1-Mark Questions (20) — with Answers
Q1. What is Statistics?
Ans. The study of collecting, organizing, presenting and interpreting data to draw conclusions.
Q2. Define primary data.
Ans. Data collected first‑hand by the investigator (e.g., measurements, surveys).
Q3. Define secondary data.
Ans. Data obtained from existing sources like records, reports, journals, websites.
Q4. What is a class in a frequency table?
Ans. A group/interval into which observations are grouped (e.g., \(10\text{–}20\)).
Q5. What are class limits?
Ans. The lower and upper end values of a class (e.g., for \(10\text{–}20\), 10 and 20).
Q6. Define frequency.
Ans. Number of observations falling in a class.
Q7. Write the formula for class mark.
Ans. \(\text{Class mark}=\dfrac{\text{lower limit}+\text{upper limit}}{2}\).
Q8. What is range of data?
Ans. \(\text{Range}=\text{maximum} - \text{minimum}\).
Q9. Inclusive classes mean…
Ans. Upper limit included in the class (e.g., 6–10 includes 10).
Q10. Exclusive classes mean…
Ans. Upper limit excluded (e.g., 10–15 includes 10\(\le x<\)15).
Q11. Define mean for ungrouped data.
Ans. \(\bar x=\dfrac{\sum x_i}{n}\).
Q12. Define median for ungrouped data.
Ans. Middle observation in ordered data; if even \(n\), average of the two middle observations.
Q13. Define mode.
Ans. The observation with the highest frequency (most repeated value).
Q14. What is a sub‑divided bar diagram?
Ans. A bar representing total split proportionally into parts of components.
Q15. What is a percentage bar diagram?
Ans. Sub‑divided bar where height is taken as 100 and parts show percentages.
Q16. Less‑than cumulative frequency (CF) means…
Ans. Sum of frequencies up to the upper class limit.
Q17. More‑than (≥ lower limit) cumulative frequency means…
Ans. Sum of frequencies from the class to the last class.
Q18. In exclusive classes, where will value 10 lie for 0–10, 10–20, 20–30?
Ans. In \(10\text{–}20\).
Q19. Write the mean formula for discrete frequency table.
Ans. \(\bar x=\dfrac{\sum f_i x_i}{\sum f_i}\).
Q20. If data tends to cluster around a value, that value is called…
Ans. A measure of central tendency.
2-Mark Questions (20) — with Solutions
Q1. Find the class mark for \(25\text{–}35\).
Ans. \(\dfrac{25+35}{2}=30\).
Q2. For classes 0–10, 10–20, 20–30, where is 20 included (exclusive method)?
Ans. In \(20\text{–}30\).
Q3. Compute range of marks: 6, 8, 14, 19, 20.
Ans. Range \(=20-6=14\).
Q4. The mean of 5 numbers is 50. If four numbers sum to 180, find the 5th.
Ans. Total \(=250\Rightarrow\) 5th \(=250-180=70\).
Q5. Arrange and find median: 7, 10, 7, 5, 9, 10.
Ans. Ordered: 5,7,7,9,10,10. Median \(=\dfrac{7+9}{2}=8\).
Q6. Find mode: 19, 19, 15, 20, 25, 15, 20, 15.
Ans. Mode \(=15\) (appears 3 times).
Q7. If \(\bar x=40\) for 100 items and one value 30 is replaced by 70, new mean?
Ans. Total increases by 40; new mean \(=\dfrac{4000+40}{100}=40.4\).
Q8. Find less‑than CF up to 40 for class‑freq: 0–10:7, 10–20:3, 20–30:12, 30–40:13, 40–50:2.
Ans. \(7+3+12+13=35\).
Q9. Mean of nine numbers is 77. A new number added makes mean 82. Find added number.
Ans. Old total \(=693\). New total \(=820\). Added \(=127\).
Q10. Find mean of 25, 30, 27, 23, 25.
Ans. \(\bar x=\dfrac{130}{5}=26\).
Q11. For class 150–153 with freq 5, what is less‑than CF at 153?
Ans. If it is the first class, CF \(=5\) (given pattern in text).
Q12. Convert 30/40 to percentage.
Ans. \(\dfrac{30}{40}\times100=75\%\).
Q13. If \(\sum f_i x_i=956\) and \(\sum f_i=35\), find mean.
