Mathematics Class 11
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13.1 Introduction

🎓 Class 11 Mathematics CBSE Theory Ch 13 — Statistics ⏱ ~15 min
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This MCQ module is based on: 13.1 Introduction

This mathematics assessment will be based on: 13.1 Introduction
Targeting Class 11 level in Statistics, with Advanced difficulty.

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13.1 Introduction

You already know measures of central tendency? — mean, median, mode — that summarise "where" data are centred. But two datasets with the same mean can be wildly different in their spread.

Consider Batsman A: scores 30, 91, 0, 64, 42, 80, 30, 5, 117, 71. Mean = 53.
Batsman B: scores 53, 46, 48, 50, 53, 53, 58, 60, 57, 52. Mean = 53.
Same mean, but A is volatile (lots of variation) while B is consistent. To capture this we need a measure of dispersion?.

Karl Pearson

1857 – 1936

English mathematician and biostatistician. Founder of modern statistics. He defined the standard deviation (1893), the correlation coefficient (1896), the chi-squared test (1900), and the term "histogram". His four-volume "Tables for Statisticians and Biometricians" was the standard reference for half a century. The "P" in Pearson's correlation r honours him.

13.2 Measures of Dispersion

Five common measures
  1. Range = max − min. Simplest; ignores all interior data.
  2. Mean deviation = average absolute deviation from the mean (or median).
  3. Variance = average squared deviation from the mean.
  4. Standard deviation = √variance.
  5. Quartile deviation (covered in Class 10) = (Q₃ − Q₁)/2.

Range

\[\text{Range}=\text{Maximum value}-\text{Minimum value}.\]

For Batsman A: range = 117 − 0 = 117. For B: 60 − 46 = 14. Range already reveals A is more dispersed than B. But range uses just two values; we want a measure that uses all the data.

13.3 Mean Deviation

Mean deviation about \(a\)
For \(n\) observations \(x_1, x_2, \ldots, x_n\), the mean deviation about a fixed value \(a\) is \[\boxed{\;\text{M.D.}(a)=\dfrac{1}{n}\sum_{i=1}^{n}|x_i-a|\;}\] The absolute value ensures positive and negative deviations don't cancel (without it, the sum is zero whenever \(a\) is the mean).
About mean vs. about median
\(\text{M.D.}(\bar x)\) is the mean deviation about the mean. \(\text{M.D.}(M)\) is the mean deviation about the median \(M\).

Fact: M.D. is minimised when computed about the median (not the mean). The median is the "L₁-best" centre. The mean is the "L₂-best" centre — and that's exactly why standard deviation (L₂ measure) goes with the mean, while mean deviation pairs naturally with median.

13.3.1 Mean Deviation for Ungrouped Data

Step-by-step
  1. Compute the mean \(\bar x\) (or median \(M\)) of the \(n\) observations.
  2. Compute deviations \(d_i=x_i-\bar x\) (or \(x_i-M\)).
  3. Take absolute values: \(|d_i|\).
  4. Mean of these absolute values: \(\text{M.D.}=\dfrac{1}{n}\sum|d_i|\).

13.3.2 Mean Deviation for Grouped Data

Frequency distribution
Discrete: values \(x_1,x_2,\ldots,x_n\) with frequencies \(f_1,f_2,\ldots,f_n\). Total observations \(N=\sum f_i\). Mean \(\bar x=\dfrac{1}{N}\sum f_i x_i\). \[\text{M.D.}(\bar x)=\dfrac{1}{N}\sum_{i=1}^{n}f_i\,|x_i-\bar x|.\] Continuous: use class mid-points \(x_i\) (and their frequencies \(f_i\)). Same formula.

Worked Examples

Example 1. Find the mean deviation about the mean of: 6, 7, 10, 12, 13, 4, 8, 12.
\(\bar x=(6+7+10+12+13+4+8+12)/8=72/8=9\). Deviations: -3, -2, 1, 3, 4, -5, -1, 3. Absolute values: 3, 2, 1, 3, 4, 5, 1, 3 — sum = 22. \(\text{M.D.}(\bar x)=22/8=2.75\).
Example 2. Find the mean deviation about the median of: 12, 3, 18, 17, 4, 9, 17, 19, 20, 15, 8, 17, 2, 3, 16, 11, 3, 1, 0, 5.
Sort: 0, 1, 2, 3, 3, 3, 4, 5, 8, 9, 11, 12, 15, 16, 17, 17, 17, 18, 19, 20. n = 20 (even). Median = (10th + 11th)/2 = (9 + 11)/2 = 10.
Deviations from 10 (absolute): 10, 9, 8, 7, 7, 7, 6, 5, 2, 1, 1, 2, 5, 6, 7, 7, 7, 8, 9, 10 — sum = 124. M.D.(M) = 124/20 = 6.2.
Example 3. Find the mean deviation about the mean for the discrete distribution: \(x_i = 5, 10, 15, 20, 25\) with \(f_i = 7, 4, 6, 3, 5\).
N = 7+4+6+3+5 = 25. \(\sum f_i x_i = 35+40+90+60+125=350\). \(\bar x=350/25=14\).
\(|x_i-14|\) = 9, 4, 1, 6, 11. \(f_i|x_i-14|\)= 63+16+6+18+55 = 158. M.D. = 158/25 = 6.32.
Example 4. Find the mean deviation about the median for the distribution: 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17. (n=12)
Sort: 10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18. Median = (13 + 14)/2 = 13.5. Absolute deviations from 13.5: 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.
Example 5. (Continuous frequency) Find the mean deviation about the mean for the data: marks 0–10, 10–20, 20–30, 30–40, 40–50 with frequencies 5, 8, 15, 16, 6.
Mid-points \(x_i\) = 5, 15, 25, 35, 45. \(\sum f_i = 50\). \(\sum f_i x_i = 25+120+375+560+270=1350\). \(\bar x=27\).
\(|x_i-27|\) = 22, 12, 2, 8, 18. \(\sum f_i|x_i-\bar x|=110+96+30+128+108=472\). M.D. = 472/50 = 9.44.

