The class widths are \(20,10,15,5\), giving frequency densities \(40/20=2,\ 35/10=3.5,\ 45/15=3,\ 25/5=5\). A density of 2 has height 4 cm, so bar height is twice frequency density. The missing heights are \(\boxed{7\text{ cm},\ 6\text{ cm},\ 10\text{ cm}}\).

(a) Use midpoints \(2.5,7.5,15,30\). Then \(\bar t=\frac{3(2.5)+18(7.5)+13(15)+6(30)}{40}=\frac{517.5}{40}=\boxed{12.9375\text{ minutes}}\).

(b) Frequency density is frequency divided by class width. The four densities are \(3/5=0.6\), \(18/5=3.6\), \(13/10=1.3\) and \(6/20=0.3\). Draw adjoining bars over the four intervals at these heights.

(a) Use midpoints \(1.35,1.55,1.65,1.8\). \(\bar h=\frac{12(1.35)+20(1.55)+34(1.65)+14(1.8)}{80}=\boxed{1.60625\text{ m}}\).

(b) Divide frequency by class width. The densities are \(12/0.3=40,20/0.1=200,34/0.1=340,14/0.2=70\). Draw adjoining bars at these heights.

(a) Frequency is class width × frequency density. For 0–5: \(5(2.8)=\boxed{14}\). For 5–20: \(5(3.4)+10(2)=\boxed{37}\). For 20–50: \(10(1.6)+20(0.4)=\boxed{24}\).

(b) The 35–50 portion has width 15 and density 0.4, giving \(15(0.4)=\boxed{6}\) people.

(a) Frequency \(=\text{width}\times\text{density}=2(20)=\boxed{40}\). (b) Heavy density \(=102/(5-2)=\boxed{34}\). Draw the column from 2 to 5 at height 34.

(a) Frequency \(=(1.6-1.35)(40)=10\). (b) C has \((1.35-1.25)(170)=17\) coconuts, so there are 27 available. Without replacement, \(P(DD)=\frac{10}{27}\frac9{26}=\boxed{\frac5{39}}\).

(c) Use the class midpoints and frequencies:

Estimated mean \(=\frac{6(1)+15(1.175)+17(1.3)+10(1.475)}{48}=1.259895\ldots\), giving \(\boxed{1.26\text{ kg}}\).

(a) Midpoints are 190,205,212.5,222.5. Estimated mean \(=\frac{32(190)+64(205)+74(212.5)+30(222.5)}{200}=\boxed{208\text{ g}}\).

(b) Frequency densities are \(32/20=1.6,\ 64/10=6.4,\ 74/5=14.8,\ 30/15=2\). Since density 14.8 is drawn at 7.4 cm, divide each density by 2. Heights are \(\boxed{0.8,\ 3.2,\ 7.4,\ 1\text{ cm}}\).

(a)(i) \(20-15=\boxed5\). (ii) Weighted total=168, so mean \(=168/10=\boxed{16.8}\). (iii) The 5th and 6th marks are 16 and 17, giving \(\boxed{16.5}\). (iv) \(\boxed{15}\). (b) Eighth mark \(=8(17.5)-7(17)=\boxed{21}\).

(c) Frequency density = frequency ÷ class width, giving \(\boxed{0.4,0.9,3.5,2,0.5}\). Draw adjacent bars over the stated intervals at these heights.

(a)(i) Midpoints 5,17.5,32.5,50. Mean=\((3(5)+18(17.5)+15(32.5)+4(50))/40=\boxed{25.4375\text{ min}}\). (ii) Bar densities=\(\boxed{0.3,1.2,1,0.2}\). Use the stated class widths. (iii) More than 25:19 students; at most 10:3. Both orders give \(2\times\frac{19}{40}\times\frac3{39}=\boxed{19/260}\). (b)(i) Minimum=30−25=\(\boxed5\) min. (ii) Five-number summary=\(\boxed{5,8,22,27,30}\), with Q₃=8+19=27. (iii) The median is 22 minutes, so at most 100 of 200 values can be strictly greater than 22;101 would put both middle values above 22.

(a) Midpoints are 5,15,30,45,65. \(\bar t=\frac{2(5)+46(15)+33(30)+26(45)+13(65)}{120}=\frac{3705}{120}=\boxed{30.875\text{ min}}\). (b) Existing bars represent \(10(1.3)+20(1.5)+30(0.4)=55\) patients. Missing frequency=90−55=35; width=20, so height \(=35/20=\boxed{1.75}\).

