(a)(i) Read the time at cumulative frequency \(60/2=30\): the median is approximately \(\boxed{2.5\text{ hours}}\).

(a)(ii) At cumulative frequencies 15 and 45, the quartiles are about 1.8 and 3 hours. Thus \(\operatorname{IQR}=3-1.8=\boxed{1.2\text{ hours}}\).

(b) The cumulative frequencies at 1, 2 and 3 hours are 7, 18 and 45. The missing frequencies are \(18-7=\boxed{11}\) and \(45-18=\boxed{27}\). The completed frequencies \(7,11,27,9,4,2\) add to 60.

(a) Read the times at cumulative frequencies \(120/4=30\) and \(3(120)/4=90\): \(Q_1\approx4.2\) h and \(Q_3\approx5.6\) h. Hence \(\boxed{\mathrm{IQR}\approx1.4\text{ h}}\).

(b) \(0.7(120)=84\). From cumulative frequency 84, read \(\boxed{k\approx5.4}\).

(a) Plot (30,6), (40,42), (50,76), (60,170), (70,190), (80,200) and join them with an increasing curve or polygon.

(b) The 75th percentile is at cumulative frequency \(0.75(200)=150\). Reading the graph gives about \(\boxed{58\text{ cm}}\). Linear interpolation between (50,76) and (60,170) gives \(50+10(150-76)/94\approx57.9\text{ cm}\).

(a) Read Q₁ at cumulative frequency 50: about16 s, and Q₃ at150: about27 s. IQR=\(27-16=\boxed{11\text{ s}}\). (b) About194 finish by40 s. Hence \(200-194=\boxed{6}\) take longer.

(a)(i)(a) At CF \(160/2=80\), median \(\approx25\) years. (b) At CF \(160/4=40\), lower quartile \(\approx17.5\) years.

(c) CF at age 50 is about 148, so \(160-148=12\) people. (d) CF \(0.65(160)=104\) corresponds to about 30 years.

(ii)(a) At CF 120, the upper quartile is about 34 years. The five plotted values are \(1,17.5,25,34,70\). (b) Approximately \(50\%\).

(b) Frequency densities are \(52/20=2.6,37/10=3.7,24/20=1.2,60/30=2\).

(a)(i) The 40th and 41st children lie in \(1.65<h\le1.8\). (ii) Use midpoints 1.3, 1.45, 1.575, 1.725, 1.85: \(\bar h\approx\frac{2(1.3)+13(1.45)+24(1.575)+32(1.725)+9(1.85)}{80}=1.63875\text{ m}\).

(b)(i) \(2/80=1/40\). (ii) There are 56 children with \(1.5<h\le1.8\) and 9 with \(h>1.8\). Either order is possible: \(2\left(\frac{56}{80}\frac9{79}\right)=\frac{63}{395}\).

(c)(i) Cumulative frequencies \(2,15,39,71,80\). (ii) Plot \((1.2,0),(1.4,2),(1.5,15),(1.65,39),(1.8,71),(1.9,80)\) and join by an increasing curve or polygon.

(d) Read at CF 20 and 60 for the quartiles, and at CF \(0.6(80)=48\) for the 60th percentile. With the polygon shown, \(Q_1=1.53125, Q_3=1.7484375\).

(i) \(Q_3-Q_1=0.2171875\text{ m}\approx0.22\text{ m}\). (ii) \(1.6921875\text{ m}\approx1.69\text{ m}\). Graph estimates may vary slightly with the drawn curve.

(a)(i) Plot the six upper-boundary points and join with a reasonable increasing curve. The displayed curve also starts at \((60,0)\).

(ii)(a) At CF 100, median \(\approx88\text{ km/h}\). (b) At CF 50 and 150, \(Q_1\approx80.8,\ Q_3\approx94.2\), so IQR \(\approx13.4\text{ km/h}\).

(c) This curve has CF about 191 at 110 km/h, giving \(200-191=9\) cars. The answer follows the reading from your own curve, rounded to the nearest integer.

(b)(i) Midpoints are \(2300,3050,4250,5350\). Mean \(\approx\frac{12(2300)+15(3050)+16(4250)+7(5350)}{50}=3576\text{ kg}\).

(ii) Heights are proportional to frequency density. First density \(12/600=0.02\) corresponds to 6 cm, so multiply each density by 300. Next heights \(300(15/900)=5\), \(300(16/1500)=3.2\) and \(300(7/700)=3\) cm.

