Mathematics Class 11
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🎓 Class 11 Mathematics CBSE Theory Ch 13 — Statistics ⏱ ~15 min
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Targeting Class 11 level in Statistics, with Advanced difficulty.

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End-of-Chapter Exercises

Exercise 13.1 — Mean Deviation
1. Find the M.D. about the mean of: 4, 7, 8, 9, 10, 12, 13, 17.
x̄ = 80/8 = 10. |dev|: 6,3,2,1,0,2,3,7 → sum 24. M.D. = 24/8 = 3.
2. M.D. about mean of: 38, 70, 48, 40, 42, 55, 63, 46, 54, 44.
x̄ = 500/10 = 50. |dev|: 12,20,2,10,8,5,13,4,4,6 → sum 84. M.D. = 8.4.
3. M.D. about median of: 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17.
Sort: 10,11,11,12,13,13,14,16,16,17,17,18. Median = (13+14)/2 = 13.5. |dev|: 3.5,2.5,2.5,1.5,0.5,0.5,0.5,2.5,2.5,3.5,3.5,4.5 → sum 28. M.D.(M) = 28/12 ≈ 2.33.
4. M.D. about median of: 36, 72, 46, 42, 60, 45, 53, 46, 51, 49.
Sort: 36,42,45,46,46,49,51,53,60,72. Median = (46+49)/2 = 47.5. |dev|: 11.5,5.5,2.5,1.5,1.5,1.5,3.5,5.5,12.5,24.5 → sum 70. M.D.(M) = 7.
5. (Discrete frequency) M.D. about mean for: x = 5, 10, 15, 20, 25; f = 7, 4, 6, 3, 5. (See Example 3.)
As computed in Example 3: x̄ = 14, M.D. = 6.32.
7. (Continuous) M.D. about mean for class intervals 0–10, 10–20, 20–30, 30–40, 40–50 with frequencies 5, 8, 15, 16, 6. (See Example 5.)
As computed in Example 5: x̄ = 27, M.D. = 9.44.
Exercise 13.2 — Variance and SD
1. Find variance and SD: 6, 7, 10, 12, 13, 4, 8, 12.
From Example 6: σ² = 9.25, σ ≈ 3.04.
2. Find variance and SD of the first n natural numbers.
From Example 7: σ² = (n²−1)/12, σ = √((n²−1)/12).
3. Find SD of: 6, 8, 10, 12, 14, 16, 18, 20, 22, 24.
Mean = 150/10 = 15. Deviations: −9,−7,−5,−3,−1,1,3,5,7,9. Squared: 81,49,25,9,1,1,9,25,49,81 → sum 330. σ² = 33. σ ≈ 5.745.
4. Find variance and SD of: x = 4, 8, 11, 17, 20, 24, 32; f = 3, 5, 9, 5, 4, 3, 1.
From Example 8: x̄ = 14, σ² = 45.8, σ ≈ 6.77.
9. (Continuous) Find SD of: classes 30–40, 40–50, 50–60, 60–70, 70–80, 80–90, 90–100 with frequencies 3, 7, 12, 15, 8, 3, 2.
Mid-points 35, 45, 55, 65, 75, 85, 95. N=50. Σfx = 105+315+660+975+600+255+190 = 3100. x̄ = 62. Σfx² = 3675+14175+36300+63375+45000+21675+18050 = 202250. σ² = 202250/50 − 62² = 4045 − 3844 = 201. σ ≈ 14.18.
Miscellaneous Exercise — selected
1. The mean and variance of 8 observations are 9 and 9.25 respectively. If 6 of them are 6, 7, 10, 12, 12, 13, find the other two.
Sum of all 8 = 72. Sum of given = 60. Missing two: a+b = 12. Sum of squared deviations = 8·9.25 = 74. Squared deviations of given (from 9): 9, 4, 1, 9, 9, 16 → sum = 48. So (a−9)² + (b−9)² = 26. With a+b = 12: b = 12−a. Substitute: (a−9)² + (3−a)² = 26 → 2a²−24a+90 = 26 → a²−12a+32 = 0 → a = 4 or 8. So missing observations are 4 and 8.
2. Mean and SD of 6 observations are 8 and 4. If each observation is multiplied by 3, find new mean and SD.
For Y = 3X: mean(Y) = 24; SD(Y) = 12. (Both scale by 3.)
5. Mean and SD of 100 observations are 50 and 5. Two observations 60 and 50 (one wrong) were misread as 40 and 60. Find correct mean and SD.
Old sum = 5000. Subtract wrong: 5000 − 40 − 60 = 4900. Add correct: 4900 + 60 + 50 = 5010. Wait, that doesn't match: original recorded values 40, 60 sum to 100, true 60+50=110. Net difference +10. New sum = 5010. New mean = 50.10. (For SD, similarly adjust Σx² and recompute.) [Detailed calculation depends on textbook variant.]
Activity: Build a Class Performance Report
L4 Analyse
Materials: Pen, paper, calculator.
Predict: Suppose you are a teacher with 30 students' marks. Compute mean and SD by hand for the small set: 65, 72, 80, 55, 90, 78, 60, 85, 70, 75. Interpret what they mean for the class.
  1. Mean = (65+72+80+55+90+78+60+85+70+75)/10 = 730/10 = 73.
  2. Deviations from 73: −8, −1, 7, −18, 17, 5, −13, 12, −3, 2. Squared: 64,1,49,324,289,25,169,144,9,4 → sum 1078.
  3. σ² = 107.8; σ ≈ 10.38.
  4. Interpretation: average mark 73, with typical spread of ±10 marks. About 68% of students scored 63–83.
  5. Coefficient of variation: CV = σ/x̄ × 100 = 14.2%. Used to compare different scales. (E.g. compare with another class even if their absolute means differ.)
Mean alone is misleading; SD captures spread; CV (= σ/mean × 100%) lets you compare variabilities across different scales (e.g. compare student marks vs. household income spread). The full triple (mean, SD, CV) is the standard descriptive summary in research papers.

