Q1. Find \(\bar{x}\) for data: \(x:\{2,3,5,10\}\) with \(f:\{4,6,3,2\}\).

Ans. \(\sum fx=8+18+15+20=61,\ N=15\Rightarrow \bar{x}=61/15=4.066\overline{6}\).

Q2. For class 10–20 with \(f=7\) find class mark.

Ans. \(x=\frac{10+20}{2}=15\).

Q3. Convert classes 30–39, 40–49 to continuous.

Ans. \(29.5\text{–}39.5,\ 39.5\text{–}49.5\) (subtract/add \(0.5\)).

Q4. Median class check: \(N=200\). If cumulative just before is \(92\) and next is \(150\), which is median class?

Ans. The latter (since \(N/2=100\in(92,150]\)).

Q5. Mode formula — identify \(f_0,f_1,f_2\).

Ans. \(f_1=\) frequency of modal class; \(f_0=\) of previous class; \(f_2=\) of next class.

Q6. If \(A=50,\ \sum f=40,\ \sum fd=-120\), find mean (assumed mean method).

Ans. \(\bar{x}=50+(-120/40)=47\).

Q7. Step-deviation: \(A=100,\ g=5,\ \sum fu=30,\ \sum f=60\). Find mean.

Ans. \(\bar{x}=100+(30/60)\cdot 5=102.5\).

Q8. In median, \(L=20,h=5,cf=18,f=10,N=60\). Compute.

Ans. \(N/2=30\Rightarrow 20+\frac{(30-18)}{10}\cdot 5=26\).

Q9. Mode with \(L=40,h=10,f_0=18,f_1=27,f_2=21\).

Ans. \(40+\frac{27-18}{54-18-21}\cdot 10=40+\frac{9}{15}\cdot 10=46\).

Q10. Write two uses of frequency polygon.

Ans. Shows shape (peaks/spread) and compares two distributions on same axes.

Q11. Central angle for 25% in pie.

Ans. \(0.25\times 360^\circ=90^\circ\).

Q12. Why use median instead of mean sometimes?

Ans. Median is robust to extreme values/outliers.

Q13. Krink mark (break) on axis is used when?

Ans. To skip a large gap near origin and keep graph compact.

Q14. Find missing \(f\): classes have \(f=\{6,?,10\}\) and total \(=20\).

Ans. Missing \(f=20-6-10=4\).

Q15. For 0–10, 10–20, 20–30 with \(f=5,8,7\), what is \(N\) and c.f. at 20–30?

Ans. \(N=20\), c.f. less than 30 \(=5+8+7=20\).

Q16. If \(\bar{x}=40\) and new data add 3 values each increased by 2, effect?

Ans. New mean = \(40+2=42\).

Q17. Which graph best displays component shares?

Ans. Pie diagram.

Q18. Class width if classes are 15–20, 20–25, ...

Ans. \(h=5\).

Q19. Identify outlier impact on mean.

Ans. Mean shifts toward outlier; median/mode less affected.

Q20. For equal widths, histogram bars should be …?

Ans. Adjacent (no gaps) and heights proportional to frequencies.

Q1. Compute mean (direct):

Classes 10–20, 20–30, 30–40, 40–50 with \(f=6,12,8,4\). Class marks \(15,25,35,45\). \(\sum fx=6\cdot15+12\cdot25+8\cdot35+4\cdot45=90+300+280+180=850,\ N=30\Rightarrow \bar{x}=850/30=28.\overline{3}.\)

Q2. Mean by assumed mean.

Same data, take \(A=25\). \(d:\{-10,0,10,20\}\). \(\sum fd=6(-10)+12(0)+8(10)+4(20)=-60+0+80+80=100\). \(\bar{x}=25+\frac{100}{30}=28.\overline{3}\).

Q3. Mean by step-deviation.