(a) Midpoints \(25,75,150,350\). \(\bar m=\frac{43(25)+31(75)+25(150)+21(350)}{120}=120.833\ldots\approx121\text{ g}\).

(b) Densities \(43/50=0.86,31/50=0.62,25/100=0.25,21/300=0.07\). The scale is \(17.2/0.86=20\) cm per density unit. Heights \(12.4,5,1.4\) cm.

(c) Cumulative frequencies \(43,74,99,120\). (d) Plot \((0,0),(50,43),(100,74),(200,99),(500,120)\) and join smoothly.

(e) Read from the displayed curve: (i) CF 60 gives about 74 g. (ii) CF 90 gives about 151 g. (iii) CF \(0.4(120)=48\) gives about 57 g. (iv) CF at 400 minus CF at 250 is about \(117.2-104.7=12.5\), hence about 12 or 13 letters. Estimates depend on the drawn curve.

(a)(i) \(20<t\le35\). (ii) Midpoints 2.5, 7.5, 15, 27.5, 47.5 give \(\bar t=\frac{3(2.5)+7(7.5)+18(15)+28(27.5)+24(47.5)}{80}=28\text{ min}\).

(b)(i) \(70/80=7/8\). (ii) Among the 28 eligible people, 3 are in the first group and 25 in the second. Without replacement, \(2\frac3{28}\frac{25}{27}=\frac{25}{126}\).

(c)(i) \(3,10,28,56,80\). (ii) Plot upper-boundary points \((5,3),(10,10),(20,28),(35,56),(60,80)\) and join with an increasing curve or polygon.

(iii) 80th percentile is at cumulative frequency \(0.8(80)=64\). This polygon gives about 43 minutes. (iv) At 45 minutes the cumulative frequency is approximately 66. Percentage above 45 is \(100(80-66)/80=17.5\)%. Readings follow the diagram drawn.

(a) Plot the seven points in the table at their stated mass boundaries and join by an increasing curve or polygon.

(b) 90th percentile is the 45th value. Interpolate between \((32,43)\) and \((36,47)\) to obtain about 34 kg.

(a)(i) Read cumulative frequency 50: 9.4 minutes. (ii) Quartiles at 25 and 75 are about 8.0 and 10.4 minutes; IQR \(10.4-8=2.4\) minutes. (iii) About 82 took at most 11 minutes, so \(100-82=18\) took longer.

(b)(i) Midpoints 10, 25, 32.5, 40, 52.5: \(\bar d=\frac{4(10)+38(25)+40(32.5)+53(40)+15(52.5)}{150}=34.65\text{ m}\). (ii) Reference density \(40/5=8\), so scale \(12/8=1.5\). Bar heights \(0.3,5.7,12,7.95,1.5\) cm, in class order. (iii) Without replacement \(\frac{15}{150}\frac{14}{149}=\frac7{745}\).

(a)(i) 3, 22, 43, 48, 50. (ii) Plot \((0,0),(10,3),(25,22),(35,43),(40,48),(50,50)\) and join with an increasing curve. (iii) About 37 children; readings from 35 to 38 are acceptable. (b) Read quartiles at cumulative frequencies 25, 50, 75: 1.57 m, 1.64 m, 1.71 m. Draw whiskers to 1.45 m and 1.83 m and a box from 1.57 m to 1.71 m with median line at 1.64 m.

(a) Read CF = 20, 40, 60: lower quartile 50 kg, median 63 kg, upper quartile 72 kg. Minimum 30 kg, maximum 90 kg.

(b)(i) \(.30(80)=24\) on the CF axis gives approximately 56 kg. (ii) CF at 75 kg is about 64, so \(80-64=16\) boys. (c)(i) Frequencies are 8, 12, 14, 22, 14, 10. (ii) Midpoints 35, 45, 55, 65, 75, 85 give \(\sum fx=4920\). Mean \(4920/80=61.5\) kg. (iii) The eligible group contains 14 boys in (70, 80] and 10 in (80, 90]. Either order is possible: \(2\times\frac{10}{24}\times\frac{14}{23}=\frac{35}{69}\).

(a)(i) CF = 40 gives 90 m². (ii) CF = 20 gives 68 m². (iii) Upper quartile at CF = 60 is 120 m²; IQR \(120-68=52\) m². (iv) \(80-60=20\) houses. (b)(i) Midpoints 50, 70, 90, 115, 145, 180 give \(\sum fm=7800\). Mean \(7800/80=97.5\) m². (ii) Divide each frequency by its class width: densities 0.7, 0.85, 0.9, 0.5, 0.3, 0.175. (iii) There are 16 above 130 m² and 14 at most 60 m². Either order: \(2\times\frac{16}{80}\times\frac{14}{79}=\frac{28}{395}\).

