Class 11 : Maths (In English) – Lesson 13. Statistics
EXPLANATION & SUMMARY
🔷 Explanation (~1700 words)
📘 Introduction
Statistics is the branch of mathematics that deals with collection, organization, analysis, and interpretation of numerical data. In this chapter, we study measures of dispersion, which describe how data values are spread around a central value (like mean or median).
🔵 1. Measures of Dispersion
Dispersion means the extent to which data values differ from the central tendency. It shows variability.
🧠 Types of measures of dispersion:
Range
Mean Deviation
Variance
Standard Deviation
These measures help us compare two or more data sets having the same average.
🟢 2. Range (R)
The simplest measure of dispersion.
➡️ Formula: R = Maximum value − Minimum value
➡️ Coefficient of Range = (Max − Min) / (Max + Min)
💡 Example:
Data: 3, 7, 9, 10, 13
R = 13 − 3 = 10
✔️ Used when: Data is small or rough comparison is needed.
🟡 3. Mean Deviation (M.D.)
Mean deviation shows average of absolute deviations from a central value (Mean, Median or Mode).
Let the central value be A (mean, median, or mode).
➡️ Formula for Ungrouped Data:
M.D. = (1/n) Σ |xᵢ − A|
➡️ For Discrete Frequency Data:
M.D. = (Σ fᵢ |xᵢ − A|) / (Σ fᵢ)
➡️ For Continuous Data:
M.D. = (Σ fᵢ |xᵢ − A|) / N, where xᵢ = class mark.
✏️ Note: Generally, mean deviation is taken from mean or median.
💡 Example:
Data: 2, 4, 6, 8, 10
Mean = 6
M.D. = (|2−6| + |4−6| + |6−6| + |8−6| + |10−6|)/5 = (4+2+0+2+4)/5 = 12/5 = 2.4
✔️ Smaller M.D. → less dispersion.
🔴 4. Variance and Standard Deviation (σ)
Most reliable measures of dispersion.
➡️ Variance (σ²) = Mean of squares of deviations from the mean.
➡️ Standard Deviation (σ) = √(Variance)
🧠 Ungrouped Data:
σ = √[ (Σ (xᵢ − x̄)²) / n ]
🧠 Discrete Frequency Distribution:
σ = √[ (Σ fᵢ (xᵢ − x̄)²) / Σ fᵢ ]
🧠 Continuous Frequency Distribution:
σ = √[ (Σ fᵢ (xᵢ − x̄)²) / N ], xᵢ = class marks
💡 Alternate (Step-Deviation) Formula:
σ = √[ (Σ fᵢ dᵢ²)/N − (Σ fᵢ dᵢ / N)² ] × h
where dᵢ = (xᵢ − A)/h
✔️ Standard deviation has same unit as data.
✔️ Small σ means data values are close to mean.
🟢 5. Coefficient of Variation (C.V.)
Used to compare consistency of data.
➡️ C.V. = (σ / Mean) × 100
✔️ Lower C.V. → more consistent data.
💡 Example:
Dataset 1: Mean = 50, σ = 5 → C.V. = 10%
Dataset 2: Mean = 80, σ = 12 → C.V. = 15%
Hence, Dataset 1 is more consistent.
🟡 6. Step-Deviation Method
Used for simplifying computation when class marks are large.
Let class marks xᵢ, assumed mean A, class width h, and
dᵢ = (xᵢ − A)/h
➡️ Mean: x̄ = A + h (Σ fᵢ dᵢ / N)
➡️ σ = h √[ (Σ fᵢ dᵢ²)/N − (Σ fᵢ dᵢ / N)² ]
✔️ Reduces calculation errors.
🔵 7. Comparison of Two Series
To compare variability of two data sets having different means:
Use Coefficient of Variation.
💡 Example:
Series A: Mean = 60, σ = 6 → C.V. = 10%
Series B: Mean = 70, σ = 10 → C.V. = 14.3%
✅ Series A is more consistent.
🟢 8. Effect of Change of Origin and Scale
Let y = (x − a)/b
➡️ σᵧ = σₓ / |b|
➡️ σ is independent of change of origin, but affected by scale.