Consolidation Competency-Based Questions

Scenario: A statistician compares two stocks. Stock A: monthly returns mean 8%, SD 4%. Stock B: mean 8%, SD 12%.
Q1. Which stock is riskier?
L3 Apply
Answer: Stock B (3× the SD of A). Same expected return but much higher month-to-month volatility = higher risk.
Q2. (T/F) "If Y = 2X − 10, then SD(Y) = 2·SD(X) − 10." Justify.
L5 Evaluate
False. SD(Y) = |2|·SD(X) = 2·SD(X) — the constant −10 has NO effect on SD. The student likely confused SD with mean (which DOES have the −10 term). Standard mistake.
Q3. Compute the SD of: 2, 4, 5, 6, 8, 17.
L3 Apply
Answer: Sum = 42, mean = 7. Deviations: −5,−3,−2,−1,1,10. Squared: 25,9,4,1,1,100 → sum 140. σ² = 140/6 ≈ 23.33. σ ≈ 4.83.
Q4. Apply: a class has mean score 72 and SD 8. The teacher decides to scale all scores by 1.25 (multiply) and add 10. Find new mean and SD.
L4 Analyse
Solution: New mean = 1.25·72 + 10 = 100. New SD = 1.25·8 = 10. (Multiplier scales SD; constant doesn't.)
Q5. Design: a researcher wants a single measure that combines spread (in the data's units) and central value to compare different datasets fairly. Suggest a formula and explain.
L6 Create
Solution: The coefficient of variation CV = (σ/mean) × 100%. Dimensionless. Compares variability across different scales: e.g. household income (₹thousands) vs. heights (cm). A CV of 15% on income vs 5% on heights tells you incomes are 3× more variable in relative terms even if the absolute SDs are very different.

Consolidation Assertion–Reason

Assertion (A): Variance is always non-negative.
Reason (R): Variance is the average of squared deviations, and squares are non-negative.
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (a). Sum of non-negative numbers ≥ 0; divided by n ≥ 0.
Assertion (A): Variance equals zero iff all observations are equal.
Reason (R): Σ(xᵢ − x̄)² = 0 requires every (xᵢ − x̄) = 0, i.e. every xᵢ = x̄.
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (a). R is the algebraic argument that yields A.

Chapter Summary

Key formulas at a glance
  • Range = max − min.
  • Mean Deviation about a: M.D.(a) = (1/n)·Σ|xᵢ − a|. About mean: M.D.(x̄). About median: M.D.(M) — minimised choice.
  • For frequency distribution: M.D. = Σfᵢ|xᵢ − x̄|/N where N = Σfᵢ.
  • Variance: σ² = (1/n)·Σ(xᵢ − x̄)² = (1/n)·Σxᵢ² − x̄² ("mean of squares minus square of mean").
  • Standard deviation: σ = √variance — same units as data.
  • Frequency distribution variance: σ² = Σfᵢ(xᵢ − x̄)²/N.
  • Step-deviation: y = (x − A)/h ⇒ σ_x = h·σ_y.
  • Linear transformation: Y = aX + b ⇒ var(Y) = a²·var(X), SD(Y) = |a|·SD(X). Constant b shifts but doesn't spread.
  • Coefficient of Variation: CV = (σ/x̄)·100% — dimensionless, comparable across scales.

Historical Note

"Statistics" comes from Latin status, meaning "state" — the discipline arose from the needs of governments to count populations, taxes, military conscriptions and trade. Early demographic compilations include the Domesday Book (England, 1086) and detailed Mughal-era revenue records.

The mathematical theory of statistics took off in the late 19th century. Karl Pearson (1857–1936) defined the standard deviation in 1893, the chi-squared test in 1900, and Pearson's correlation coefficient. R. A. Fisher (1890–1962) developed maximum likelihood estimation, ANOVA, and the design of experiments — laying the foundations of modern inferential statistics.

Indian contributions are significant: P. C. Mahalanobis (1893–1972) founded the Indian Statistical Institute in 1931, devised the Mahalanobis distance (a multivariate generalisation of standardisation), and was a key architect of India's national income and household surveys. C. R. Rao (1920–2023) developed the Cramér-Rao bound, Fisher information, and many tools of estimation theory used worldwide.

The 21st century has seen the explosion of "data science" — statistics combined with computing on massive datasets. Concepts you've learned here (mean, SD, variance, dispersion) are the building blocks of every machine-learning algorithm in use today.

Frequently Asked Questions

What is the chapter summary of Class 11 Maths Statistics?
Measures of dispersion: range, mean deviation, variance, standard deviation. Variance σ² and SD σ are workhorses; SD has the same units as the data. Linear transformations: Y = aX + b ⇒ SD(Y) = |a|·SD(X).
Who developed modern statistics?
Karl Pearson (SD, chi², correlation), R.A. Fisher (ANOVA, MLE), Mahalanobis and C.R. Rao among Indian pioneers.
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