(a) Midpoints: \(5,20,37.5,60\). Estimated mean \(=\frac{13(5)+35(20)+36(37.5)+6(60)}{90}=\frac{2475}{90}=27.5\) minutes.

(b) Frequency densities are \(13/10=1.3\), \(35/20=1.75\), \(36/15=2.4\), \(6/30=0.2\). A density of 1.75 gives height 14, so the scale is 8 cm per unit. The other heights are \(10.4,19.2,1.6\) cm, in interval order.

(c) Among the 42 eligible trains, 36 have delays at most 45 minutes and 6 have longer delays. Without replacement, \(P=\frac{36}{42}\frac6{41}+\frac6{42}\frac{36}{41}=\frac{72}{287}\).

(a) \(\frac{200+51+2n+n+1}{4}=3n\). Thus \(252+3n=12n\), giving \(\boxed{n=28}\).

(b) Use midpoints \(10,30,50,70\). \(\bar m=\frac{29(10)+67(30)+55(50)+49(70)}{200}=\frac{8480}{200}=\boxed{42.4\text{ g}}\).

(a) The largest frequency is 14, corresponding to \(\boxed{1\text{ book}}\).

(b) The median lies between the 20th and 21st values. The cumulative frequencies are \(5,19,29,37,38,40\), so both middle values are 2: \(\boxed{2\text{ books}}\).

(c) \(\bar x=\frac{0(5)+1(14)+2(10)+3(8)+4(1)+5(2)}{40}=\frac{72}{40}=\boxed{1.8\text{ books}}\).

(a) The greatest frequency is 34, so \(\boxed{60<x\le80}\).

(b) Use midpoints \(10,30,50,70,90\). \(\bar x=\frac{4(10)+20(30)+31(50)+34(70)+11(90)}{100}=\frac{5560}{100}=\boxed{55.6}\).

(a)(i) Class midpoints are \(30,50,80\). Thus \(\bar t\approx\frac{30(30)+40(50)+30(80)}{100}=\frac{5300}{100}=\boxed{53\text{ s}}\).

(a)(ii) Frequency density is frequency divided by class width. The bar heights are \(30/20=1.5,40/20=2,30/40=0.75\). Draw the remaining bars over \((40,60]\) and \((60,100]\).

(b)(i) Plot \((20,0),(40,12),(60,36),(80,60),(100,74),(120,80)\) and join with a smooth increasing curve.

(b)(ii)(a) At cumulative frequency \(80/2=40\), read \(\boxed{\text{median}\approx63\text{ s}}\).

(b)(ii)(b) At cumulative frequency \(80/4=20\), read \(\boxed{Q_1\approx47\text{ s}}\).

(a) Use midpoints \(170,182.5,187.5,195,205\). The weighted total is \(7(170)+12(182.5)+31(187.5)+70(195)+20(205)=26942.5\). Mean \(=26942.5/140=192.446428\ldots\), giving \(\boxed{192\text{ cm}}\) to 3 significant figures.

(b) Choose without replacement: \(\frac{70}{140}\frac{69}{139}=\boxed{69/278}\).

(c) Frequency densities are \(\boxed{0.35,2.4,6.2,7,2}\), from frequency divided by class width. Draw each bar over its class interval.

Use class midpoints 7.5,11,16. Weighted total \(=3(7.5)+24(11)+23(16)=654.5\). Estimated mean \(=654.5/50=\boxed{13.09\text{ cm}}\).

Use midpoints 110,125,140,170. Weighted total \(=32(110)+55(125)+64(140)+49(170)=27685\). Estimated mean \(=27685/200=\boxed{138.425\text{ cm}}\).

(a)(i) \(\boxed{9.3\text{ s}}\). (ii) Q₁=8.2,Q₃=11.6, so IQR=\(\boxed{3.4\text{ s}}\). (iii) Fastest time 6 s, slowest 13 s. Difference=\((195/6-195/13)\times3.6=\boxed{63\text{ km/h}}\). (b)(i) Cumulative frequencies 7,20,39,60,80 put the 40 th and 41 st in \(\boxed{420<d\le450}\). (ii) Midpoints 275,350,410,435,475 give \((7(275)+13(350)+19(410)+21(435)+20(475))/80=\boxed{411.25\text{ km}}\). (iii) Heights are proportional to frequency/width. The first density=7/50=0.14, so scale=2.8/0.14=20. Heights=\(20(13/100),20(19/20),20(21/30)=\boxed{2.6,19,14\text{ cm}}\). (iv) Include both orders: \(2\times\frac{20}{80}\times\frac7{79}=\boxed{7/158}\).

(a) Mode=\(\boxed5\) (highest frequency 14). Cumulative frequencies 10,16,27,40,54,60 put both middle values at 4: median=\(\boxed4\). Weighted mean=\((10+12+33+52+70+36)/60=\boxed{3.55}\). (b)(i) Midpoints 25,40,62.5 give \((10(25)+72(40)+18(62.5))/100=\boxed{42.55\text{ km/h}}\). (ii) First density=10/10=1, so scale 3 cm per density unit. Other heights=\(3(72/20)=\boxed{10.8\text{ cm}}\), \(3(18/25)=\boxed{2.16\text{ cm}}\).