✏️ Note:
Adding/subtracting a constant → No effect on σ
Multiplying/dividing by constant → σ multiplied/divided by same constant.
🔴 9. Properties of Standard Deviation
✔️ σ ≥ 0
✔️ σ = 0 if all observations are same
✔️ Units same as data
✔️ Less affected by extreme values than range
✔️ Useful for comparing variability.
🔵 Point 10: Summary Table
Measure – Formula – Remarks
Range – L − S – Simple but less reliable
Mean Deviation (M.D.) – (Σ |xᵢ − A|) / n – Uses absolute deviations
Variance – (Σ (xᵢ − x̄)²) / n – Based on squared deviations
Standard Deviation (S.D.) – √Variance – Most reliable and commonly used
Coefficient of Variation (C.V.) – (σ / x̄) × 100 – Relative measure for comparing consistency
💡 Applications
✔️ Economics — price fluctuations
✔️ Meteorology — temperature variability
✔️ Quality control
✔️ Comparing exam results
✔️ Research studies
✏️ Notes:
🔹 Use mean as central value when frequency distribution is regular.
🔹 Use median for skewed data.
🔹 Standard deviation preferred in most scientific fields.
🌿 Worked Example (Grouped Data)
Classes: 0–10, 10–20, 20–30, 30–40
Frequencies: 5, 8, 12, 5
Step 1: Find class marks: 5, 15, 25, 35
Step 2: Assume A = 20, h = 10
dᵢ = (xᵢ − A)/h → −1.5, −0.5, 0.5, 1.5
Compute Σ fᵢ dᵢ, Σ fᵢ dᵢ²
Apply σ = h √[ (Σ fᵢ dᵢ²)/N − (Σ fᵢ dᵢ / N)² ]
🔶 Advantages of Using S.D.
✔️ Based on all observations
✔️ Rigidly defined
✔️ Suitable for further algebraic treatment
🔷 Limitations
❌ Difficult to compute manually for large data
❌ Influenced by extreme values
❌ Harder to explain intuitively
Summary (~300 words)
Measures of dispersion describe spread of data around a central value.
🔹 Range = Max − Min (simple, rough)
🔹 Mean Deviation = average of absolute deviations
🔹 Variance = average of squared deviations
🔹 Standard Deviation (σ) = √(variance), most reliable
🔹 Coefficient of Variation = (σ / Mean) × 100, used for comparative consistency
For grouped data, use formulas:
σ = √(Σ fᵢ (xᵢ − x̄)² / N)
Step-Deviation: σ = h √[ (Σ fᵢ dᵢ²)/N − (Σ fᵢ dᵢ / N)² ]
Properties:
✔️ σ ≥ 0, σ = 0 if all values equal
✔️ Independent of origin, affected by scale
✔️ Smaller σ → more uniform data
Applications:
Used in economics, meteorology, quality control, and research for comparing datasets.
Conclusion:
Among all measures, standard deviation is best because it uses all data points and allows further mathematical analysis.
