.9) minutes b 7~th score se 13.3 min. So, any time less than 13.3 minutes. c 2 finished within 12 minutes and 25 finished within 15 min .'. 23 finished between 12 and 15 minutes. d = peA) P(E). Ql::::; 5 runners out of 30 finished within 13 minutes. Pea runner finished in less than 13 min) = fa- = ~ - peA) P(E). El Mathematics SL - Exam Preparation & Pradice Guide (3rd edition) 15 J.L= 310 mL, (J"= 5 mL ii The relationship has changed from moderate to very strong. a P(300 < X < 310) ~ 0.477 .'. 47.7% lie between 300 mL and 310 mL. b P(X? •• + 32.0 a y R:::2.43x 20 304) ~ 0.885 b If Y 88.5% are at least 304 mL. = 70 R:::2.43x 38 R::: 2.43x 70, e P(X < 300) R:::0.0228 + 32 .'. x ~ 15.6 So, we estimate that Tony revised for 15.6 hours. 16 For English, P(X > 80) R:::0.266 .'. 26.6% scored higher than Ashleigh a e The y-intercept (32%) is the estimate of the result for a student who did not do any revision. ii For History, P(X > 72) R:::0.212 .'. 21.2% scored higher than Ashleigh b z-score for English 80; The gradient of the line indicates that the result will increase by 2.43% for each additional hour studied. 75 = 0.625 21 72 - 60 = 0.800 15 Ashleigh achieved a higher standard in History. z-score for History .'. 17 Let X be the number he correctly guesses, so X ~ B(10, 0.25) a X ~ N(J.L, (J"2) Now > 90) = 0.12 P(X ~ 90) = 0.88 = 0.88 P(X P ( X ; J.L~ 90; J.L) P(Robert fails) = P(X ~ 2) R:::0.526 e = J.L) 0.88 < P ( X ; J.L < 60; < 60 P(X = 3) R:::0.250 5) R:::0.0781 (J" P (Z = 1- P(X ~ 4) R:::1 - 0.9219 22 90 - J.LR:::1.1750(J" .... (1) P(X P( answers exactly 8 correctly) P(achieves a C or better) = P(X? 90 - J.LR:::1.1750 Also, b = (z ~ 90; P If Robert gets the 5 known correct then there are 10 answers he guesses. = 60) J.L) ; . J.L) J.L=O, (J"=1, and X~N(0,12). a P(X ~ 1) R:::0.841 0.20 b P( -0.5 = 0.20 e P(X = 0.20 ~ X ~ 0.5) R:::0.383 > 2) R:::0.0228 we can expect that 0.0228 x 850 R::: 19 speeds measured will exceed the actual speed by more than 2 kmh-1. 60 - J.L';'R::: -0.8416 U J.L=40, (J" 60 - J.LR::: -0.8416(J" .... (2) a Solving (1) and (2): + J.LR::: Adding, 30 P(X > 45) R:::0.159 < X < 50) R:::0.819 b We need to find k such that P(X>k)=O.l 0.8416(J" R:::2.0166(J" P(X 30 (J"R:::-R:::14.88 2.0166 and and X~N(40,52). ii P(35 90 - J.LR::: 1.1750(J" -60 (J"=5, ~ k) = 0.9 .'. k R:::46.4 40 90 - J.LR:::1.1750 x 14.88 So, the minimum 46.4 cm. J.LR:::72.5 Thus : J.LR:::72.5 and (J"~ 14.9 k • x length of the longest 10% of fish is 24 J.L= 500, (J"= 2.5 18 p=t=0.8 a P(X < 495) R:::0.0228 Let X be the number of people sampled who oppose the traffic lights. X ~ B(20, 0.8) a P(X = 25 16) R:::0.218 b P(X? = 16) 1 - P(X R:::1- ~ 15) 0.370 R:::0.630 19 . air b ii y R:::-0.555x R:::-0.647 Mathematics + 71.3 The outlier is (50, 12). With the outlier removed: i r R:::-0.986 ii y R:::-0.637x e & Pradice Guide (3rd edition) J.L= 38.4, a (J"= 4.6 i P(X > + 80.5 43.5) ~ 0.134 ii P(X ~ 36.4) R:::0.332 iii P(30 ~ X ~ 40) R:::0.602 b We need to find k such that P(X > k) P(X ~ k) .'