CHAPTER 12 – STATISTICS (Class 9, NCERT/CBSE)
Part A — 1-Mark Questions (20) with Solutions
Q1. What is a bar graph?
A bar graph is a pictorial display using bars of equal width with equal gaps; bar heights are proportional to values.
Q2. When do we use a histogram instead of a bar graph?
For grouped frequency distributions with continuous classes (no gaps between bars).
Q3. Define class-mark (mid-point) for a class \(a\!-\!b\).
\( \dfrac{a+b}{2}\).
Q4. What is frequency density when class widths vary?
Frequency density \(=\dfrac{\text{frequency}}{\text{class width}}\). Rectangle height in such histograms is proportional to this.
Q5. Why is there no gap between bars in a histogram?
Because classes are continuous and cover every value in the range.
Q6. Name three graphical representations you learned.
Bar graphs, Histograms, Frequency polygons.
Q7. Convert 40–50 to a continuous class if previous class is 30–39.
Make equal adjustments: \(39.5\!-\!49.5\).
Q8. What points are joined to draw a frequency polygon from a histogram?
Mid-points of the tops of adjacent histogram rectangles (and two extra zeros at ends).
Q9. What does the area of a histogram rectangle represent?
It is proportional to the class frequency.
Q10. If a class is 10–20 with frequency 8, what is class-mark?
\( \dfrac{10+20}{2}=15\).
Q11. What is a ‘kink’ (break) on an axis?
A symbol showing that the scale starts after skipping a portion (e.g., classes not beginning at 0).
Q12. Which axis usually shows the variable in bar graphs?
Usually the horizontal (x-) axis shows categories; vertical (y-) axis shows values.
Q13. Write the formula to adjust unequal class width to an equivalent width \(w_0\).
Adjusted height \(= \text{frequency}\times\dfrac{w_0}{\text{actual width}}\).
Q14. State one use of a frequency polygon.
To compare two distributions on the same axes neatly.
Q15. What is a grouped frequency distribution?
Data arranged into non-overlapping class intervals with corresponding frequencies.
Q16. Identify the class width of 55.5–60.5.
\(60.5-55.5=5\).
Q17. For class 0–20, frequency 7, what is adjusted height for width \(10\)?
\(7\times \dfrac{10}{20}=3.5\).
Q18. What is meant by ‘continuous classes’?
Classes with adjoining boundaries matching (e.g., 10–20 and 20–30) or with boundary corrections like 19.5–29.5.
Q19. Why add two ‘zero-frequency’ classes when finishing a frequency polygon?
To close the polygon and keep its area equivalent to the histogram’s area.
Q20. Can a bar graph be drawn for qualitative categories?
Yes (e.g., favourite colour), unlike histograms which need quantitative classes.
Part B — 2-Mark Questions (20) with Solutions
Q1. Find class-marks for classes 30–40, 40–50, 50–60.
They are \(35,\ 45,\ 55\) respectively.
Q2. Convert the classes 5–9, 10–14, 15–19 into continuous classes.
Subtract \(0.5\) from lower limits and add \(0.5\) to upper limits: \(4.5\!-\!9.5,\ 9.5\!-\!14.5,\ 14.5\!-\!19.5\).
Q3. A class 0–20 has width 20 and frequency 8. What height should you draw if the standard width is 10?
Adjusted height \(=8\times \dfrac{10}{20}=4\).
Q4. For the table: 10–20:4, 20–30:6, 30–50:10. Compute adjusted heights for width \(10\).
Heights: \(4, 6, 10\times \frac{10}{20}=5\).
Q5. Why is a histogram unsuitable for discrete categorical data?
Because histogram implies continuity; discrete categories require separated bars (bar graph).
Q6. Explain why equal gaps are kept between bars in bar graphs but not in histograms.
Gaps indicate discrete categories; no gaps in histograms indicate continuous numerical ranges.
Q7. From classes 100–110, 110–130, 130–140 with frequencies 5, 10, 5, find class-marks and plot points for a polygon (just coordinates).
Class-marks: \(105, 120, 135\). Points: \((105,5),(120,10),(135,5)\) plus two zeros at \(95,145\).
Q8. The first class is 0–10. Where do you plot the left zero point for the polygon?
At the class-mark of \((-10)\!-\!0\), i.e., \(-5\) on the x-axis with frequency \(0\).
Q9. Why should scales be mentioned under a drawn graph?
To interpret bar heights/widths correctly and avoid misreading values.
Q10. If classes are 0–20, 20–40 (freq 6, 9), draw the frequency polygon coordinates.
Marks: \(10,30\). Points: \((10,6),(30,9)\) with zeros at \((-10,0)\) and \((50,0)\).
Q11. You have a histogram with unequal widths; what stays proportional to frequency?
Rectangle area (height × width), so height must represent frequency density.