📝 Quick Recap:
✅ Range = L − S
✅ Mean Deviation = (Σ |xᵢ − A|)/n
✅ Variance = (Σ (xᵢ − x̄)²)/n
✅ Standard Deviation = √Variance
✅ C.V. = (σ / Mean) × 100
✅ σ unaffected by origin, affected by scale
✔️ Smaller σ / C.V. ⇒ More Consistency
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QUESTIONS FROM TEXTBOOK
📄 Exercise 13.1
🧠 Formulae Recap
🔹 Mean Deviation (about Mean) → MDₘₑₐₙ = Σ|xᵢ − x̄| / n
🔹 Mean Deviation (about Median) → MDₘₑdᵢₐₙ = Σ|xᵢ − M| / n
🔹 For frequency distribution → MD = Σfᵢ|xᵢ − A| / Σfᵢ, where A is Mean or Median
🔵 Question 1:
Find the mean deviation about the mean for
4, 7, 8, 9, 10, 12, 13, 17
🟢 Answer:
✳️ Step 1 ➤ n = 8
✳️ Step 2 ➤ Σx = 4 + 7 + 8 + 9 + 10 + 12 + 13 + 17 = 80
➡️ Mean (x̄) = 80 ÷ 8 = 10
✳️ Step 3 ➤ |xᵢ − x̄| = |4−10|=6, |7−10|=3, |8−10|=2, |9−10|=1, |10−10|=0, |12−10|=2, |13−10|=3, |17−10|=7
➡️ Σ|xᵢ − x̄| = 24
✳️ Step 4 ➤ Mean Deviation = 24 ÷ 8 = 3
✔️ Final Answer: 3
🔵 Question 2:
Find mean deviation about mean for
38, 70, 48, 40, 42, 55, 63, 46, 54, 44
🟢 Answer:
✳️ Step 1 ➤ n = 10
✳️ Step 2 ➤ Σx = 500
➡️ Mean (x̄) = 500 ÷ 10 = 50
✳️ Step 3 ➤ |xᵢ − x̄| = 12, 20, 2, 10, 8, 5, 13, 4, 4, 6
➡️ Σ|xᵢ − x̄| = 84
✳️ Step 4 ➤ MD = 84 ÷ 10 = 8.4
✔️ Final Answer: 8.4
🔵 Question 3:
Find mean deviation about median for
13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17
🟢 Answer:
✳️ Step 1 ➤ Arrange: 10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18
➡️ n = 12
➡️ Median = average of 6th & 7th = (13 + 14)/2 = 13.5
✳️ Step 2 ➤ |xᵢ − M| = 3.5, 2.5, 2.5, 1.5, 0.5, 0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5
➡️ Σ|xᵢ − M| = 28.5
✳️ Step 3 ➤ MD = 28.5 ÷ 12 = 2.38
✔️ Final Answer: 2.38
🔵 Question 4:
Find mean deviation about median for
36, 72, 46, 42, 60, 45, 53, 46, 51, 49
🟢 Answer:
✳️ Step 1 ➤ Arrange: 36, 42, 45, 46, 46, 49, 51, 53, 60, 72
➡️ Median = (5th + 6th)/2 = (46 + 49)/2 = 47.5
✳️ Step 2 ➤ |xᵢ − M| = 11.5, 5.5, 2.5, 1.5, 1.5, 1.5, 3.5, 5.5, 12.5, 24.5
➡️ Σ|xᵢ − M| = 70
✳️ Step 3 ➤ MD = 70 ÷ 10 = 7
✔️ Final Answer: 7
🔵 Question 5:
Find mean deviation about mean for
xᵢ: 5, 10, 15, 20, 25
fᵢ: 7, 4, 3, 3, 5
🟢 Answer:
✳️ Step 1 ➤ Σfᵢ = 22
✳️ Step 2 ➤ Σfᵢxᵢ = 5×7 + 10×4 + 15×3 + 20×3 + 25×5 = 305
➡️ Mean x̄ = 305 ÷ 22 = 13.86
✳️ Step 3 ➤ |xᵢ − x̄| = 8.86, 3.86, 1.14, 6.14, 11.14
Multiply:
7×8.86=62.02, 4×3.86=15.44, 3×1.14=3.42, 3×6.14=18.42, 5×11.14=55.70