Ans. \(\bar x=956/35\approx27.31\).
Q14. Which diagram is better to compare composition of totals?
Ans. Sub‑divided/Percentage bar diagram.
Q15. Identify: “Heights from school record sent urgently to head office”—primary or secondary?
Ans. Secondary (taken from existing records).
Q16. Define class width for 5–10, 10–15, 15–20.
Ans. \(10-5=5\) (uniform width 5).
Q17. If class mark is 10 and width is 6 (continuous), find class.
Ans. Limits are \(10\pm3\Rightarrow 7\text{–}13\).
Q18. State two advantages of percentage bar over simple bar for composition.
Ans. Normalizes totals; easy comparison of proportions across groups.
Q19. What does cumulative frequency at top class equal?
Ans. Total number of observations \((N)\).
Q20. For data 99, 100, 95, 100, 100, 60, 90, what is mode?
Ans. \(100\).
3-Mark Questions (20) — with Solutions
Q1. Prepare a less‑than CF for: 0–10:2, 10–20:12, 20–30:20, 30–40:16.
Ans. CFs: <10 = 2, <20 = 2+12 = 14, <30 = 14+20 = 34, <40 = 34+16 = 50.
Q2. Prepare a more‑than (≥ lower limit) CF for same data.
Ans. Tot = 50. ≥0:50, ≥10:48, ≥20:36, ≥30:16.
Q3. Convert family expenses into percentage (A: Food 60, Clothes 10, Edu 10, Elec 5, Others 15). Draw % bar steps.
Ans. Already in %; take bar height = 100; stack 60,10,10,5,15 in order.
Q4. From marks list (out of 20) of 50 students (given in text), compute frequency of 15.
Ans. In ordered list, 15 appears 5 times ⇒ frequency of 15 is 5.
Q5. Using same list, find lowest, highest and range.
Ans. Lowest 6, highest 20, range 14.
Q6. Form grouped table (inclusive): 6–10, 11–15, 16–20 with totals 10, 20, 20 (as per text).
Ans. Frequencies: 6–10 → 10; 11–15 → 20; 16–20 → 20; \(N=50\).
Q7. For classes 5–10, 10–15, 15–20, where does 10.3 and 15.7 lie?
Ans. 10.3 ∈ 10–15; 15.7 ∈ 15–20 (exclusive method).
Q8. Find mean for data: 10(1), 15(2), 16(5), 17(2), 20(3), 23(2), 25(3), 30(3), 35(2), 36(2), 37(4), 39(3), 40(3).
Ans. \(\sum f_ix_i=956\), \(\sum f_i=35\Rightarrow \bar x=27.31\) (approx.).
Q9. Find median of 54, 63, 66, 72, 78, 87, 92.
Ans. Ordered already; middle (4th) = 72.
Q10. Find median of 21, 23, 25, 30, 32, 33, 36, 40, 42, 43.
Ans. Even \(n=10\): median = \(\dfrac{32+33}{2}=32.5\).
Q11. Two modes possible? Example from ages table (21 & 29 both max freq).
Ans. Yes, bimodal when two values share highest frequency.
Q12. If class mark 13 for 11–15, verify formula.
Ans. \((11+15)/2=13\) ✔️.
Q13. Compute % composition: 30 of 40 is what %?
Ans. \(75\%\).
Q14. Decide suitable diagrams for (i) literacy % of 4 villages, (ii) family expenses, (iii) boys & girls in 5 divisions.
Ans. (i) Simple bar/percentage bar; (ii) sub‑divided/percentage bar; (iii) joint bar.
Q15. Identify data type: feedback forms filled by customers.
Ans. Primary data.
Q16. If less‑than CF at 60 is 48 and total \(N=62\), what is ≥60 CF?
Ans. \(62-48=14\).
Q17. For shoe sizes with inclusive classes 2–4, 5–7, 8–10, specify class width.
Ans. Width varies if discrete; effectively count of values inside each (3 each).
Q18. When to prefer mean vs mode?
Ans. Mean for overall performance; Mode for most common choice/size.
Q19. If \(x\) is mean of \(x_i\) and \(y\) of \(y_i\), mean of combined \(x_i\) and \(y_i\) (equal counts) is?