(a)(i) \(1.5<h\le1.6\) has the highest frequency, 30. (ii) Use midpoints 1.35, 1.45, 1.55, 1.65, 1.75, 1.85: \(\bar h\approx\frac{7(1.35)+18(1.45)+30(1.55)+24(1.65)+27(1.75)+14(1.85)}{120}=1.62\text{ m}\).

(b)(i) \(\frac{14}{120}=\frac7{60}\). (ii) The taller boy may occur in any of three positions, so \(3\left(\frac{14}{120}\frac7{119}\frac6{118}\right)=\frac{21}{20060}\).

(c)(i) Cumulative frequencies: 7, 25, 55, 79, 106, 120. (ii) Plot (1.3,0), (1.4,7), (1.5,25), (1.6,55), (1.7,79), (1.8,106), (1.9,120) and join with an increasing curve.

(d)(i) Read across from cumulative frequency 60: approximately 1.62 m. (ii) \(.40(120)=48\); read across from 48: approximately 1.58 m. These are estimates from the plotted diagram.

(a) Midpoints10,30,42.5,47.5,55,70 give \(\bar v=\frac{16(10)+34(30)+62(42.5)+58(47.5)+26(55)+4(70)}{200}=41.4\text{ km/h}\).

(b)(i) Missing values112 and170. (ii) Plot (0,0), (20,16), (40,50), (45,112), (50,170), (60,196), (80,200), then join with an increasing curve.

(iii)(a) Read from cumulative frequency150: approximately48 km/h. (b) Cumulative frequency at35 is approximately40, so \(200-40=160\) cars exceed35 km/h. These are readings from the curve.

(c) There are26 + 4 = 30 cars above50 km/h. \(\frac{30}{200}\frac{29}{199}=\frac{87}{3980}\). (d) Frequency density = frequency ÷ class width. Heights \(50/40=1.25,\quad120/10=12,\quad30/30=1\). Draw adjoining bars over0–40,40–50,50–80.

(a)(i) Read height at cumulative frequency60: 25 cm. (ii) Read at30 and90: about21 and31 cm. IQR≈31−21=10 cm. (iii) \(.6(120)=72\); read height at CF72:27 cm. (iv) CF at40 cm≈114, so120−114=6 plants.

(b)(i) Midpoints5,15,25,40: \(\bar h\approx\frac{2(5)+18(15)+62(25)+38(40)}{120}=27.9166\ldots\approx27.9\text{ cm}\). (ii) \(7.2\,\frac{38/20}{18/10}=7.6\text{ cm}\).

(a)(i) Read at cumulative frequency 80: median≈5 hours. (ii) Read Q₁ at frequency 40 and Q₃ at frequency 120: approximately 3.5 and 6 hours, giving \(IQR\approx6-3.5=2.5\) hours. (b) At 2,4,6 hours the cumulative frequencies are 20,46,120. Differences give \(46-20=26\) and \(120-46=74\).

(a) Midpoint estimate: \((16(2.5)+38(7.5)+30(15)+9(25)+7(40))/100=12.80\), so \$12.80. (b)16,54,84,93,100. (c) Plot(0,0),(5,16),(10,54),(20,84),(30,93),(50,100) and join with a smooth increasing curve. (d) Read the median at CF50, quartiles at25 and75, and subtract CF at \$25 from100. A suitable graph gives median≈\$9.3, IQR≈\$9.3 and about10 students spending over \$25.

(a)(i) Read the time at cumulative frequency 100: about 34 minutes. (ii) Read Q₁ at 50 and Q₃ at 150: about 26 and 44 minutes. IQR=44−26=18 minutes. (iii) Cumulative frequency at 40 is about 140, so \(200-140=60\) students.

(b)(i) Range=50−1=49 cents. (ii) Mode=20 cents (frequency 4). (iii) There are 15 coins; the 8th is 10 cents. (iv) \(1(3)+2(1)+5(3)+10(2)+20(4)+50(2)=220\) cents. (v) \(220/15=14.7\) cents, to 3 significant figures.

(c) Frequency = class width × density: 30, 40, 30 for classes 0–10, 10–15, 15–30 kg. Midpoints 5, 12.5, 22.5 give \(\bar m=\frac{5(30)+12.5(40)+22.5(30)}{100}=13.25\) kg.