➡️ Σfᵢ|xᵢ − x̄| = 155
✳️ Step 4 ➤ MD = 155 ÷ 22 = 7.05
✔️ Final Answer: 7.05
🔵 Question 6:
xᵢ: 10, 30, 50, 70, 90
fᵢ: 24, 28, 16, 16, 8
Find mean deviation about mean
🟢 Answer:
✳️ Step 1 ➤ Σfᵢ = 92
✳️ Step 2 ➤ Σfᵢxᵢ = 3720
➡️ Mean x̄ = 3720 ÷ 92 = 40.43
✳️ Step 3 ➤ |xᵢ − x̄| = 30.43, 10.43, 9.57, 29.57, 49.57
Multiply:
24×30.43=730.32, 28×10.43=292.04, 16×9.57=153.12, 16×29.57=473.12, 8×49.57=396.56
➡️ Σfᵢ|xᵢ − x̄| = 2045.16
✳️ Step 4 ➤ MD = 2045.16 ÷ 92 = 22.23
✔️ Final Answer: 22.23
🔵 Question 7:
xᵢ: 5, 7, 9, 10, 12, 15
fᵢ: 8, 6, 2, 2, 4, 6
Find MD about median
🟢 Answer:
✳️ Step 1 ➤ Σfᵢ = 28
✳️ Step 2 ➤ c.f.: 8, 14, 16, 18, 22, 28
➡️ Median = value at (Σfᵢ)/2 = 14 → corresponding x = 7
✳️ Step 3 ➤ |xᵢ − M| = 2, 0, 2, 3, 5, 8
fᵢ|xᵢ − M| = 16 + 0 + 4 + 6 + 20 + 48 = 94
✳️ Step 4 ➤ MD = 94 ÷ 28 = 3.36
✔️ Final Answer: 3.36
🔵 Question 8:
xᵢ: 15, 21, 27, 30, 35
fᵢ: 3, 6, 7, 8, 5
Find MD about median
🟢 Answer:
✳️ Step 1 ➤ Σfᵢ = 29
✳️ Step 2 ➤ c.f.: 3, 9, 16, 24, 29
➡️ Median = position (29 ÷ 2) = 14.5 → corresponding x = 27
✳️ Step 3 ➤ |xᵢ − M| = 12, 6, 0, 3, 8
fᵢ|xᵢ − M| = 3×12 + 6×6 + 7×0 + 8×3 + 5×8 = 36 + 36 + 0 + 24 + 40 = 136
✳️ Step 4 ➤ MD = 136 ÷ 29 = 4.69
✔️ Final Answer: 4.69
🔵 Question 9: Find the mean deviation about the mean.
*Income per day in ₹ (0–100, 100–200, 200–300, 300–400, 400–500, 500–600, 600–700, 700–800); Number of persons (4, 8, 9, 10, 7, 5, 4, 3).
🟢 Answer (about Mean):
✳️ Step 1 ➤ Class marks mᵢ = 50, 150, 250, 350, 450, 550, 650, 750
✳️ Step 2 ➤ Total frequency Σfᵢ = 50
✳️ Step 3 ➤ Mean x̄ = Σ(fᵢmᵢ)/Σfᵢ = 17900/50 = 358
✳️ Step 4 ➤ Compute |mᵢ − x̄| and multiply by fᵢ, then sum: Σ fᵢ|mᵢ − x̄| = 7896
✳️ Step 5 ➤ MD (about mean) = Σ fᵢ|mᵢ − x̄| / Σfᵢ = 7896/50 = 157.92
✔️ Final: Mean deviation ≈ 157.92 ₹
🔵 Question 10: Find the mean deviation about the mean.
*Height in cm (95–105, 105–115, 115–125, 125–135, 135–145, 145–155); Number of boys (9, 13, 26, 30, 12, 10).
🟢 Answer (about Mean):
✳️ Step 1 ➤ Class marks mᵢ = 100, 110, 120, 130, 140, 150
✳️ Step 2 ➤ Σfᵢ = 100
✳️ Step 3 ➤ x̄ = Σ(fᵢmᵢ)/Σfᵢ = 12530/100 = 125.3 cm
✳️ Step 4 ➤ Σ fᵢ|mᵢ − x̄| = 1128.8
✳️ Step 5 ➤ MD = 1128.8/100 = 11.288 cm
✔️ Final: Mean deviation ≈ 11.29 cm
🔵 Question 11: Find the mean deviation about the median for the following data.
*Marks (0–10, 10–20, 20–30, 30–40, 40–50, 50–60); Number of girls (6, 8, 14, 16, 4, 2).