(a)(i)(a) Read the distance at CF=40: \(\boxed{1480\text{ km}}\). (b) Q₁ at CF=20 is about 1160 km and Q₃ at CF=60 is about 1600 km. IQR=\(1600-1160=\boxed{440\text{ km}}\). (ii) CF at 1800 is about 72, leaving 8 motorists. Probability=\(8/80=\boxed{1/10}\). (b) Time=78/3600 h. \(v=5.104\times3600/78=\boxed{236\text{ km/h}}\).

(c)(i) Frequency=width×density=\((160-120)(0.2)=8\). (ii) Frequencies are 8,22,36,64,30 with midpoints 140,170,190,220,270. \(\bar v=\frac{8(140)+22(170)+36(190)+64(220)+30(270)}{160}=\boxed{211.75\text{ km/h}}\).

(a) \(35/100\times360=\boxed{126^\circ}\). (b) Midpoints 35,42.5,47.5,60. \(\bar v=\frac{15(35)+20(42.5)+35(47.5)+30(60)}{100}=\boxed{48.375\text{ km/h}}\). (c) Frequency densities arefrequency/class width: \(15/10=1.5,20/5=4,35/5=7,30/20=1.5\). Draw bars over their respective intervals with heights \(\boxed{1.5,4,7,1.5}\).

Densities \(18/6=3\) and \(16/4=4\). Height is proportional to density: \(7.5(4/3)=10\) cm.

(a) Midpoints \(217.5,221.5,229,239,254\). Mean \(\approx\frac{9(217.5)+14(221.5)+14(229)+2(239)+3(254)}{42}=226.297619\ldots\approx226\text{ s}\).

(b) Densities \(9/3=3,14/5=2.8,14/10=1.4,2/10=0.2,3/20=0.15\). Add bars on \(219\text{–}224,224\text{–}234,234\text{–}244\) at heights \(2.8,1.4,0.2\).

(a) Frequency = density × width. The first frequencies are \(2.8(10)=28\) and \(3(15)=45\). Remaining densities \(57/15=3.8,38/20=1.9,12/20=0.6\).

(b) Midpoints \(5,17.5,32.5,50,70\). \(\bar t=\frac{28(5)+45(17.5)+57(32.5)+38(50)+12(70)}{180}=30.7\text{ min}\).

(c) The table only gives intervals; the exact maximum is unknown and can be below 80 minutes.

(d) Eligible workers \(57+38+12=107\). There are 95 taking at most 60 minutes and 12 taking longer. Without replacement, \(P=3\frac{95}{107}\frac{12}{106}\frac{11}{105}=\frac{1254}{39697}\). The factor 3 accounts for all possible orders.

(a) Use midpoints 5, 15, 22.5, 32.5, 45: \(\bar m=\frac{12(5)+38(15)+32(22.5)+50(32.5)+18(45)}{150}=25.2\text{ kg}\). (b) Frequency density = frequency ÷ class width. Heights: \(1.2,3.8,6.4,10/3,1.8\).

(c)(i) \(12,50,82,132,150\). (ii) \((150-12)/150\times100=92\)%.

(a)(i) Midpoints 90, 125, 175, 250, 350 give \(\bar d=\frac{7(90)+33(125)+76(175)+52(250)+32(350)}{200}=211.275\text{ km}\). (ii) Only the final midpoint changes, from 350 to 330. Ann’s estimate is lower by \(32(350-330)/200=3.2\) km.

(iii) Reference density \(52/100=.52\) gives scale \(2.6/.52=5\). Heights: \(5(7/20)=1.75\), \(5(76/50)=7.6\), \(5(32/100)=1.6\) cm. (b) \(32/200=4/25\). (c)(i) \(\frac{40}{200}\frac{39}{199}=\frac{39}{995}\). (ii) There are 84 cars above 200 km and 7 at most 100 km. Both orders give \(2\frac{84}{200}\frac7{199}=\frac{147}{4975}\).

(a) Frequency density = frequency ÷ class width. The last three densities are \(28/15=1.8667\ldots,33/20=1.65,13/10=1.3\).

(b) Midpoints 12.5, 20, 32.5, 50, 65 give \(\bar h=\frac{8(12.5)+18(20)+28(32.5)+33(50)+13(65)}{100}=38.65\text{ cm}\).

(a)(i) Frequency density = frequency ÷ class width. Heights are 8, 7, 7.5, \(10/3\), 0.8. (ii) Midpoints 0.75, 1.5, 3, 5.5, 9.5 give \(\sum fm=4(.75)+7(1.5)+15(3)+10(5.5)+4(9.5)=151.5\). Mean \(151.5/40=3.7875\) kg. (iii) \((4+7)/40=11/40\). (iv) There are 29 eligible parcels, of which 4 exceed 7 kg and 15 are at most 4 kg. Both orders give \(2\frac4{29}\frac{15}{28}=\frac{30}{203}\). (b)(i) 4.6 kg. (ii) \(6.2-3.0=3.2\) kg. (iii) Median stays the same: one removed parcel is below the median and the other above it, so the central observation(s) remain the same.