Q12. Mention one advantage of polygons over histograms.
Two or more distributions can be compared clearly on the same axes.
Q13. If you start a histogram at 30.5 instead of 0, how do you show that on the axis?
Use a kink/break near the origin on the horizontal axis.
Q14. Explain ‘making classes continuous’ with 20–29, 30–39.
Write as \(19.5\!-\!29.5,\ 29.5\!-\!39.5\) so adjacent boundaries match without gaps.
Q15. For 6 classes of equal width, how many vertices does a frequency polygon have (including zeros)?
\(6+2=8\) vertices.
Q16. Given 0–10: 7, 10–20: 12, 20–30: 9. Which class has the highest frequency?
10–20 (frequency 12).
Q17. State two essential labels every statistical graph must have.
Title; labelled axes with units (and scale).
Q18. If a class width halves while frequency stays same, what happens to frequency density?
It doubles: \( \text{FD}=\dfrac{f}{w}\).
Q19. Provide the boundary-corrected form of 145–153.
\(144.5\!-\!153.5\).
Q20. A frequency polygon passes through \((25,8)\) and \((35,12)\). What are the corresponding class-marks?
25 and 35 are class-marks (mid-points) of those classes.
Part C — 3-Mark Questions (20) with Solutions
Q1. Convert the following to continuous classes and compute class-marks: 10–14:6, 15–19:8, 20–24:10.
Continuous: \(9.5\!-\!14.5,\ 14.5\!-\!19.5,\ 19.5\!-\!24.5\). Class-marks: \(12,\ 17,\ 22\).
Q2. Create adjusted heights for width \(w_0=5\): 0–10:7, 10–15:9, 15–25:14.
Widths: 10, 5, 10. Heights: \(7\!\times\! \frac{5}{10}=3.5,\ 9\!\times\!\frac{5}{5}=9,\ 14\!\times\!\frac{5}{10}=7\).
Q3. The classes 30.5–35.5:9, 35.5–40.5:6, 40.5–45.5:15 are given. List the coordinates for a frequency polygon.
Class-marks: \(33,\ 38,\ 43\). Points: \((33,9),(38,6),(43,15)\) plus zeros at \((28,0)\) and \((48,0)\).
Q4. Explain why the histogram in which widths vary must use frequency density and not raw frequencies for heights.
Area \(=\) height × width must be ∝ frequency. If width varies, taking height = frequency would distort areas; using \( \text{FD}=\dfrac{f}{w} \) keeps area \(=\) FD×width \(=f\).
Q5. Data: 0–20:7, 20–30:10, 30–40:10, 40–50:20, 50–60:20, 60–70:15, 70–100:8. Compute adjusted heights for base width 10.
Heights: \(3.5,10,10,20,20,15, 8\times \frac{10}{30}=2.\overline{6}\).
Q6. A teacher recorded weekly costs of living indices: 140–150:5, 150–160:10, 160–170:20, 170–180:9, 180–190:6, 190–200:2. Find class-marks and list polygon points.
Class-marks: \(145,155,165,175,185,195\). Points: \((145,5),(155,10),(165,20),(175,9),(185,6),(195,2)\) with zeros at \(135\) and \(205\).
Q7. Make the following classes continuous and indicate the ‘kink’ position on x-axis: 118–126, 127–135, 136–144,…
Continuous: \(117.5\!-\!126.5,\ 126.5\!-\!135.5,\ 135.5\!-\!144.5,\dots\). Since they start at \(117.5\ne 0\), place a kink before \(117.5\) on x-axis.
Q8. Two sections have marks (0–10: A=3,B=5), (10–20: A=9,B=19), (20–30: A=17,B=15), (30–40: A=12,B=10), (40–50: A=9,B=1). Compare performances.
Section B has more students in lower classes (0–20), Section A has more in higher classes (20–50). Hence A’s overall performance is better.
Q9. For age groups 1–2:5, 2–3:3, 3–5:6, 5–7:12, 7–10:9, 10–15:10, 15–17:4, compute frequency densities.
Widths: \(1,1,2,2,3,5,2\). FD: \(5,3,3,6,3,2,2\) respectively (units per year).
Q10. In a histogram with class width 5 throughout, what is the relation between rectangle height and frequency?
Heights are directly proportional to frequencies (since width is constant).
Q11. The class 55.5–60.5 has frequency 2. What is its class-mark and boundary type?
Class-mark \(=58\). This is a continuous-type class with real boundaries.
Q12. Explain why the area under a frequency polygon equals the area of its histogram.
Each histogram rectangle can be split into two congruent triangles that assemble to the polygon’s trapeziums; areas match pairwise.
Q13. Correct the statement: “Maximum number of leaves are 153 mm long” given the modal class 145–153 has frequency 12.
Incorrect. Frequency refers to the entire class 144.5–153.5 (after correction), not exactly 153 mm.