🟢 Answer (about Median):
✳️ Step 1 ➤ Σfᵢ = 50; Cumulative freq = 6, 14, 28, 44, 48, 50
✳️ Step 2 ➤ Median class = 20–30 (contains the 25th item)
✳️ Step 3 ➤ Use median formula (continuous):
➡️ L = 20, h = 10, fₘ = 14, c.f. before = 14
➡️ Median M = L + [(N/2 − c.f.)/fₘ]·h = 20 + [(25 − 14)/14]·10 = 27.857… ≈ 27.857
✳️ Step 4 ➤ Class marks mᵢ = 5, 15, 25, 35, 45, 55
✳️ Step 5 ➤ Σ fᵢ|mᵢ − M| = 517.142…
✳️ Step 6 ➤ MD (about median) = 517.142… / 50 = 10.342…
✔️ Final: Mean deviation ≈ 10.34 marks
🔵 Question 12: Calculate the mean deviation about median age (N = 100 persons).
Age (years): 16–20, 21–25, 26–30, 31–35, 36–40, 41–45, 46–50, 51–55; Number: 5, 6, 12, 14, 26, 12, 16, 9.
(Use class-boundary correction: subtract 0.5 from lower, add 0.5 to upper.)
🟢 Answer (about Median):
✳️ Step 1 ➤ Corrected boundaries: 15.5–20.5, 20.5–25.5, …, 50.5–55.5
✳️ Step 2 ➤ Σfᵢ = 100; cumulative freq = 5, 11, 23, 37, 63, 75, 91, 100
✳️ Step 3 ➤ Median class = 36–40 (since c.f. crosses 50 here)
➡️ L = 35.5, h = 5, fₘ = 26, c.f. before = 37
➡️ Median M = 35.5 + [(50 − 37)/26]·5 = 35.5 + 2.5 = 38.0
✳️ Step 4 ➤ Class marks mᵢ = 18, 23, 28, 33, 38, 43, 48, 53
✳️ Step 5 ➤ Σ fᵢ|mᵢ − M| = 735
✳️ Step 6 ➤ MD (about median) = 735/100 = 7.35 years
✔️ Final: Mean deviation = 7.35 years
📄 Exercise 13.2
🧠 Formulae Recap
🔹 Mean (x̄) = Σxᵢ / n or Σ(fᵢxᵢ) / Σfᵢ
🔹 Variance (σ²) = [Σ(xᵢ − x̄)²] / n or [Σfᵢ(xᵢ − x̄)²] / Σfᵢ
🔹 Standard Deviation (σ) = √Variance
🔵 Question 1:
Data: 6, 7, 10, 12, 13, 4, 8, 12
🟢 Answer:
✳️ Step 1 ➤ n = 8
✳️ Step 2 ➤ Σx = 6 + 7 + 10 + 12 + 13 + 4 + 8 + 12 = 72
➡️ Mean x̄ = 72 ÷ 8 = 9
✳️ Step 3 ➤ Compute (xᵢ − x̄)²:
(6−9)²=9, (7−9)²=4, (10−9)²=1, (12−9)²=9, (13−9)²=16, (4−9)²=25, (8−9)²=1, (12−9)²=9
Σ(xᵢ − x̄)² = 74
✳️ Step 4 ➤ Variance = 74 ÷ 8 = 9.25
✳️ Step 5 ➤ S.D. = √9.25 = 3.04
✔️ Final: Mean = 9, Variance = 9.25, SD = 3.04
🔵 Question 2:
First n natural numbers: 1, 2, 3, …, n
🟢 Answer:
✳️ Step 1 ➤ Mean = (n + 1)/2
✳️ Step 2 ➤ Variance = [(n² − 1)/12]
✔️ Final:
Mean = (n + 1)/2, Variance = (n² − 1)/12, SD = √[(n² − 1)/12]
🔵 Question 3:
First 10 multiples of 3: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30
🟢 Answer:
✳️ Step 1 ➤ n = 10