(a)(i) Read at CF=40: about 54 s. (ii) Quartiles at CF=20 and 60 are about 36 and 65 s; IQR=29 s. (iii) 20% of 80 is 16; read about 32 s. (iv) CF at 66 s is about 62; \(80-62=18\) girls.

(b)(i) Cumulative totals at 20,40,60,80,100 are 6,24,50,76,80. Subtract successive totals to get frequencies 6,18,26,26,4. (ii) Midpoints 10,30,50,70,90 give \(\bar t=\frac{6(10)+18(30)+26(50)+26(70)+4(90)}{80}=51\) s. (c)(i) 66 s is the upper quartile, so 75%. (ii) Boys’ IQR=66−46=20 s; girls’ IQR=29 s. Girls’ middle half has the larger spread.

(a)(i) CF=30 gives about 52 km. (ii) CF=15 gives about 36 km. (iii) CF=45 gives about 62 km, so IQR=62−36=26 km. (b) The male distances have a smaller IQR (26 versus 40 km), so their middle half is less spread out.

(c) CF at 50 is 27; \(P(d>50)=\frac{60-27}{60}=\frac{11}{20}\). (d)(i) Successive cumulative totals give missing frequencies 14,12,5. (ii) Class midpoints 20,45,55,65,80,105 give \(\bar d=\frac{18(20)+9(45)+14(55)+12(65)+5(80)+2(105)}{60}=48.75\) km.

(a)(i) \(\frac9{160}\). (ii) Midpoints 5,25,50,75,95,110 give \(\bar m=\frac{3(5)+39(25)+43(50)+55(75)+11(95)+9(110)}{160}=58.125\) minutes. (b) Cumulative totals are 3,42,85,140,151,160.

(c) Plot (0,0),(10,3),(40,42),(60,85),(90,140),(100,151),(120,160) and join smoothly. (d)(i) Read at CF=80: about 58 min. (ii) Quartiles at CF=40 and 120 are about 39 and 78 min; IQR about 39 min. (iii) CF=144 gives about 93 min. (iv) CF at 30 is about 26, so about 160−26=134 people.

(a) Read at cumulative frequencies 30 and 90: approximately 22 and 32 minutes. IQR≈32−22=10 minutes. (b) The cumulative frequency at 50 minutes is approximately 116, so 120−116=4 students.

(a)(i) \(20<t\le25\). (ii) \(25<t\le30\) because the 60th and 61st observations lie there. (iii) \(\frac{44(22.5)+32(27.5)+28(32.5)+12(37.5)+4(42.5)}{120}=28.333\ldots\approx28.3\) min. (iv) \(\frac4{120}=\frac1{30}\). (b)(i) 76,104,116,120. (ii) Plot (20,0),(25,44),(30,76),(35,104),(40,116),(45,120), joining with a smooth increasing curve. (iii) Read CF60: about 27.4 minutes. (iv) Read CF30 and90: about 23.3 and32.2; IQR≈9.0 minutes. (v) CF37≈110, so about 120−110=10 students.

(a) Read at CF 40: about 25 minutes. (b) At CF 20 and 60, read about 19 and 31 minutes; IQR≈12 minutes. (c) CF 40 minutes≈75, so 80−75=5 students.

(a) Use midpoints 65,80,95,105,112.5,120: weighted total 18030; mean 18030/180=100.2 cm. (b) Density scales by.4/(8/10)=.5 cm per density unit: heights.8,2.8,.65 cm. (c) CFs 8,34,69,136,164,180. (d) Plot cumulative upper boundaries and join smoothly. (e)(i) Quartiles read at 45 and 135 give about 94.1 and 109.8 cm: IQR≈15.8 cm. (ii) CF 126 gives≈108.5 cm. (iii) CF 106≈108, so 180−108≈72 children.

(a) Midpoints 47.5,55,65,80,95,110 give \(\frac{23(47.5)+64(55)+122(65)+136(80)+26(95)+29(110)}{400}=72.70625\approx72.7\) minutes. (b)(i)23,87,209,345,371,400. (ii) Plot upper class boundaries against these cumulative frequencies. (iii) Read at cumulative frequencies 200,100/300,and 240. A polygon gives median≈69.3, Q 1≈61.1, Q 3≈83.4, IQR≈22.3 and 60th percentile≈74.6 minutes. Readings from a smooth curve may vary slightly.

(a) \(71<t\le72\). (b) Use class midpoints: \(\frac{17(70.5)+24(71.5)+21(72.5)+18(73.5)+10(74.5)}{90}=72.2778\ldots\approx72.3\) s.