Q14. Compute total frequency from the lamp lifetime table: 300–400:14, 400–500:56, 500–600:60, 600–700:86, 700–800:74, 800–900:62, 900–1000:48.
Total \(=14+56+60+86+74+62+48=400\).
Q15. From Q14, how many lamps have life > 700 hours?
\(74+62+48=184\) lamps.
Q16. For surnames classes 1–4:6, 4–6:30, 6–8:44, 8–12:16, 12–20:4, which class contains most surnames?
Class 6–8 (frequency 44).
Q17. State two differences between bar graphs and histograms.
Bar graphs for discrete/categorical data with gaps; histograms for continuous numeric classes with no gaps and areas ∝ frequencies.
Q18. Give one reason pictures/graphs are preferred to long tables.
They allow quick visual comparison and trend spotting at a glance.
Q19. If all class widths are doubled but frequencies unchanged, how must heights change to represent the same histogram areas?
Heights must be halved (since area = height × width must stay ∝ frequency).
Q20. Why are class intervals written as 30–40, 40–50 (and not overlapping)?
To avoid duplication; boundary equality assigns a value to exactly one class after boundary correction.
Exercise 12.1 — Perfect Solutions
| Question | Solution (MathJax) |
|---|---|
| 1. Female fatality causes data (percentages). (i) Graphically represent (ii) major cause (iii) two factors. |
(i) Draw a bar graph with causes on x-axis and % on y-axis (scale e.g., 1 cm = 5%). (ii) Major cause: Reproductive health conditions (31.8%). (iii) Possible factors (illustrative): inadequate maternal health care & nutrition; lack of access to timely medical facilities & awareness. |
| 2. Girls per 1000 boys in various sections. (i) Bar graph (ii) conclusions. |
(i) Bars for each section with values: SC 940, ST 970, Non SC/ST 920, Backward 950, Non-backward 920, Rural 930, Urban 910. (ii) Conclusions: Highest in ST (970); lowest in Urban (910). Backward districts (950) > Non-backward (920). Urban–rural gap present (910 vs 930). |
| 3. Seats won: A 75, B 55, C 37, D 29, E 10, F 37. (i) Bar graph (ii) winner? | (i) Draw bars for each party. (ii) Maximum seats: Party A (75). |
| 4. Leaf lengths (mm): 118–126:3, 127–135:5, 136–144:9, 145–153:12, 154–162:5, 163–171:4, 172–180:2. |
(i) Make classes continuous: \(117.5\!-\!126.5,\ 126.5\!-\!135.5,\ 135.5\!-\!144.5,\ 144.5\!-\!153.5,\ 153.5\!-\!162.5,\ 162.5\!-\!171.5,\ 171.5\!-\!180.5\). Draw histogram (equal width). (ii) Another suitable graph: Frequency polygon (or ogive if cumulative, though not required here). (iii) Not correct to claim “maximum leaves are exactly 153 mm”; the highest class is 144.5–153.5 (frequency 12), not a single exact value. |
| 5. Neon lamp lifetimes (hours): 300–400:14, 400–500:56, 500–600:60, 600–700:86, 700–800:74, 800–900:62, 900–1000:48. |
(i) Histogram with equal widths (100 hours). (ii) Lamps with life > 700 hours \(=\ 74+62+48=184\). |
| 6. Two sections A & B (marks 0–50 given). Draw frequency polygons and compare. | Make classes continuous if needed; plot class-marks vs frequencies for both on same axes. Comparison: Section B has more students in 0–20; Section A has more in 20–50. Overall, Section A performs better. |
| 7. Runs in first 60 balls (teams A & B). Draw frequency polygons (make classes continuous). | Convert intervals to continuous (e.g., 1–6 → 0.5–6.5, etc.), compute class-marks (3.5, 9.5, …, 57.5), plot A and B on same axes and join. Visual comparison shows phases of dominance across overs. |
| 8. Park age groups (yrs) & children: 1–2:5, 2–3:3, 3–5:6, 5–7:12, 7–10:9, 10–15:10, 15–17:4. Draw histogram. | Unequal widths ⇒ use frequency density. Widths \(=\ 1,1,2,2,3,5,2\). FD \(=\ 5,3,3,6,3,2,2\). Plot rectangles with base = class width and height = FD. |
| 9. Surnames by letters: 1–4:6, 4–6:30, 6–8:44, 8–12:16, 12–20:4. | (i) Histogram; widths vary (3,2,2,4,8), so use frequency density if comparing heights strictly. (ii) Class with maximum surnames: 6–8 (44). |
Reminder: In all histograms with unequal class widths, set a base width (e.g., \(w_0=2\) or \(10\)) and plot heights as \( \text{frequency}\times \dfrac{w_0}{\text{actual width}} \) so rectangle areas remain proportional to frequencies.
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