With \(g=10\), \(u=\{-1,0,1,2\}\). \(\sum fu=6(-1)+12(0)+8(1)+4(2)=-6+0+8+8=10\). \(\bar{x}=A+(10/30)\cdot 10=25+3.\overline{3}=28.\overline{3}\).

Q4. Find median.

Classes 0–10,10–20,20–30,30–40 with \(f=5,9,12,4\). \(N=30,\ N/2=15\). c.f.: 5,14,26 → median class 20–30. \(L=20,h=10,cf=14,f=12\Rightarrow 20+\frac{15-14}{12}\cdot 10=20+\frac{10}{12}=20+0.833\ldots=20.833\ldots\).

Q5. Find mode.

Classes 5–15,15–25,25–35,35–45 with \(f=8,15,12,5\). Modal class 15–25. \(L=15,h=10,f_1=15,f_0=8,f_2=12\Rightarrow 15+\frac{7}{30-8-12}\cdot 10=15+\frac{7}{10}\cdot 10=22\).

Q6. Convert to continuous & give mid-points: 20–24, 25–29, 30–34.

Continuous: \(19.5\text{–}24.5,\ 24.5\text{–}29.5,\ 29.5\text{–}34.5\). Mid-points: \(22,27,32\).

Q7. Read median from ‘less than’ c.f. table: c.f. are 8, 20, 36, 50, 65; classes 0–10,…40–50; find median.

\(N=65,\ N/2=32.5\Rightarrow\) median class 20–30. \(L=20,h=10,cf=20,f=16\Rightarrow 20+\frac{32.5-20}{16}\cdot 10=27.8125\).

Q8. For pie: parts \(12,18,30,40\). Find angles.

Total \(=100\). Angles: \(43.2^\circ,64.8^\circ,108^\circ,144^\circ\) respectively.

Q9. Histogram note.

Make classes continuous, choose equal scale, draw adjacent rectangles of width \(h\) and height proportional to \(f\).

Q10. Frequency polygon key points.

Plot \((\text{mid},f)\) for each class plus two end points at \(f=0\); join with straight lines.

Q11. Decide when to use mean/median/mode.

Mean: overall average; Median: skewed data/outliers; Mode: most popular item.

Q12. Show that \(\sum (x_i-\bar{x})=0\) for grouped data.

\(\sum f(x-\bar{x})=\sum fx-\bar{x}\sum f=\sum fx-N\bar{x}=0\).

Q13. If two classes tie as modal, what to do?

Data may be bimodal; report both modal classes or use graphical inspection.

Q14. From pie, sector angle \(=126^\circ\) out of total 900 people—find count.

\(\frac{126}{360}\cdot 900=315\).

Q15. Build ‘less than’ c.f. from \(f=\{6,9,4,1\}\).

c.f. \(=\{6,15,19,20\}\).

Q16. Mean of \(\{40,42,43,45,47,48\}\) via \(A=43\).

Deviations \(-3,-1,0,2,4,5\), sum \(=7\). \(\bar{x}=43+7/6=44\frac{1}{6}\).

Q17. Compute class mark list for 0–5,5–10,10–15,15–20.

Marks: \(2.5,7.5,12.5,17.5\).

Q18. If \(f\) all doubled, effect on mean/median/mode?

No change (weights scale, positions unchanged).

Q19. Why make classes continuous before median/mode?

To ensure no gaps so interpolation in formulas is valid.

Q20. For 6 classes, how many polygon vertices including zeros?

8 points (6 mids + 2 end zeros).

Q1. Time used for study per day (hrs): 0–2:7, 2–4:18, 4–6:12, 6–8:10, 8–10:3. Find mean (direct).

Ans. Class marks \(x:1,3,5,7,9\). \(\sum fx=1\cdot7+3\cdot18+5\cdot12+7\cdot10+9\cdot3=218,\ N=50\Rightarrow \bar{x}=\frac{218}{50}=4.36\ \text{hrs}.\)

Q2. Toll (₹): 300–400:80, 400–500:110, 500–600:120, 600–700:70, 700–800:40. Find mean (assumed mean).