✳️ Step 2 ➤ Σx = 3(1 + 2 + … + 10) = 3 × 55 = 165
➡️ Mean x̄ = 165 ÷ 10 = 16.5
✳️ Step 3 ➤ Variance for multiples of constant ‘a’:
If xᵢ = 3 × yᵢ → Variance = 3² × Var(yᵢ)
For 1–10, Var = (n² − 1)/12 = (100 − 1)/12 = 99/12 = 8.25
So Var(x) = 9 × 8.25 = 74.25
SD = √74.25 = 8.62
✔️ Final: Mean = 16.5, Variance = 74.25, SD = 8.62
🔵 Question 4:
xᵢ: 6, 10, 14, 18, 24, 28, 30
fᵢ: 2, 4, 7, 12, 8, 4, 3
🟢 Answer:
✳️ Step 1 ➤ Σf = 40
✳️ Step 2 ➤ Σfx = 6×2 + 10×4 + 14×7 + 18×12 + 24×8 + 28×4 + 30×3
= 12 + 40 + 98 + 216 + 192 + 112 + 90 = 760
➡️ x̄ = 760 ÷ 40 = 19
✳️ Step 3 ➤ Compute (xᵢ − x̄)² and fᵢ(xᵢ − x̄)²:
(6−19)²=169×2=338
(10−19)²=81×4=324
(14−19)²=25×7=175
(18−19)²=1×12=12
(24−19)²=25×8=200
(28−19)²=81×4=324
(30−19)²=121×3=363
Σf(x−x̄)² = 1736
✳️ Step 4 ➤ Variance = 1736 ÷ 40 = 43.4
SD = √43.4 = 6.59
✔️ Final: Mean = 19, Variance = 43.4, SD = 6.59
🔵 Question 5:
xᵢ: 92, 93, 97, 98, 102, 104, 109
fᵢ: 3, 2, 3, 2, 6, 3, 3
🟢 Answer:
✳️ Step 1 ➤ Σf = 22
✳️ Step 2 ➤ Σfx = 92×3 + 93×2 + 97×3 + 98×2 + 102×6 + 104×3 + 109×3
= 276 + 186 + 291 + 196 + 612 + 312 + 327 = 2200
➡️ x̄ = 2200 ÷ 22 = 100
✳️ Step 3 ➤ (xᵢ − x̄)²:
(92−100)²=64×3=192
(93−100)²=49×2=98
(97−100)²=9×3=27
(98−100)²=4×2=8
(102−100)²=4×6=24
(104−100)²=16×3=48
(109−100)²=81×3=243
Σf(x−x̄)² = 640
✳️ Step 4 ➤ Variance = 640 ÷ 22 = 29.09
SD = √29.09 = 5.39
✔️ Final: Mean = 100, Variance = 29.09, SD = 5.39
🔵 Question 6:
Find mean and standard deviation using short-cut method
xᵢ: 60, 61, 62, 63, 64, 65, 66, 67, 68
fᵢ: 2, 1, 12, 29, 25, 12, 10, 4, 5
🟢 Answer:
✳️ Step 1 ➤ Σf = 100
✳️ Step 2 ➤ Choose A = 64 (Assumed Mean), h = 1
✳️ Step 3 ➤ u = (xᵢ − A)/h = −4, −3, −2, −1, 0, 1, 2, 3, 4
Σfu = (2)(−4)+(1)(−3)+(12)(−2)+(29)(−1)+(25)(0)+(12)(1)+(10)(2)+(4)(3)+(5)(4)
= −8−3−24−29+0+12+20+12+20 = 0
➡️ Mean x̄ = A + (Σfu / Σf)h = 64 + 0 = 64
✳️ Step 4 ➤ Σfu² = 2×16 + 1×9 + 12×4 + 29×1 + 25×0 + 12×1 + 10×4 + 4×9 + 5×16
= 32 + 9 + 48 + 29 + 0 + 12 + 40 + 36 + 80 = 286
✳️ Step 5 ➤ Variance = [Σfu² / Σf − (Σfu/Σf)²] × h² = 286 ÷ 100 = 2.86
SD = √2.86 × 1 = 1.69
✔️ Final: Mean = 64, Variance = 2.86, SD = 1.69
🔵 Question 7:
Classes (width = 30): 0–30, 30–60, 60–90, 90–120, 120–150, 150–180, 180–210
Frequencies (fᵢ): 2, 3, 5, 10, 3, 5, 2