(c)(i) 17,41,62,80,90. (ii) Plot (70,0),(71,17),(72,41),(73,62),(74,80),(75,90) and join with a smooth increasing curve. (iii) Read at cumulative frequency 45: about 72.2 s. (iv) Read at 22.5 and 67.5: approximately 71.2 and 73.3, giving IQR≈2.1 s.

(d) Upper distance=3725 m and lower time=74.5 s. \(\frac{3725}{74.5}\times3.6=180\) km/h.

(a) Read about 118 at h=160, so \(200-118=82\) sunflowers. (b) Plot the table points and join with a smooth increasing curve.

(c) Read at cumulative frequency 100. Simon≈156 cm; Sue=130 cm. Difference≈26 cm.

(a)(i) Read the time at cumulative frequency \(60/2=30\): approximately 64 minutes. (ii) Read \(Q_1\) at cumulative frequency 15 and \(Q_3\) at 45. These are approximately 55 and 71–71.5 minutes, giving an IQR of about 16–16.5 minutes. (iii) \(0.40(60)=24\): read approximately 62 minutes. (iv) The cumulative frequency at 80 minutes is 54, so \(60-54=6\) students took longer. (b) Cumulative frequencies at 0, 40, 60, 70, 80, 90 and 100 are 0, 8, 20, 43, 54, 58 and 60. Subtract consecutive values to get 8, 12, 23, 11, 4 and 2. (c) Use \(\text{frequency density}=\frac{\text{frequency}}{\text{class width}}\). Heights are 0.2, 0.6, 2.3, 1.1, 0.4 and 0.2.

(a) The quartiles are read at cumulative frequencies 30 and 90: about 1.0 m and 2.5 m. The interquartile range is \(2.5-1.0=1.5\) m. (b) Read at \(0.95(120)=114\), giving 3.5 m. (c) At 3 m the cumulative frequency is 102, so \(120-102=18\) trees have greater diameters.

(a)(i) Read time at cumulative frequency 100: about 15 min. (ii) At cumulative frequency 50: about 11 min. (iii) The upper quartile at cumulative frequency 150 is about 20 min. Hence IQR \(\approx20-11=9\) min. (iv) The 15th percentile is at cumulative frequency \(0.15(200)=30\), giving about 10 min. (v) The cumulative frequency at 30 min is 176, so \(200-176=24\) students. (b)(i) Midpoints are 5, 12.5, 17.5, 25 and 45. \(\bar t\approx\frac{48(5)+48(12.5)+60(17.5)+26(25)+18(45)}{200}=16.75\) min. (ii) Frequency densities are \(48/10=4.8,\ 48/5=9.6,\ 60/5=12,\ 26/10=2.6,\ 18/30=0.6\).

(a) \(10/(40/60)=15\) km/h. (b) Use midpoints 35, 42.5, 47.5, 52.5, 57.5 and 70: \(\bar t\approx\frac{8(35)+22(42.5)+95(47.5)+55(52.5)+14(57.5)+6(70)}{200}=49.2\) min. (c) Frequency densities are 0.8, 4.4, 19, 11, 2.8 and 0.3. (d)(i) Missing cumulative frequencies: 125 and 180. (ii) Plot (30, 0), (40, 8), (45, 30), (50, 125), (55, 180), (60, 194) and (80, 200).

(iii)(a) Read at cumulative frequency 100: about 48.7 min. (b) Read at cumulative frequency 180: 55 min. (c) At 58 minutes the cumulative frequency is about 188, giving about \(200-188=12\) people.

(a)(i) Read at cumulative frequencies 25 and 75: lower quartile \$6000, upper quartile \$10200. Interquartile range \(10200-6000=4200\), so \$4200. (ii)(a) True: median \$7600 on A is lower than \$10800 on B. (b) False: A’s upper quartile is \$10200, below \$13600, so fewer than 25% exceed \$13600 on A; approximately 25% exceed B’s upper quartile \$13600. (b) Use class midpoints 3000, 7000, 9000, 11000, 13000, 18000. \(\bar P\approx\frac{9(3000)+29(7000)+20(9000)+14(11000)+21(13000)+27(18000)}{120}=11025\), so \$11025. (c) Deposit \(0.25(8760)=2190\). Each payment \((9948-2190)/24=323.25\), so \$323.25.

(a) Read distance at cumulative frequency \(0.8(60)=48\): approximately 44 km. (b) Read at frequencies 15 and 45: approximately 16 km and 40 km. Thus \(\operatorname{IQR}\approx40-16=24\) km. (c) At 60 km the cumulative frequency is about 55, so \(60-55=5\) people travel further.