Ans. \(x:350,450,550,650,750\). Take \(A=550\). \(d=x-A=\{-200,-100,0,100,200\}\). \(\sum f=420,\ \sum fd=80(-200)+110(-100)+120(0)+70(100)+40(200)=-12000\). \(\bar{x}=A+\frac{\sum fd}{\sum f}=550-28.571\ldots=\mathbf{₹\,521.43}\) (approx).

Q3. Milk sold (L): 1–2:17, 2–3:13, 3–4:10, 4–5:7, 5–6:3. Mean (direct).

Ans. \(x:1.5,2.5,3.5,4.5,5.5\). \(\sum fx=141.0,\ N=50\Rightarrow \bar{x}=2.82\ \text{L}.\)

Q4. Orange production (₹ thousand): 25–30:20, 30–35:25, 35–40:15, 40–45:10, 45–50:10. Mean (assumed mean).

Ans. \(x:27.5,32.5,37.5,42.5,47.5\). \(A=32.5\). \(d=\{-5,0,5,10,15\}\). \(\sum f=80,\ \sum fd=225\Rightarrow \bar{x}=32.5+225/80=\mathbf{35.3125}\ (₹\,\text{thousand}).\)

Q5. Funds by 120 workers (₹): 0–500:35, 500–1000:28, 1000–1500:32, 1500–2000:15, 2000–2500:10. Mean (step-deviation).

Ans. \(x:250,750,1250,1750,2250\). \(A=1250,\ g=250,\ u=(x-A)/g=\{-4,-2,0,2,4\}\). \(\sum fu=35(-4)+28(-2)+32(0)+15(2)+10(4)=-126,\ \sum f=120\). \(\bar{x}=1250+(-126/120)\cdot 250=\mathbf{₹\,987.50}.\)

Q6. Weekly wages (₹): 1000–2000:25, 2000–3000:45, 3000–4000:50, 4000–5000:30. Mean (step-deviation).

Ans. \(x:1500,2500,3500,4500\). \(A=3500,\ g=500,\ u=\{-4,-2,0,2\}\). \(\sum fu=25(-4)+45(-2)+50(0)+30(2)=-130,\ N=150\). \(\bar{x}=3500+(-130/150)\cdot 500=\mathbf{₹\,3066.67}\) (approx).

Q1. Hours worked: 8–10:150, 10–12:500, 12–14:300, 14–16:50. Median?

Ans. \(N=1000,\ N/2=500\). c.f.: 150, 650, 950… Median class 10–12. \(L=10,h=2,cf=150,f=500\Rightarrow 10+\frac{500-150}{500}\cdot 2=\mathbf{11.4\ hr}.\)

Q2. Mango yield (per tree): 50–100:33, 100–150:30, 150–200:90, 200–250:80, 250–300:17.

Ans. \(N=250,\ N/2=125\). c.f.: 33,63,153… Median class 150–200. \(L=150,h=50,cf=63,f=90\Rightarrow 150+\frac{125-63}{90}\cdot 50=\mathbf{184.44}.\)

Q3. Vehicle speeds (km/h): 60–64:10, 64–69:34, 70–74:55, 75–79:85, 79–84:10, 84–89:6. Median?

Ans. Make continuous: \(59.5\text{–}64.5,\dots\). \(N=200,\ N/2=100\). c.f.: 10,44,99,184… Median class \(74.5\text{–}79.5\). \(L=74.5,h=5,cf=99,f=85\Rightarrow 74.5+\frac{1}{85}\cdot 5=\mathbf{74.56\ km/h}\) (approx).

Q4. Bulbs (thousands): 30–40:12, 40–50:35, 50–60:20, 60–70:15, 70–80:8, 80–90:7, 90–100:8.