(a) The 10 th and 11 th observations both lie in the 2-coins group, so median=\(\boxed{2}\). (b) Total coins \(=0(3)+1(1)+2(7)+3(8)+6(1)=45\). Mean \(=45/20=\boxed{2.25}\).

(a)(i) Mode \(7\text{ s}\). (ii) Cumulative frequencies \(3,35,54,83,94,100\) place both middle students at \(8\text{ s}\).

(iii) \(\bar t=\frac{6(3)+7(32)+8(19)+9(29)+10(11)+11(6)}{100}=8.31\text{ s}\).

(iv) 46 times are at least 9: \(\frac{46}{100}\frac{45}{99}=\frac{23}{110}\).

(b) Range \(158-105=53\text{ cm}\). IQR \(142-122=20\text{ cm}\).

(c)(i) Class midpoints \(80,155,250\). \(\bar m\approx\frac{80(60)+155(99)+250(41)}{200}=151.975\text{ g}\).

(ii) Frequency densities \(60/60=1,\quad99/90=1.1,\quad41/100=0.41\).

(a)(i) Midpoints \(45,53,60,67\). \(\bar m\approx\frac{45(126)+53(520)+60(154)+67(200)}{1000}=55.87\text{ g}\).

(ii) \(\frac{154+200}{1000}=\frac{177}{500}\).

(b)(i) \$25 000. (ii) \(2.473\times10^4\) dollars.

(c)(i) Upper bounds are 505 hens and 330 eggs per hen. Product \(505(330)=166650\) eggs is an upper bound.

(ii) Use the lower bound 790 for the larger farm and upper bound 505 for the smaller: \(790-505=285\). If restricting to actual integer numbers of hens, the smaller farm has at most 504, giving the sharper bound 286.

(a)(i) \(20-15=5\). (ii) 17 has the highest frequency. (iii) Cumulative frequencies \(1,4,23,34,44,50\) place both the 25th and 26th marks at 18.

(iv) \(\bar x=\frac{15+3(16)+19(17)+11(18)+10(19)+6(20)}{50}=17.88\).

(b)(ii) The 6th ordered score is 21. (iii) Excluding the median, quartiles are the 3rd and 9th ordered scores, 20 and 30. IQR \(30-20=10\).

Known frequency sum 32 and weighted sum 251. \(\frac{251+7x}{32+x}=7.6\). Thus \(251+7x=243.2+7.6x\), \(.6x=7.8\), and \(x=13\).

(a)(i) \(600(60/360)=100\). (ii) \(600(27/360)=45\). (iii) Cricket angle \(125/600\times360=75\)°. Swimming angle \(360-105-60-27-75=93\)° (155 people). (b) Maximum \(.6+8.1=8.7\) cm, upper quartile \(3.4+4.1=7.5\) cm. Plot the five values 0.6, 3.4, 5.2, 7.5, 8.7. (c)(i) \(6-1=5\). (ii) Highest frequency 25 occurs at score 2, so mode 2. (iii) Cumulative frequencies 16, 41, 58, 77, 85, 100 place both the 50th and 51st values at score 3, so median 3. (d) Midpoints 15, 22.5, 27.5, 40, 65 give \(\sum ft=1960\). Mean \(1960/50=39.2\) s.

\(\bar x=\frac{5(8)+6(5)+7(11)+8(7)+9(5)+10(4)}{40}=\frac{288}{40}=7.2\).

Use midpoints 10, 22.5, 30, 42.5 and 75. \(\bar x\approx\frac{21(10)+16(22.5)+6(30)+10(42.5)+7(75)}{60}=\frac{1700}{60}=28.33\). Estimated mean = \$28.33.

(a) Disagree: women have the higher median (about 100 minutes compared with 80). Disagree: women have the larger interquartile range (about 150 minutes compared with 110).

(b)(i) Midpoints 30,80,130,190,270. \(\bar t\approx\frac{41(30)+24(80)+23(130)+8(190)+4(270)}{100}=87.4\text{ min}\). (ii)(a) Read across from cumulative frequency 60: about 90 minutes. (b) 3 h = 180 min; cumulative frequency is about 92, so \(100-92=8\) children. (iii) Heights are proportional to frequency density: \(10.8\,\frac{8/60}{24/40}=2.4\text{ cm}\).

(a)(i) Midpoints25,75,112.5,137.5,175: \(\bar p\approx\frac{2(25)+5(75)+4(112.5)+6(137.5)+3(175)}{20}=111.25\text{ g}\). (ii)2,7,11,17,20. (iii)3/20.