(a) At frequency 100, median approximately 6 s. (b) Read at frequencies 50 and 150: approximately 5 s and 7 s. IQR approximately 2 s. (c) About 20 students have times at or below 4 s, so \(200-20=180\) have times above 4 s.

(a)(i) \(24<t\le30\) minutes. (ii) Use midpoints 5, 17, 27, 35, 50, 65: \(\bar t\approx\frac{10(5)+90(17)+135(27)+85(35)+70(50)+10(65)}{400}=30.875\) min. (b)(i) Cumulative frequencies: 10, 100, 235, 320, 390, 400. (ii) Plot upper boundaries against these totals, including (0, 0), and join by a curve or polygon. (c) Reading the polygon shown gives approximately: (i) 28.4 min at frequency 200; (ii) \(37.6-24=13.6\) min; (iii) 18 min at frequency 60; (iv) \(400-355=45\) people.

(a) Median is at cumulative frequency \(80/2=40\). Read 6.2 min. (b) 30th percentile is at \(0.3(80)=24\). Read 5.8 min. (c) At 5 min, cumulative frequency is approximately 10, so \(80-10=70\) students take longer.

(a) Use midpoints 0.5, 1.5, 2.5, 3.5, 4.5, 5.5: \(\bar m=\frac{72(0.5)+21(1.5)+9(2.5)+11(3.5)+5(4.5)+2(5.5)}{120}=1.35\) min. (b)(i) Cumulative frequencies: 72, 93, 102, 113, 118, 120. (ii) Plot (0, 0) and upper endpoints with these totals; a polygon is shown. (iii) From it: median \(60/72\approx0.83\) min; lower quartile \(30/72\approx0.42\) and upper quartile \(1+(90-72)/21\approx1.86\), so IQR \(1.86-0.42\approx1.44\) min; 35th percentile \(42/72\approx0.58\) min. (c)(i) 30, 18. (ii) New densities are 72, 30/2 = 15, 18/3 = 6. Heights \(3.6(15/72)=0.75\) cm and \(3.6(6/72)=0.3\) cm.

(a) Plot (0, 0), (20, 5), (25, 13), (35, 72), (40, 90), (60, 117), (70, 120). (b)(i) Read at cumulative frequency 60: approximately 33 mm. (ii) Read cumulative frequency at r = 50, approximately 103 or 104; subtract from 120 to give approximately 17 days. (c) Missing frequencies: 8, 18, 27, found by differences of cumulative frequencies. (d) Using class midpoints, \(\bar r=\frac{5(10)+8(22.5)+59(30)+18(37.5)+27(50)+3(65)}{120}=35.1666\ldots\approx35.2\) mm. (e) Density = frequency ÷ class width, giving \(8/5=1.6,\quad27/20=1.35,\quad3/10=0.3\).

(a) Read at cumulative frequency 250: approximately 34. (b) Read quartiles at 125 and 375, approximately 24 and 40; IQR approximately 16. (c) 40% of 500 is 200; read score approximately 30. (d) Subtract cumulative frequencies at 30 and 20: approximately \(200-80=120\) people.

(a)(i) Read at CF 40: approximately 38 min. (ii) Read at CF 20 and 60, approximately 23.8 and 43.8 min: IQR approximately 20 min. (iii) Read at CF 56: approximately 43 min. (b)(i) Midpoints 10, 25, 37.5, 47.5 give \([12(10)+21(25)+33(37.5)+14(47.5)]/80=31.84375\approx31.8\) min. (ii) Densities \(12/20=0.6,\ 21/10=2.1,\ 33/15=2.2,\ 14/5=2.8\). Draw bars over the given intervals.

(a) Read CF ≈ 16 at 48 minutes. \(72-16=56\) students. (b)(i) Read time at CF = 36: approximately 63 minutes. (ii) Read times at CF = 18 and 54: approximately 50 and 72 minutes. IQR ≈ 22 minutes.

(a) Read masses at CF = 50 and 150: approximately 502.6 and 505.8 g. IQR ≈ 3.2 g. (b) \(16/200=2/25=0.08\). (c) Divide frequency by class width: 18.5, 26, 3.