Ans. \(N=105,\ N/2=52.5\). c.f.: 12,47,67… Median class 50–60. \(L=50,h=10,cf=47,f=20\Rightarrow 50+\frac{52.5-47}{20}\cdot 10=\mathbf{52.75}.\)

Q1. Fat% vs milk (L): 2–3:30, 3–4:70, 4–5:80, 5–6:60, 6–7:20. Mode?

Ans. Modal class 4–5. \(L=4,h=1,f_1=80,f_0=70,f_2=60\Rightarrow 4+\frac{10}{160-70-60}\cdot 1=\mathbf{4.33\%}.\)

Q2. Electricity units: 0–20:13, 20–40:50, 40–60:70, 60–80:100, 80–100:80, 100–120:17.

Ans. Modal 60–80. \(L=60,h=20,f_1=100,f_0=70,f_2=80\Rightarrow 60+\frac{30}{200-70-80}\cdot 20=\mathbf{72}.\)

Q3. Milk to hotels (L): 1–3:7, 3–5:5, 5–7:15, 7–9:20, 9–11:35, 11–13:18.

Ans. Modal 9–11. \(L=9,h=2,f_1=35,f_0=20,f_2=18\Rightarrow 9+\frac{15}{70-20-18}\cdot 2=\mathbf{9.9375\ L}.\)

Q4. Patients’ ages: 0–5:38, 5–9:32, 10–14:50, 15–19:36, 20–24:24, 25–29:20.

Ans. Modal 10–14. \(L=10,h=5,f_1=50,f_0=32,f_2=36\Rightarrow 10+\frac{18}{100-32-36}\cdot 5=\mathbf{12.8125\ yrs}.\)

Q1. Heights (cm): 135–140:4, 140–145:12, 145–150:16, 150–155:8. Draw histogram.

Ans. Classes are continuous (width \(=5\)). On X-axis plot classes; Y-axis frequencies. Draw adjacent rectangles of heights \(4,12,16,8\). (Mobile tip: label mid-points \(137.5,142.5,147.5,152.5\) under bars.)

Q2. Jowar yield per acre (quintal): 2–3:30, 4–5:50, 6–7:55, 8–9:40, 10–11:20.

Ans. Make classes continuous if needed (e.g., \(1.5\text{–}3.5,\dots\)). Draw bars with heights \(30,50,55,40,20\).

Q3. Investment (₹ thousand): 10–15:30, 15–20:50, 20–25:60, 25–30:55, 30–35:15.

Ans. Equal widths; bars with heights \(30,50,60,55,15\).

Q4. Prep time (min): 60–80:14, 80–100:20, 100–120:24, 120–140:22, 140–160:16.

Ans. Draw bars for each class with the listed heights. Use a krink mark if axis needs skipping.

Q1. From polygon (fig 6.6 in text), answer:

Ans. (1) Max students at class 60–70 (peak). (2) Zero at 20–30 and 90–100. (3) Class-mark for 50 students \(=55\Rightarrow\) class 50–60. (4) Class-mark 85 \(\Rightarrow\) class 80–90 (limits 80,90). (5) Students in 80–90 \(\approx 35\).

Q2. Electricity bill (₹): 0–200:240, 200–400:300, 400–600:450, 600–800:350, 800–1000:160. Draw polygon.

Ans. Plot points at \((100,240),(300,300),(500,450),(700,350),(900,160)\) and add \(( - ,0)\) at \(x= -100\) & \((1100,0)\); join sequentially.

Q3. Result %: 30–40:7, 40–50:33, 50–60:45, 60–70:65, 70–80:47, 80–90:18, 90–100:5.

Ans. Points: \((35,7),(45,33),(55,45),(65,65),(75,47),(85,18),(95,5)\) plus end zeros at \(25,105\); join to form polygon.

Q1. Blood donors by age (20–25:80, 25–30:60, 30–35:35, 35–40:25). Draw pie — find angles.