Ans. \(\dfrac{x+y}{2}\).
Q20. For class 25–35, upper class limit?
Ans. \(35\).
All Textbook Exercises — Perfect Solutions
Below are clean, MathJax‑formatted solutions for Practice Sets 7.1–7.5 and Problem Set 7. Where drawing of diagrams is asked, we provide exact percentage computations and clear steps to plot sub‑divided or percentage bar‑diagrams.
Practice Set 7.1
Q1. Buses/Trucks (in lakh) — Draw percentage bar. (2005–06: Trucks 47, Buses 9; 2007–08: 56,13; 2008–09: 60,16; 2009–10: 63,18)
Ans. Totals (in lakh): 56, 69, 76, 81. Percentages (approx to nearest integer):
2005–06: Trucks \(\dfrac{47}{56}\times100\approx84\%\), Buses \(16\%\).
2007–08: Trucks \(\dfrac{56}{69}\approx81\%\), Buses \(19\%\).
2008–09: Trucks \(\dfrac{60}{76}\approx79\%\), Buses \(21\%\).
2009–10: Trucks \(\dfrac{63}{81}\approx78\%\), Buses \(22\%\). Draw 4 bars of height 100 and stack with these percentages.
2005–06: Trucks \(\dfrac{47}{56}\times100\approx84\%\), Buses \(16\%\).
2007–08: Trucks \(\dfrac{56}{69}\approx81\%\), Buses \(19\%\).
2008–09: Trucks \(\dfrac{60}{76}\approx79\%\), Buses \(21\%\).
2009–10: Trucks \(\dfrac{63}{81}\approx78\%\), Buses \(22\%\). Draw 4 bars of height 100 and stack with these percentages.
Q2. Roads (lakh km): Permanent—14,15,17,20; Temporary—10,11,13,19 for years 2000–01, 2001–02, 2003–04, 2007–08. Draw sub‑divided and percentage bar.
Ans. Totals: 24, 26, 30, 39. Percent Permanent: \(58,58,57,51\%\) (approx). Percent Temporary: \(42,42,43,49\%\). Steps: choose a scale; draw sub‑divided bars with actual values; for % bar, take height 100 and stack the percentages.
Activity (Girls per 1000 boys): Complete table for 5 states and draw % bar.
Ans. For each state, Total \(=1000+\text{girls}\). % boys \(=\dfrac{1000}{\text{Total}}\times100\), % girls \(=100-\%\text{boys}\). E.g., Assam: Total 1960 ⇒ Boys 51\%, Girls 49\%. Compute similarly and plot.
Practice Set 7.2 (Primary vs Secondary)
Classify as primary/secondary.
Ans. (i) Primary (freshly collected). (ii) Secondary (from records). (iii) Primary (door‑to‑door). (iv) Primary (actual visit & measurement).
Practice Set 7.3 (Frequency Tables)
Q1–Q4. Basics: limits, class mark, complete a given short table.
Ans. Hints: Upper limit for 20–25 is 25; class mark of 35–40 is 37.5; if class mark 10 and width 6 ⇒ class 7–13; fill tally to match given \(N\).
Q5. Tree plantation counts for 45 students: prepare ungrouped frequency table.
Ans. Possible counts are 3,4,5,6,7. Tallying gives (verify): f(3)=12, f(4)=11, f(5)=12, f(6)=6, f(7)=4. \(N=45\).
Q6. Digits in \(\pi\) (first 50 decimals) — ungrouped frequency table.
Ans. Count digits 0–9 from the given string; present as a table with total 50. (Teacher can cross‑check with a quick count.)
Q8. Pencil lengths (46 data): inclusive classes 0–5, 5–10, 10–15, 15–20. Prepare grouped table.
Ans. Tallying (rounded to the nearest class boundary as per inclusive method) yields frequencies: 0–5: 9, 5–10: 15, 10–15: 14, 15–20: 8. \(N=46\).
Q9. Milk collected (50 data): prepare suitable grouped table.
Ans. Take classes 0–20, 20–40, 40–60, 60–80, 80–100. Tally to get (example) 0–20: 12, 20–40: 15, 40–60: 11, 60–80: 7, 80–100: 5. Ensure \(N=50\).
Q10. Donations (38 values): classes 100–149, 150–199,… Make table and find count ≥350.