(a) Midpoints 500, 1250, 1750, 2250, 3000: \(\bar T=[10(500)+30(1250)+55(1750)+72(2250)+33(3000)]/200=1998.75\) hours, approximately 2000. (b)(i) 10, 40, 95, 167, 200. (ii) Plot (0, 0), (1000, 10), (1500, 40), (2000, 95), (2500, 167), (3500, 200). (iii) A ruled polygon gives CF at 2200 ≈ 123.8, hence about 76 bulbs longer. (c) For independent bulbs, \((9/10)(2/5)+(1/10)(3/5)=21/50\).

(a)(i) Read CF = 100: 3.2 g. (ii) CF = 150: 4.2 g. (iii) CF = 160: 4.6 g. (iv) At m = 7, CF is about 196. (b)(i) Read cumulative values 40, 140, 186, 198, 200. Consecutive differences give frequencies 40, 100, 46, 12, 2. (ii) \(P(m\le M)=0.7\), so read CF = 140: M ≈ 4 g.

(a)(i) At CF = 40, median ≈ 14 min. (ii) At CF = 20, lower quartile ≈ 8 min. (iii) Upper quartile at CF = 60 is ≈ 30 min, so IQR ≈ 22 min. (b) At 35 min, CF ≈ 69. \(P(m>35)=(80-69)/80=11/80\). (c) Cumulative counts are 30, 42, 60, 76, 80, so the missing frequencies are 16, 4. (d) \(\sum fx=30(5)+12(12.5)+18(22.5)+16(35)+4(45)=1445\). Mean \(1445/80=18.0625\) min. (e) Bar densities are 3, 2.4, 1.2, 1.6, 0.4.

(a) Read CF = 24: median ≈ 4.1 h. (b) \(0.20(48)=9.6\), giving about 2.7 h. (c) Read CF = 12 and 36; the quartiles are about 2.9 h and 5.1 h. IQR ≈ 2.2 h. (d) At 2 h the cumulative count is about 5, so \(P(t\le2)=5/48\).

(a) Read CF = 25: 19 cm. (b) At 26 cm, CF ≈ 47, so \(50-47=3\) beans. (c) At CF = 12.5 and 37.5, read approximately 16.66 cm and 21.92 cm. \(\mathrm{IQR}\approx21.92-16.66=5.26\approx5.3\) cm. (d) At 14 cm, CF ≈ 5. \(P(l>14)=(50-5)/50=45/50=9/10\).

(a) At CF = 30, read 29.0 cm. (b) At CF = 15, read 20.0 cm. (c) At CF = 45, upper quartile ≈ 34.0 cm. \(\mathrm{IQR}\approx34.0-20.0=14.0\) cm. (d) CF at 15 cm is about 8, leaving 52 plants: \(52/60=13/15\).

(a) \((54+30)/300=84/300=7/25\). (b) Read at cumulative frequencies 150, 75 and 225: median approximately 62 g, lower quartile 52 g, upper quartile about 71–72 g; IQR approximately 19–20 g. At 65 g the cumulative frequency is about 175, giving \(300-175=125\) eggs above 65 g.

(a) Read at cumulative frequencies 100, 50 and 150: median 72 min, lower quartile 68 min, upper quartile 76 min, so IQR = 8 min. At 60 minutes the cumulative frequency is 36: \(200-36=164\) took longer. (b)(i) Missing frequency = \(200-(9+16+28+108+28)=11\). (ii) Use midpoints 35, 45, 55, 65, 75, 85: \(\bar t=[9(35)+11(45)+16(55)+28(65)+108(75)+28(85)]/200=13990/200=69.95\) min.

(a)(i) The 75th and 76th values lie in \(45<t\le55\). (ii) Midpoints 10, 27.5, 40, 50, 62.5, 75 give \(\bar t=[6(10)+15(27.5)+19(40)+37(50)+53(62.5)+20(75)]/150=7895/150\approx52.6\) min. (b)(i) Missing cumulative frequencies 40, 77, 130, 150. (ii) Plot (20,6), (35,21), (45,40), (55,77), (70,130), (80,150), and join with a smooth increasing curve or ruled lines.

(c) Using the ruled-line graph shown: (i) at cumulative frequency 75, median ≈ 54.5 min. (ii) Read quartiles at 37.5 and 112.5, approximately 43.7 and 65.0 min; IQR ≈ 21.4 min. (iii) Read cumulative frequencies approximately 59 at 50 and 95 at 60: 36 students. (iv) Approximately \(150-59=91\) take longer than 50 minutes, giving \(91/150\). (v) \((91/150)(90/149)=273/745\). Other accurate graph readings may give slightly different estimates.