Ans. Total \(=200\). Angles: \(80/200\cdot 360=144^\circ,\ 60/200\cdot 360=108^\circ,\ 35/200\cdot 360=63^\circ,\ 25/200\cdot 360=45^\circ.\)

Q2. Marks (Eng 50, Mar 70, Sci 80, Math 90, SS 60, Hin 50). Angles?

Ans. Total \(=400\). Angles: Eng \(45^\circ\), Mar \(63^\circ\), Sci \(72^\circ\), Math \(81^\circ\), SS \(54^\circ\), Hin \(45^\circ\). (Sum \(=360^\circ\)).

Q3. Trees planted by Std 5–10: 40,50,75,50,70,75. Angles?

Ans. Total \(=360\). Angles: \(40:50:75:50:70:75\Rightarrow\) \(40/360\cdot 360=40^\circ,\ 50^\circ,\ 75^\circ,\ 50^\circ,\ 70^\circ,\ 75^\circ.\)

Q4. Fruit demand %: Mango 30, Sweet lime 25, Apples 20, Cheeku 10, Oranges 15. Angles?

Ans. \(108^\circ, 90^\circ, 72^\circ, 36^\circ, 54^\circ\) respectively.

Q5. Workers pie (fig 6.13). If total \(=10{,}000\): (i) construction? (ii) admin? (iii) production %?

Ans. Read sectors: suppose angles (as typical) Construction \(=72^\circ\Rightarrow \frac{72}{360}\cdot 10000=2000\). Admin sector angle (given) → count accordingly. Production percentage equals its sector’s degree ÷ \(360^\circ\times 100\%\). (Use actual figure labels on handout.)

Q6. Investments pie (fig 6.14). If ‘Shares’ \(=₹2000\) and shares sector \(=20\%\), find totals and amounts.

Ans. Total \(=2000/0.20=₹10000\). Deposit in bank (say \(=30\%\)) → \(₹3000\). Difference (Immovable \(-\) Mutual) (e.g., \(35\%-10\%=25\%\)) → \(₹2500\). Post (e.g., \(5\%\)) → \(₹500\). (Insert exact percentages from your pie diagram.)

Q1. 40% are blood group O. Sector angle?

Ans. \(\dfrac{40}{100}\cdot 360^\circ=\mathbf{144^\circ}\).

Q2. Cement ₹45{,}000 is a \(75^\circ\) sector. Total cost?

Ans. \(\frac{75}{360}=\frac{45000}{T}\Rightarrow T=\frac{45000\cdot 360}{75}=₹\mathbf{216000}\).

Q3. Cumulative frequencies help to find …?

Ans. Median.

Q4. In \(u\)-method, \(u_i=\ ?\)

Ans. \(u_i=\dfrac{x_i-A}{g}\).

Q5. Distance per litre groups 12–14:11, 14–16:12, 16–18:20, 18–20:7. Median group?

Ans. \(N=50,\ N/2=25\). c.f.: 11,23,43 → median class \(16\text{–}18\).

Q6. Trees per student 1–3:7, 4–6:8, 7–9:6, 10–12:4. Polygon point for class 4–6?

Ans. Class mark \(=5\Rightarrow (5,8)\).

Q7. Farmers’ income (k₹): 20–30:10, 30–40:11, 40–50:15, 50–60:16, 60–70:18, 70–80:14. Mean?

Ans. Class marks \(25,35,45,55,65,75\). \(\sum fx=10\cdot25+11\cdot35+15\cdot45+16\cdot55+18\cdot65+14\cdot75= (250+385+675+880+1170+1050)=\mathbf{4410}\). \(N=84\Rightarrow \bar{x}=\mathbf{52.5}\ k₹.

Q8. Bank loans (k₹): 40–50:13, 50–60:20, 60–70:24, 70–80:36, 80–90:7. Mean?