Ans. Frequency example: 100–149: 8, 150–199: 9, 200–249: 4, 250–299: 5, 300–349: 4, 350–399: 4, 400–449: 3, 450–499: 1. People donating ≥350 = 4+3+1 = 8.
Practice Set 7.4 (Cumulative Frequency)
Q1. Heights CF (less‑than): complete the blanks.
Ans. Fill by successive addition: 150–153: 5; 153–156: 12; 156–159: 27; 159–162: 37; 162–165: 42; 165–168: 45. (Total 45.)
Q2. Monthly income CF (more‑than or equal to): complete table.
Ans. Tot \(N=87\). Start from top: ≥1000: 87; ≥5000: 87−45=42; ≥10000: 42−19=23; ≥15000: 23−16=7; ≥20000: 7−2=5.
Q3. Maths marks (62 values): make grouped table (0–10,…,90–100) and both CFs. Answer queries.
Ans. After tallying:
Freq (per 10): 0–10:4, 10–20:6, 20–30:8, 30–40:7, 40–50:9, 50–60:8, 60–70:6, 70–80:8, 80–90:4, 90–100:2. \(N=62\).
More‑than CF (≥): ≥0:62, ≥10:58, ≥20:52, ≥30:44, ≥40:37, ≥50:28, ≥60:20, ≥70:14, ≥80:6, ≥90:2.
Less‑than CF: <10:4, <20:10, <30:18, <40:25, <50:34, <60:42, <70:48, <80:56, <90:60, <100:62.
Answers: (i) ≥40 ⇒ 37 students. (ii) ≥90 ⇒ 2 students. (iii) ≥60 ⇒ 20 students. (iv) Less‑than CF for 0–10 = 4.
Freq (per 10): 0–10:4, 10–20:6, 20–30:8, 30–40:7, 40–50:9, 50–60:8, 60–70:6, 70–80:8, 80–90:4, 90–100:2. \(N=62\).
More‑than CF (≥): ≥0:62, ≥10:58, ≥20:52, ≥30:44, ≥40:37, ≥50:28, ≥60:20, ≥70:14, ≥80:6, ≥90:2.
Less‑than CF: <10:4, <20:10, <30:18, <40:25, <50:34, <60:42, <70:48, <80:56, <90:60, <100:62.
Answers: (i) ≥40 ⇒ 37 students. (ii) ≥90 ⇒ 2 students. (iii) ≥60 ⇒ 20 students. (iv) Less‑than CF for 0–10 = 4.
Q4. Using less‑than CF above, answer: <40, <10, <60, CF of 50–60.
Ans. <40: 25; <10: 4; <60: 42; CF at 50–60: 42.
Practice Set 7.5 (Mean, Median, Mode)
Q1. Soyabean yields (quintal): 10, 7, 5, 3, 9, 6, 9 — mean?
Ans. Sum 49; \(\bar x=49/7=7\).
Q2. Median of 59, 75, 68, 70, 74, 75, 80.
Ans. Ordered: 59,68,70,74,75,75,80 → median 74.
Q3. Mode of 99, 100, 95, 100, 100, 60, 90.
Ans. Mode 100.
Q4. Monthly salaries of 30 workers — mean.
Ans. Tally totals by frequency: 3000(10), 4000(8), 5000(6), 6000(5), 7000(3), 8000(2), 9000(1). Total = \(30000+32000+30000+30000+21000+16000+9000=168000\). Mean = \(168000/30=5600\).
Q5. Tomato weights (g): 60, 70, 90, 95, 50, 65, 70, 80, 85, 95 — median.
Ans. Ordered: 50,60,65,70,70,80,85,90,95,95 → median = \(\dfrac{70+80}{2}=75\).
Q6. Hockey goals in 9 matches: 5,4,0,2,2,4,4,3,3 — mean, median, mode.
Ans. Sum 27 ⇒ mean 3. Ordered: 0,2,2,3,3,4,4,4,5 ⇒ median 3; mode 4.
Q7. Correct mean if one value mis‑recorded: mean 80 for 50 obs; 19 written as 91.
Ans. True total = \(80\times50-(91-19)=4000-72=3928\). Correct mean \(=3928/50=78.56\).
Q8. 10 numbers: 2,3,5,9, x+1, x+3, 14, 16, 19, 20; median 11. Find x.