13.3.3 Limitations of Mean Deviation

Why mean deviation is rarely used in advanced statistics
  • The absolute-value function \(|x-a|\) is not differentiable at \(x=a\), making calculus manipulations awkward.
  • Mean deviation cannot be combined nicely (e.g. M.D. of a sum ≠ sum of M.D.s).
  • It does not appear in standard sampling distributions, so inferential statistics (confidence intervals, hypothesis tests) is built on standard deviation instead.
For these reasons, Section 13.4 introduces variance and standard deviation, which dominate practical statistics.
Activity: Compare Mean Deviations About Mean and Median
L4 Analyse
Materials: Pen, paper, calculator.
Predict: For data 1, 2, 3, 4, 100, will M.D. about mean be larger or smaller than M.D. about median? Why?
  1. Mean = (1+2+3+4+100)/5 = 22. M.D. about mean = (21+20+19+18+78)/5 = 156/5 = 31.2.
  2. Median = 3 (middle of sorted 1,2,3,4,100). M.D. about median = (2+1+0+1+97)/5 = 101/5 = 20.2.
  3. M.D. about median is SMALLER (20.2 vs 31.2). The outlier 100 pulls the mean far from the bulk of the data, inflating M.D. about mean.
  4. General fact: \(\sum|x_i-a|\) is minimised when \(a\) = median. So M.D. about median ≤ M.D. about mean, always.
The median is the "L₁-optimal" centre — minimises sum of absolute deviations. The mean is the "L₂-optimal" centre — minimises sum of squared deviations. Outliers affect L₂ much more than L₁ (because squaring amplifies large numbers). This is why robust statistics often prefers median-based measures when outliers are suspected.

Competency-Based Questions

Scenario: Two basketball players' point totals over 8 games. Player A: 18, 22, 20, 21, 19, 20, 21, 19. Player B: 5, 35, 12, 28, 22, 10, 30, 18.
Q1. Compute mean for both players.
L3 Apply
Answer: A: total 160, mean 20. B: total 160, mean 20. Same mean!
Q2. Compute M.D. about mean for both — which is more consistent?
L4 Analyse
A: deviations 2,2,0,1,1,0,1,1 → sum 8 → M.D. = 1. B: deviations 15,15,8,8,2,10,10,2 → sum 70 → M.D. = 8.75. Player A is far more consistent.
Q3. (T/F) "Mean deviation about the mean is always the smallest mean deviation." Justify.
L5 Evaluate
False. M.D. is minimised about the MEDIAN, not the mean. Activity above shows this.
Q4. Find M.D. about mean for the discrete distribution: x = 2, 5, 7, 8, 10, 12, 15, 17, 20; f = 2, 4, 5, 9, 11, 7, 5, 4, 3.
L3 Apply
Solution: N=50; Σfx = 4+20+35+72+110+84+75+68+60 = 528. Mean = 528/50 = 10.56. Compute Σf|x−10.56| → ≈ 158. M.D. ≈ 3.16. (Detailed arithmetic left for student practice.)
Q5. Design: a teacher wants to compare three classes' performance variability with a single measure. Why might M.D. be a poor choice for further inference (e.g. comparing class A vs class B with a hypothesis test)?
L6 Create
Solution: M.D. lacks a clean sampling distribution — it doesn't combine with the central limit theorem the way variance does. For inferential statistics (t-tests, ANOVA), we need standard deviation, not M.D. So while M.D. is a fine descriptive summary, it's a dead-end for inference. Hence Part 2's pivot to variance/SD.

Assertion–Reason Questions

Assertion (A): Sum of deviations from the mean is always zero.
Reason (R): By definition of mean, \(\sum(x_i-\bar x)=\sum x_i - n\bar x = n\bar x - n\bar x=0\).
(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 proof of A.
Assertion (A): Mean deviation about median ≤ Mean deviation about mean.
Reason (R): The median minimises Σ|xᵢ − a| over all choices of a.
(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 optimisation property; A is the comparison statement.
Assertion (A): Range is the simplest measure of dispersion.
Reason (R): Range uses only two extreme values and ignores all middle data.
(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). Simplicity comes precisely from using only two values.

Frequently Asked Questions

What is dispersion in statistics?
Dispersion measures how spread out the data are from a central value.
What is range?
Range = max − min. Simplest dispersion measure.
What is mean deviation?
Average of absolute deviations from a fixed value (mean or median).
How do you compute mean deviation for grouped data?
M.D.(x̄) = Σ fᵢ |xᵢ − x̄| / Σ fᵢ.
Why use absolute values in mean deviation?
Without them, deviations from the mean sum to zero (positives cancel negatives).
What are the limitations of mean deviation?
Non-differentiable absolute value, no clean sampling theory — limits inferential use.
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