Ans. Marks \(45,55,65,75,85\). \(\sum fx=585+1100+1560+2700+595=\mathbf{6540}\). \(N=100\Rightarrow \bar{x}=\mathbf{65.40}\ k₹.\)

Q9. Weekly wages (₹): 0–2000:15, 2000–4000:35, 4000–6000:50, 6000–8000:20. Mean?

Ans. \(x:1000,3000,5000,7000\). \(\sum fx=15000+105000+250000+140000=\mathbf{510000}\). \(N=120\Rightarrow \bar{x}=\mathbf{₹\,4250}.\)

Q10. Aid (k₹): 50–60:7, 60–70:13, 70–80:20, 80–90:6, 90–100:4. Mean?

Ans. \(x:55,65,75,85,95\). \(\sum fx=385+845+1500+510+380= \mathbf{3620}\). \(N=50\Rightarrow \bar{x}=\mathbf{72.4}\ k₹.\)

Q11. Bus distances (km): 200–210:40, 210–220:60, 220–230:80, 230–240:50, 240–250:20. Median?

Ans. \(N=250,\ N/2=125\). c.f.: 40,100,180 → median class 220–230. \(L=220,h=10,cf=100,f=80\Rightarrow 220+\frac{25}{80}\cdot 10=\mathbf{223.125}.\)

Q12. Prices vs demand: <20:140, 20–40:100, 40–60:80, 60–80:60, 80–100:20. Median price class?

Ans. Convert to grouped with frequencies \(f:\{140,100,80,60,20\}\). \(N=400,\ N/2=200\). c.f.: 140,240 → median class 20–40.

Q13. Demand for sweet (g): 0–250:10, 250–500:60, 500–750:25, 750–1000:20, 1000–1250:15. Mode?

Ans. Modal 250–500. \(L=250,h=250,f_1=60,f_0=10,f_2=25\Rightarrow 250+\frac{50}{120-10-25}\cdot 250=250+\frac{50}{85}\cdot 250\approx \mathbf{397.06\ g}.\)

Q14. Histogram: electricity use (units) 50–70:150, 70–90:400, 90–110:460, 110–130:540, 130–150:600, 150–170:350.

Ans. Draw adjacent bars for each class with given heights (equal width \(=20\)).

Q15. Handloom days to weave (No. workers): 8–10:5, 10–12:16, 12–14:30, 14–16:40, 16–18:35, 18–20:14. Frequency polygon?

Ans. Plot \((9,5),(11,16),(13,30),(15,40),(17,35),(19,14)\) and end zeros at \(7,21\); join in order.

Q16. Experiment time (min): 20–22:8, 22–24:16, 24–26:22, 26–28:18, 28–30:12, 30–32:14. Draw histogram & polygon.

Ans. Histogram bars of listed heights. Polygon through midpoints \((21,8),(23,16),(25,22),(27,18),(29,12),(31,14)\) plus endpoints at \(19,33\) with \(0\).

Q17. Blood donors age (yrs): 20–24:38, 25–29:46, 30–34:35, 35–39:24, 40–44:15, 45–49:12. Frequency polygon?

Ans. Points \((22,38),(27,46),(32,35),(37,24),(42,15),(47,12)\) plus zeros at \(20,50\); join.

Q18. Average rainfall (cm) in 150 towns: 0–20:36, 20–40:48, 40–60:?, 60–80:?, 80–100:? (as per text) — draw polygon.

Ans. Use given full table; plot \((10,36),(30,48),\dots\) with end zeros; join. (Add actual remaining frequencies from your page.)

Q19. Vehicles pie (fig 6.15): compute each central angle and total vehicles if two-wheelers are 1200 and sector is 48%.

Ans. Angles = given % × \(360^\circ\). Total \(=1200/0.48=\mathbf{2500}\) vehicles.

Q20. Sports pie (fig 6.16) with total 1000 students. Find numbers for Cricket, Football, Others by reading sector %.

Ans. Multiply each sector % by 1000; e.g., \( \text{Cricket} = (\%\_\text{cricket})\times 10\).