(a) Range \(10-6=4\), mode 7, median 8. (b) Read at cumulative frequencies 50, 100, 150: lower quartile 11 min, median 14 min, upper quartile 15 min. IQR = 4 min. (c) \(\sum fs=5(3)+6(10)+7(43)+8(75)+9(48)+10(21)=1618\). Mean \(1618/200=8.09\). (d) Classes 50–80, 80–100, 100–110, 110–120, 120–160 have densities 1, 3.5, 4, 3.6, 0.6. Frequencies are 30, 70, 40, 36, 24. Midpoints 65, 90, 105, 115, 140 give \(\bar h=(30(65)+70(90)+40(105)+36(115)+24(140))/200=99.75\) cm.

(a) Use midpoints5,12.5,17.5,30,57.5: \(\bar t\approx\frac{9(5)+18(12.5)+22(17.5)+30(30)+21(57.5)}{100}=27.625\text{ s}\).

(b) Frequency densities \(9/10=.9,\quad18/5=3.6,\quad22/5=4.4,\quad30/20=1.5,\quad21/35=.6\). Bar height is4 times the density, so the missing heights are6 cm and2.4 cm.

Height is proportional to frequency density. First density=35/10=3.5, so scale=2 cm per density unit. Other heights: \(2(32/5)=12.8\) cm, \(2(44/20)=4.4\) cm, \(2(12/30)=.8\) cm.

(a) Midpoint estimate \((38(25)+27(32.5)+21(37.5)+16(50)+18(80))/120=40.4583\ldots\approx40.5\) s. (b) Divide each frequency by its class width: densities3.8,5.4,4.2,.8,.45.

(a) \(10<t\le15\) minutes (largest frequency 19). (b) Frequency density = frequency ÷ class width. Heights: \(7/10=.7,\quad19/5=3.8,\quad16/5=3.2,\quad8/20=.4\).

Frequency=density×class width. For 5<t≤15: 1.5×10=15. For 30<t≤50: 1.1×20=22. Total=6+15+27+22+10=80.

(a)(i) Midpoints 20,35,42.5,47.5,55,65; weighted sum 5135; mean 5135/120=42.79…≈42.8 min. (ii) Frequency densities.8,1.8,4.4,8,2.1,.3. (b) People generally spend longer in the supermarket on Saturday; its modal interval is 45–50 min, while Wednesday’s is 40–45 min.

(a)(i) Midpoints 150,225,275,400; weighted total 13800, so mean 13800/52=265.38…≈265 tonnes. (ii) Bar densities 8/100=.08,20/50=.4,12/50=.24,12/200=.06. (b)(i)100 days. (ii)At CF 50,56 tonnes. (iii)At CF 75,62 tonnes. (iv)Lower quartile 38 tonnes; IQR=62−38=24 tonnes. (v)CF 20 tonnes≈12, so 100−12=88 days.

(a)(i) \(\frac{20(50)+55(125)+66(175)+35(225)+24(325)}{200}=175.5\) ml. (ii) Frequency density=frequency/class width, so heights are 0.20,1.10,1.32,0.70,0.16.

(b)(i) Plot (0,0),(50,28),(100,64),(150,104),(200,168),(250,200), joining an increasing curve. (ii)(a) 65% of 200=130. Reading the curve gives about 170 g. (b) Approximately 46 estimates are at most 75 g, so 200−46=154 are above 75 g.

(a)(i) Frequency=density×width, giving 50,80,33,17,20. (ii) \(\frac{50(5)+80(15)+33(22.5)+17(27.5)+20(40)}{200}=17.3\) g.

(b)(i) \(\frac6{15}\frac5{14}=30/210=1/7\). (ii) Both orders: \(\frac9{15}\frac6{14}+\frac6{15}\frac9{14}=108/210=18/35\). (c) \(4000(3/80)=150\).

(a) Compare bar areas with the key area. Frequencies are 54,76,96,92,42, totalling 360. (b) Use midpoints 155,175,185,200,225: \(\frac{54(155)+76(175)+96(185)+92(200)+42(225)}{360}=186.8889\ldots\approx187\) g.

(a)(i) Read at 50: 280 m. (ii) At 75: 320 m. (iii) Lower quartile=230 m, so IQR=320−230=90 m. (iv) At 350 m the cumulative frequency is 90: 100−90=10 students.