. The gradient of the line is steeper (has become more negative). SL - Exam Preparation b 0.0228 x 10000 R:::228 bottles will require extra sauce. = 0.9 = 0.1 k R:::32.5 k 38.4 So, 90% will reduce their oxygen consumption than 32.5 mL. by more a 26 P(X < 56) = 0.8 p( Z < 56; J.L) iii Of the 779 patients who were 50 or older, 469 were male. means that = 0.8 32 Let X be the number of allergic reactions. :. X B(5000, 0.1) rv < P(X 56 - J.L~ 0.8416(4) then from a, = 470) 56 - J.L~ 3.366 :. 27 J.L= 1020, a P(X a < = J.L~ 52.6 P(X ~ 469) ~ 0.0743 33 15 The length X cm of a chopstick is normally distributed with mean J.Lcm and standard deviation 0.5 cm. < p(X-J.L 0.5 p(z< 2.28% of the containers overflow. so n = 10, P = 0.88. B(10, 0.88) rv a P(X b 24-J.L) 0.5 =0.01 J.L~ 25.2 cm 34 = 8) ~ 0.233 = ? 8) 1 - P(X If X cm is the length of a steel rod then X is normally distributed with mean 13.8 cm. Now P (X < 13.2) = 0.015 P ( X -a13.8 ~ 7) < ~1-0.109 rv B(6, 0.05) so n = ~ 2) So, the manufacturer will have to pay double money back on 0.223% of boxes. ~ 1 - 0.99777 ~ 0.00223 P(x) = P(X Defective Not defective Total 37 581 618 Corn Pineapple 24 617 641 Total 61 1198 1259 = x) = f4(x 3 = b E(X) ~ 0.0485 1~~9 ~ P(corn I defective) ~ a 0.05 0.607 iii 37 out of 618 corn tins are defective. :. P(defective I corn) ~ < 50 ? 50 Total Male 136 469 Female 155 310 605 465 Total 291 779 1070 31 7 631 8 ~ = Xi) + 2(~) + 3(f4) 50 24 =212 36 *~ P(X = 1(-k) ii 37 of the 61 defective tins are corn. .'. L: Xi i=1 61 out of the 1259 tins are defective. P(defective) x E {I, 2, 3} = f4(9) = f4 a 1259 tins .'. + 6), P(2) = f4(8) = ~ _ b ~ -2.17009 -k a P(l) = f4(7) = P(3) 30 0.015 a ~ 0.276 cm 35 P(more than 2) = -0.6 a 6, PT 0.05. The manufacturer must pay dquble money back if more than two items are defective. ~ 1 - P(X = 0.015 13.2 ~ 13.8) p( Z < -~.6) ~ 0.891 29 X 0.01 24 - J.L ~ -2.326 0.5 24 - J.L~ -1.163 A student passes if they pass 8 or more of the tests. P(X = =0.Q1 So, in a sample of 1500, we expect 0.6563 x 1500 ~ 984 containers will hold between 1 litre and 1.03 litres. X 24) 24-J.L) 0.5 c P(1000 ~ X ~ 1030) ~ 0.6563 28 < P (X Now 1000) ~ 0.0912 b Containers that overflow have more than 1050 mL added to them. Now P(X > 1050) ~ 0.0228 :. 0.602 ~!~ ~0.667 I female) = .'. P(50 or older So, a score of 56 is 0.842 standard deviations from the mean. =4 = ~~~~ or older) iv Of the 465 female patients, 310 were 50 or older. 56 - J.L ~ 0.8416 a 56 - J.L~ 0.8416a b If a I 50 :. P(male + k + 0.5 + 0.3 = 1 k + 0.85 = 1 k = 0.15 b E(X) = 0.0599 = = L: xip, 0.15 = 2.05 a 1070 patients 37 0.3 + 1(0.15) + 2(0.5) + 3(0.3) + 1 + 0.9 0(0.05) + 0.2 + m + n = 1 .'