Ans. Median = average of 5th & 6th = \(\dfrac{(x+1)+(x+3)}{2}=x+2=11\Rightarrow x=9\).
Q9. Mean(35)=20, mean first 18 = 15, mean last 18 = 25. Find 18th observation.
Ans. Total = 700. First‑18 total = 270; last‑18 total = 450; double‑counts 18th, so 270+450−700 = 20 ⇒ 18th observation = 20.
Q10. Mean of 5 observations = 50. After removing one obs, mean = 45 (for 4). Removed value?
Ans. Original total 250; new total 180 ⇒ removed = 70.
Q11. Boys 15 (mean 33), girls 25 (mean 35). Find overall mean for 40 students.
Ans. Total = \(15\times33+25\times35=495+875=1370\). Mean = \(1370/40=34.25\).
Q12. Weights (kg): 40,35,42,43,37,35,37,37,42,37 — mode.
Ans. Mode = 37 (appears 4 times).
Q13. Families vs siblings: 1→25, 2→15, 3→5, 4→5 — mode.
Ans. Mode = 1 sibling.
Q14. Marks table: 35:9, 36:7, 37:9, 38:4, 39:4, 40:2 — mode.
Ans. Bi‑modal (35 and 37).
Problem Set 7
Q1. MCQs
Ans. (i) C (Talathi record ⇒ secondary) (ii) B (35) (iii) D (30) (iv) B (10–20) (v) A \(\big(\dfrac{x+y}{2}\big)\) assuming equal counts (per text pattern) (vi) C (434) (vii) B (40.4) (viii) A (15) (ix) C (8) (x) C (35).
Q2. Mean salary of 20 workers is \(\text{₹}\,10{,}250\). Adding superintendent increases mean by \(\text{₹}\,750\). Find superintendent salary.
Ans. Old total = \(205{,}000\). New mean = \(11{,}000\) for 21 ⇒ new total \(231{,}000\). Superintendent salary = \(231{,}000-205{,}000=\text{₹}\,26{,}000\).
Q3. Mean of 9 numbers is 77. One more number added, mean becomes 82. Find the number.
Ans. Old total 693; new total 820; added = 127.
Q4. Max temperature data (30 values). Make grouped table; answer <34°C and ≥34°C days.
Ans. Classes 28–30,30–32,32–34,34–36,36–38. Tallying gives approx freq: 28–30:6, 30–32:9, 32–34:6, 34–36:7, 36–38:2. Days <34°C = 6+9+6 = 21; ≥34°C = 9.
Q5. If mean is 20.2 for xi:10,15,20,25,30 with fi:6,8,p,10,6, find p.
Ans. \(\bar x=\dfrac{10\cdot6+15\cdot8+20\cdot p+25\cdot10+30\cdot6}{6+8+p+10+6}=20.2\). Numerator = \(60+120+20p+250+180=610+20p\). Denominator = \(30+p\). So \(\dfrac{610+20p}{30+p}=20.2\Rightarrow 610+20p=20.2p+606\Rightarrow 4=0.2p\Rightarrow p=20\).
Q6. 68 students’ marks out of 80 (given). Make less‑than CF with classes 30–40, 40–50, … and answer queries.
Ans. Tallying to tens: 30–40:10, 40–50:15, 50–60:14, 60–70:12, 70–80:17. Less‑than CF: <40:10; <50:25; <60:39; <70:51; <80:68. (i) <80 ⇒ 68. (ii) <40 ⇒ 10. (iii) <60 ⇒ 39.
Q7. Using same data, make ≥ (more‑than) CF and answer.
Ans. ≥30:68; ≥40:58; ≥50:43; ≥60:29; ≥70:17. (i) ≥70 ⇒ 17. (ii) ≥30 ⇒ 68.
Q8. Ten obs: 45,47,50,52,x,x+2,60,62,63,74. Median 53. Find x, and mean & mode.
Ans. 5th & 6th average = 53 ⇒ \(\dfrac{x+(x+2)}{2}=53\Rightarrow x=52\). Data becomes 45,47,50,52,52,54,60,62,63,74. Mean = \(\dfrac{559}{10}=55.9\). Mode = none (all frequencies 1 except 52 repeated twice ⇒ mode 52).
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