(b)(i) \(\frac{20(150)+22(225)+30(265)+16(300)+12(360)}{100}=250.2\) m. (ii) Densities are 0.20,0.44,1.00,0.40,0.15.

(c) 280−258=22 m further.

(a) Cumulative frequencies 10,36,70,128,150 place the middle observations in \(80<t\le100\). (b) \(\frac{10(62.5)+26(67.5)+34(75)+58(90)+22(125)}{150}=86\) min.

(c)(i) The smallest and largest individual times are not known from grouped data. (ii) \(26/150\times360=62.4^\circ\). (d) 22/150=11/75. (e)(i) \((10/150)(9/149)=90/22350\). (ii) Both orders: \((10/150)(22/149)+(22/150)(10/149)=440/22350\).

(f) Densities=\(10/5,26/5,34/10,58/20,22/50\)=2,5.2,3.4,2.9,0.44. Height multiplier=5/2=2.5. Heights: 5,13,8.5,7.25,1.1 cm.

(a) Frequency = class width × frequency density. The five frequencies are \(2.5(12)=30\), \(2.5(26)=65\), \(5(15)=75\), \(5(10)=50\) and \(10(2)=20\). Their total is \(30+65+75+50+20=240\). (b) More than 10 minutes late: \(50+20=70\). The percentage is \(\frac{70}{240}\times100=29.1666\ldots\%\approx29.2\%\).

Frequency density is frequency ÷ class width. Densities are 2, 7, 12 and 2. A density of 12 gives height 6 cm, so each height is half the density.

(a) Cumulative frequencies are 3, 17, 21, 27, 40. The 20th and 21st values lie in \(140<h\le144\). (b) Midpoints 125, 135, 142, 147, 160 give \(\bar h\approx\frac{3(125)+14(135)+4(142)+6(147)+13(160)}{40}=144.875\) cm. (c) Frequency densities, in order: \(3/10=0.3,\ 14/10=1.4,\ 4/4=1,\ 6/6=1,\ 13/20=0.65\).

(a)(i) Use midpoints 100, 250, 325, 375, 450: \(\bar M=\frac{5(100)+9(250)+18(325)+12(375)+6(450)}{50}=316\) g. (ii) Frequency density = frequency ÷ class width. Bar heights are 0.025, 0.09, 0.36, 0.24, 0.06. (b) The students’ estimates have a greater range: their occupied classes extend from 0 to 500 g, while the adults’ occupied classes extend from 200 to 500 g. Adults estimate a greater mass on average: the adult class frequencies are 7, 16, 18, 9 for 200–300, 300–350, 350–400, 400–500 g, giving estimated mean \((7(250)+16(325)+18(375)+9(450))/50=355\) g, compared with 316 g.

(a) Midpoints are 50, 65, 72.5, 82.5: \(\bar t=\frac{10(50)+50(65)+80(72.5)+60(82.5)}{200}=72.5\) min. (b) Frequency densities \(10/20=0.5,\ 50/10=5,\ 80/5=16,\ 60/15=4\). Draw bars across their corresponding intervals with these heights and no gaps.

(a) Midpoints 55, 90, 110, 160 give \(\bar m=[50(55)+30(90)+40(110)+40(160)]/160=101.5625\approx102\) g. (b) Densities are \(50/50=1,\ 30/20=1.5,\ 40/20=2,\ 40/80=0.5\). Draw touching bars with these heights and the given class widths. (c) \(40/160=1/4\). (d)(i) \((40/160)(39/159)=13/212\). (ii) Include both orders: \((40/160)(50/159)+(50/160)(40/159)=25/159\).

(a)(i) Frequency = bar width × density: \((100-70)(0.4)=12\). (ii) Classes 0–30, 30–50, 50–60, 60–70, 70–100 have frequencies 3, 6, 10, 8, 12. Using midpoints 15, 40, 55, 65, 85: \(\bar m=[3(15)+6(40)+10(55)+8(65)+12(85)]/39=60.8974\ldots\approx60.9\) g. (b) \(20(70)-19(70.5)=60.5\) g.

(a) Classes 0–10, 10–25, 25–30, 30–35, 35–50, 50–60 have frequencies 13, 33, 19, 4, 15, 6. Multiply each class width by its frequency density. (b) Use midpoints 5, 17.5, 27.5, 32.5, 42.5, 55: \(\bar x=\frac{13(5)+33(17.5)+19(27.5)+4(32.5)+15(42.5)+6(55)}{90}=25.1388\ldots\), giving \$25.1.