. m+n = 0.5 .'. n = 0.5 - m b 605 of the 1070 patients were male. :. P(male) = 6 1 0°;0 ~ ii 291 of the 1070 patients were younger than 50. :. P(younger than 50) = i/o ~ 2 1 E(X) Now 0.565 0.272 L: x.p, = 0(0.3) .'. 1.55 .'. 1.35 = 2m .'. El = + 1(0.2) + 2m + 3n + 3(0.5 - m) m = 0.15 Mathematics SL - Exam Preparation & Pradice Guide (3rd edition) 38 a r;:::::-0.832 b As the number of tomatoes in a bag increases, the median weight of tomatoes in the bag decreases. = - 1'(x) Now = There is a moderate negative linear correlation between the variables. c d If x = 13, then y;::::: -8.21(13) + 224 ii If x = 20, then y;::::: -8.21(20) Iim + 224 So, the median weight is approximately lim (7 - 2x - h) h->O = 7- 2x 60.2 g. .'. the point of contact is (1, 6). Now The estimate in i is therefore more 1'(1) 7 - 2(1) = y-6=5 x-I = 0.613 J1.) = 0.613 P ( Z ~ 15 ~ 13) which is y - 6 = 5x - 5 or = 0.613 2 + h) dy = lirn f(x dx h->O = lim [-2(x 2 = lim -2(x P (X ~ 24) = 0.035 h->O = 0.035 = lim 24 - J1.;::::: -1.812 a 24. - J1.;::::: -1.8120- = lim .... (1) p(z< 33-;J1.) + 2xh + h2) h 0 + 2X2 0 4xh - 2h2 ~ + h) -2j{(2x 0 + h) {as h # O} dy -=-4(-1)=4 dx the tangent has gradient 4. b When =0.738 .'. x=-l, c When x=-l, .'. .... (2) y=-2(-1)2+3=1 the point of contact is (-1, 1). Since the gradient of the tangent is 4, the gradient of the (1) - (2) gives (24 - J1.)- (33 - J1.);::::: -1.812a .'. -9;::::: -2.449a .'. normal is - - 0.6372a i. So, the equation of the normal is y -1 1 x-(-l) -"4 a;:::::3.67 + 1.812(3.67) ;:::::30.7 So, the mean J1.;::::: 30.7 and the standard deviation 4y - 4 = - (x a se 3.67. .'. 4y = -x .'. y=-4- + 1) +3 -x+3 d The normal meets the curve where SOLUTIONS TO TOPIC 6 (CALCULUS) -x+3 -4HO CALCULATORS f(x)=7x-x2 x - 9 -8X2 2 +3 + 12 0 (8x-9)(x+1)=0 .'. f(x + h) = 7(x + h) - (x + h)2 = 7x + 7h - (x2 + 2xh + h2) = 7x - =-2x +3= = -x 8x2 a + 3] = -4x = 0:738 33 - J1.;::::: 0.6372 a 33 - J1.;::::: 0.6372a J1.;::::: 24 [-2x h h->O "'. 2 + h? + 3]h A:? - h->O 33) = 0.262 P (X < 33) +1 - f(x) h->O = lim -2(2x P(X? Also, 5x h h->O P ( Z ~ 24 -; J1.) = Y y=f(x)=-2x2+3 a Let 2 - ;:::::0.287 a a;:::::6.97 40 5 = the tangent has equation P(X ~ 15) P ( X ; J1.~ 15; {as h#O} IX b f(1)=7(1)-12=6 e The estimate in i is an interpolation whereas the estimate 39 h lim X(7-·2x-h) = + 7h When - X2 - 2xh - h2 x =~' y = x = -1 or ~ _2(~)2 +3 =-2(*)+3 15 32 So, the normal meets the curve again at (~' Mathematics + X2 h h->O 118 g. - X2 - 2xh - h2 - 7x 7h - 2xh - h2 h->O = ;:::::60.2 in ii is an extrapolation. reliable. + 7h 7x lim ;:::::118 So, the median weight is approximately - f(x) h h->O = + 224 y;::::: -8.21x + h) f(x Iim h----tO SL - Exam Preparation & Pradice Guide (3rd edition) El H).