(b) First density=32/4=8; height1.6 gives height/density=0.2. Frequencies are \((2/.2)(2)=20\) and \((1.2/.2)(1)=6\). (c)(i) \((179+7x)/(45+x)=4.28\Rightarrow179+7x=192.6+4.28x\Rightarrow x=5\). (ii)3 books has the largest frequency16. (iii) There are50 students; the25th and26th values both lie in the4-book group. Median4.

(a)(i) 4 visits, the largest frequency (7). (ii) \(\frac{0(5)+1(5)+2(6)+3(6)+4(7)+5(3)+6(6)+7(2)}{40}=\frac{128}{40}=3.2\). (b) \(\frac3{40}\times360^\circ=27^\circ\).

(a)(i) Positive. (ii) A representative best-fit line is y≈0.5x. (iii) At x=4, y≈2. (b) Mode=0; the 12th and13th observations are both1, so median=1. Mean=\(\frac{0(10)+1(8)+2(3)+3(2)+5(1)}{24}=\frac{25}{24}\approx1.04\). (c)(i) \(\frac{78(49)+180(57)+162(71)}{420}=\frac{25584}{420}\approx60.9\) g. (ii) Densities=78/6=13,180/10=18,162/18=9; draw adjacent bars over (46,52],(52,62],(62,80].

Use class midpoints 15, 30, 45, 55 and 75. \(\bar h\approx\frac{7(15)+15(30)+27(45)+13(55)+8(75)}{70}=\frac{3085}{70}=44.0714\ldots\approx44.1\) cm.

(a)(i) Read the median at cumulative frequency 100: 400 m³. (ii) At frequency 50, the lower quartile is 350 m³. (iii) The upper quartile, at frequency 150, is 420 m³, so the interquartile range is \(420-350=70\) m³. (iv) At 300 m³ the cumulative frequency is 30; \(200-30=170\) estimates exceed 300 m³.

(b)(i) Use midpoints 40, 80, 125 and 200: \(\bar A\approx\frac{32(40)+64(80)+80(125)+24(200)}{200}=\frac{21200}{200}=106\) m². (ii) Frequency densities are \(32/40=0.8\), \(64/40=1.6\), \(80/50=1.6\) and \(24/100=0.24\). Draw bars over the corresponding class intervals.

(iii) \(80+24=104\) estimates exceed 100 m². Selection is without replacement: \(\frac{104}{200}\times\frac{103}{199}=\frac{10712}{39800}=\frac{1339}{4975}\).

(a)(i) \(0.05^2=0.0025\). (ii) Use the complement of both rejected: \(1-0.0025=0.9975\). (b) There are four positions for the rejected coin: \(4(0.05)(0.95)^3=0.171475\approx0.171\). (c) Use midpoints 225, 275, 325, 375, 425 and 475. \(\bar a\approx\frac{13(225)+19(275)+27(325)+56(375)+62(425)+23(475)}{200}=\frac{75200}{200}=376\), giving \$376. (d)(i) The first bar has width 100 and density 0.16, so \(100(0.16)=16\) cars. (ii) The last two bars contain \(50(0.8)+100(0.26)=66\) cars. Percentage \(=\frac{66}{200}\times100=33\%\).

(a) Total pets \(0(1)+1(2)+2(3)+3(5)+4(7)+5(3)+6(3)=84\). Mean \(84/24=3.5\). (b) New total \(25(3.44)=86\), so Jennifer has \(86-84=2\) pets.

(a)(i) \((582+18p)/(33+p)=17.7\Rightarrow582+18p=584.1+17.7p\Rightarrow p=7\). (ii) 40 motorists; the 20th and 21st entries are both 17 litres. (b)(i) \(320(1.28/6.4)=64\) dollars. (ii) \(60(320/480)=40\) litres. (iii) \(64/40=1.60\) dollars/litre. (c) Upper bounds 32.5 m/s and 305 s: \(d_{\rm upper}=32.5(305)/1000=9.9125\) km.

(a) \(10<h\le13\) contains the 100th and 101st trees. (b) Midpoints are 4, 8, 11.5, 15, 18 and 19.5. \(\sum fx=23(4)+47(8)+45(11.5)+38(15)+32(18)+15(19.5)=2424\). Mean \(2424/200=12.12\) m. (c) 23, 70, 115, 153, 185, 200.

(a) Total frequency = 75 and total marks = 1065. \(\bar x=1065/75=14.2\). Median is the 38th observation, in the mark-14 group: 14. Largest frequency is 16, at mark 13, the mode. (b)(i) Missing frequencies 21, 30, 15. (ii) Widths 20, 20, 10, 10; divide frequency by width for densities 1.05, 1.5, 1.5, 0.9. (c) \([10(2.5)+12(3)+4n]/(22+n)=3.1\Rightarrow61+4n=68.2+3.1n\Rightarrow0.9n=7.2\Rightarrow n=8\).

(a)(i) 4, the most frequent size. (ii) Both the 20th and 21st values are 5, so 5. (iii) \([2+3+4(17)+5(12)+6(6)+7(3)]/40=190/40=4.75\). (b) Add the two totals: \((190+3n)/(40+n)\).