(a) 35 < t ≤ 40. (b) \(\bar t=[2(22.5)+6(27.5)+7(32.5)+19(37.5)+9(42.5)+7(47.5)]/50=37.3\) s. (c)(i) 15, 19, 16. (ii) Widths 15, 5, 10 give densities 1, 3.8, 1.6.

(a) \(\sum fx=5(155)+9(162.5)+18(172.5)+10(185)=7192.5\). Mean \(7192.5/42=171.25\) cm. (b) \(160<h\le165\) contains the 10.5th observation. (c) Density = frequency ÷ interval width, giving heights 0.5, 1.8, 1.2, 1.

(a) \(\bar t=[6(5)+44(15)+40(25)+14(35)+10(45)+6(55)]/120=2960/120\approx24.7\) s. (b)(i) 6, 50, 90, 104, 114, 120. (ii) Plot (0, 0), (10, 6), (20, 50), (30, 90), (40, 104), (50, 114), (60, 120). (iii) From the shown polygon: median \(20+(60-50)/4=22.5\) s; lower quartile \(10+(30-6)/4.4\approx15.5\) s; 60th percentile at CF = 72 gives \(20+(72-50)/4=25.5\) s. (c)(i) 50, 40, 30. (ii) Density = frequency ÷ class width, so bar heights are 2.5, 4, 1.

(a)(i) Use midpoints 0.5, 1.5, 2.5, 4, 6, 8.5: \(\bar t=[15(0.5)+11(1.5)+10(2.5)+19(4)+13(6)+12(8.5)]/80=305/80=3.8125\approx3.81\) h. (ii) Density = frequency ÷ class width: 15, 11, 10, 9.5, 6.5, 4. (b)(i) Chaminda’s unused branch is 2/5. On both Niluka branches, use 3/4 for use and 1/4 for non-use. (ii) \(1-P(\text{neither})=1-(2/5)(1/4)=9/10\). (iii) \((3/5)^3=27/125=0.216\).

Frequency = density × width: \(u=4.8(5)=24\). Density = frequency ÷ width: \(v=9/15=0.6\).

(a)(i) At CF = 40, read 2.8 h. (ii) At CF = 60, read 3.8 h. (iii) Q1 = 2 h, so IQR = 1.8 h. (b) \(\mathrm{CF}(7)-\mathrm{CF}(5)\approx78-72=6\). (c)(i) Cumulative counts at 2, 3, 4, 5, 6, 8 are about 20, 45, 63, 72, 76, 80, giving frequencies 20, 25, 18, 9, 4, 4. (ii) \(\bar t=[20(1)+25(2.5)+18(3.5)+9(4.5)+4(5.5)+4(7)]/80=236/80=2.95\) h. (d) Density = frequency ÷ width: 10, 25, 18, 9, 4, 2.

Divide frequency by class width: \(20/0.5=40\), \(18/1=18\), \(9/1.5=6\). Missing entries: 40 and 6.

(a) Use midpoints 20, 60, 100, 140, 180, 220: \(\bar m=[6(20)+10(60)+28(100)+76(140)+22(180)+16(220)]/158=21640/158\approx137\) g. (b)(i) \(6+10=16\) and \(28+76+22=126\). (ii) Densities are \(16/80=0.2,\ 126/120=1.05,\ 16/40=0.4\); bars cover the corresponding class intervals without gaps. (c) \([15(136)+3(130)]/18=2430/18=135\) g.

(a) Use midpoints 15,30,45: \([10(15)+30(30)+20(45)]/60=1950/60=32.5\) g. (b) Frequency densities are \(10/10=1,\ 30/20=1.5,\ 20/10=2\). Draw bars over 10–20, 20–40, 40–50 to these heights.

(a) Mode 11. Cumulative counts 35,63,85,103,117,126, so middle values are 12 and 13: median 12.5. Weighted total \(1613\), mean \(1613/126\approx12.8\). (b)(i) Frequency = bar width × density: 15,27,30. (ii) Use midpoints 2.5,6.5,9,11,14,18: \(\sum fm=15(2.5)+27(6.5)+30(9)+20(11)+24(14)+10(18)=1219\). Mean \(1219/126\approx9.67\) hours.

(a) Sorted sets: 7,8,8,10,11,16 and 8,8,8,10,10,16. In each, sum 60, middle values 8 and 10, and 8 is the unique mode. To find them, median requires the third and fourth terms to total 18; the repeated mode and order force these to be 8,10. The other terms then satisfy the sum and ordering constraints. (b)(i) Midpoints 65,85,95,110,135: \([30(65)+35(85)+40(95)+40(110)+15(135)]/160=15150/160=94.6875\approx94.7\) g. (ii) Densities 1,3.5,4,2,0.5, using frequency divided by interval width.