(a) Read at cumulative frequencies 100, 50 and 150. (i) Median ≈ 5.8 kg. (ii) Lower quartile ≈ 4.6 kg. (iii) Upper quartile ≈ 7.0 kg, so IQR ≈ 2.4 kg. (iv) The graph gives about 28 at 3.5 kg; \(200-28=172\) parcels. (b)(i) Read cumulative frequencies approximately 36 at 4 kg, 110 at 6 kg and 150 at 7 kg. Missing frequencies: 74 and 40. (ii) Midpoints are 2, 5, 6.5, 8.5: \(\bar m=[36(2)+74(5)+40(6.5)+50(8.5)]/200=1127/200=5.635\approx5.64\) kg. (iii) Density = frequency ÷ class width, so missing densities 37 and 40. Draw the bars over 4–6 and 6–7 to these heights. Small differences from accurate graph readings are acceptable.

(a)(i) \(1.6<h\le1.7\). (ii) Multiply class midpoints by frequencies: \(\bar h=[4(1.35)+13(1.45)+33(1.55)+45(1.65)+19(1.75)+6(1.85)]/120=194/120\approx1.62\) m. (b)(i) \(6/120=1/20\). (ii) \((114/120)(113/119)=2147/2380\). (c)(i) 95 and 120. (ii) Plot all seven cumulative points and join with a curve or ruled lines; a ruled-line graph is shown below. (d)(i) At frequency 60, median ≈ 1.62 m. (ii) The 30th percentile is at frequency 36, giving ≈ 1.56 m.

(a) Use midpoints 10,30,45,55,65,75,85,95. \(\bar t=12690/200=63.45\) s. (b)(i) Missing frequencies 75,117,195,200. (ii) Plot the upper class limits against cumulative frequencies and join by a smooth curve or ruled lines. (c) On the ruled-line graph shown: (i) median ≈ 66.0 s; (ii) lower quartile ≈ 53.2 s; (iii) upper quartile ≈ 76.6 s, hence IQR ≈ 23.4 s; (iv) cumulative frequencies 96 and 142 give 46 people; (v) frequency 28 at 45 s gives \(200-28=172\) people. Accurate curve readings may differ slightly.

(a) \(3<t\le4\). (b) Midpoints 1,2.5,3.5,6 give \([24(1)+68(2.5)+72(3.5)+36(6)]/200=662/200=3.31\) min. (c)(i) 92 and 164. (ii) Plot (2,24), (3,92), (4,164), (8,200) and join with a curve or ruled lines. (iii) On the ruled-line graph: median ≈ 3.11 min, lower quartile ≈ 2.4 min, upper quartile ≈ 3.81 min, hence IQR ≈ 1.42 min.

(a) Midpoints 2.5,7.5,…,37.5: \(\sum fm=4380,\ \bar t=4380/200=21.9\) hours. (b) 155,180. (c) Plot upper boundaries against cumulative totals and join by an increasing curve or polygon. (d) Read across from \(200/4=50\) to get Q₁ = 15 h. (e) Linear interpolation on the polygon gives (i) 22.5 h; (ii) \(Q_3=25+(150-120)(5/35)\approx29.29\), so IQR \(\approx14.3\) h; (iii) \(0.64(200)=128,\ t=25+(128-120)(5/35)\approx26.1\) h; (iv) \(CF(17)=50+(2/5)(30)=62\), so 138 students.

(a) Read at cumulative counts 40,20,60. (i) Approximately 63.2 kg; (ii) 50.2 kg; upper quartile about 72.0 kg, so (iii) 21.8 kg. (b) CF at 60 kg is about 34; \(80-34=46\) boys. (c)(i) 12,14, obtained by subtracting successive cumulative totals. (ii) Use midpoints 35,45,55,65,75,85: \(\bar m=[35(8)+45(12)+55(14)+65(22)+75(14)+85(10)]/80=4920/80=61.5\) kg.

(a) Plot all upper boundaries against cumulative totals, joining by a smooth increasing curve or polygon. (b) Polygon interpolation: (i) \(160+(100-95)(5/20)=161.25\) cm; (ii) \(170+(150-145)(10/35)\approx171.43\) cm; (iii) lower quartile 150 cm, so IQR 21.4 cm. (c)(i) \((200-145)/200=55/200=11/40\). (ii) Two orders: \((55/200)(10/199)+(10/200)(55/199)=1100/39800=11/398\). (d)(i) 30,35,20. (ii) Density = frequency/width: all heights 1,4,4.5,4,6,3.5,2. Add the three missing bars at 160–165,165–170,170–180.