🟢 Answer (using short-cut method):
✳️ Step 1 ➤ Class midpoints (xᵢ) = 15, 45, 75, 105, 135, 165, 195
✳️ Step 2 ➤ Σf = 30, choose A = 105, h = 30
✳️ Step 3 ➤ Compute uᵢ = (xᵢ − A)/h = −3, −2, −1, 0, 1, 2, 3
✳️ Step 4 ➤ Multiply by frequencies:
fᵢuᵢ = (2)(−3)+(3)(−2)+(5)(−1)+(10)(0)+(3)(1)+(5)(2)+(2)(3)
➡️ Σfᵢuᵢ = −6 − 6 − 5 + 0 + 3 + 10 + 6 = 2
✳️ Step 5 ➤ Mean x̄ = A + (Σfᵢuᵢ / Σfᵢ) × h = 105 + (2/30) × 30 = 107
✳️ Step 6 ➤ Compute uᵢ² and fᵢuᵢ²:
uᵢ² = 9,4,1,0,1,4,9
fᵢuᵢ² = 2×9 + 3×4 + 5×1 + 10×0 + 3×1 + 5×4 + 2×9
= 18 + 12 + 5 + 0 + 3 + 20 + 18 = 76
✳️ Step 7 ➤ Variance = [Σfᵢuᵢ² / Σfᵢ − (Σfᵢuᵢ / Σfᵢ)²] × h²
= [76/30 − (2/30)²] × 30²
= [2.533 − 0.004] × 900 = 2.529 × 900 = 2276.1
✳️ Step 8 ➤ SD = √2276.1 = 47.7
✔️ Final:
Mean = 107, Variance = 2276.1, Standard Deviation = 47.7
🔵 Question 8:
Classes: 0–10, 10–20, 20–30, 30–40, 40–50
Frequencies (fᵢ): 5, 8, 15, 16, 6
🟢 Answer (using short-cut method):
✳️ Step 1 ➤ Class midpoints (xᵢ) = 5, 15, 25, 35, 45
✳️ Step 2 ➤ Σf = 50, choose A = 25, h = 10
✳️ Step 3 ➤ uᵢ = (xᵢ − A)/h = −2, −1, 0, 1, 2
✳️ Step 4 ➤ fᵢuᵢ = 5(−2)+8(−1)+15(0)+16(1)+6(2)
➡️ Σfᵢuᵢ = −10 − 8 + 0 + 16 + 12 = 10
✳️ Step 5 ➤ Mean x̄ = A + (Σfᵢuᵢ / Σfᵢ) × h
= 25 + (10/50) × 10 = 27
✳️ Step 6 ➤ Compute uᵢ², fᵢuᵢ²
uᵢ² = 4,1,0,1,4
fᵢuᵢ² = 5×4 + 8×1 + 15×0 + 16×1 + 6×4 = 20 + 8 + 0 + 16 + 24 = 68
✳️ Step 7 ➤ Variance = [Σfᵢuᵢ² / Σfᵢ − (Σfᵢuᵢ / Σfᵢ)²] × h²
= [68/50 − (10/50)²] × 10²
= [1.36 − 0.04] × 100 = 1.32 × 100 = 132
✳️ Step 8 ➤ SD = √132 = 11.49
✔️ Final:
Mean = 27, Variance = 132, Standard Deviation = 11.49
🔵 Question 9
Find the mean, variance and standard deviation using short-cut method.
Heights (cm): 70–75, 75–80, 80–85, 85–90, 90–95, 95–100, 100–105, 105–110, 110–115
Frequencies (fᵢ): 3, 4, 7, 7, 15, 9, 6, 6, 3
🟢 Answer (short-cut method):
✳️ Step 1 ➤ Class midpoints xᵢ = 72.5, 77.5, 82.5, 87.5, 92.5, 97.5, 102.5, 107.5, 112.5
✳️ Step 2 ➤ Choose A = 92.5, width h = 5, total Σfᵢ = 60
✳️ Step 3 ➤ uᵢ = (xᵢ − A)/h = −4, −3, −2, −1, 0, 1, 2, 3, 4 divided by (actually values) → −4? −3?
➡️ Correct uᵢ (with our A, h): −4? No, compute exactly → −4?
➡️ Final uᵢ list (exact): −4? (we only use sums below)
✳️ Step 4 ➤ Multiply by frequencies
➡️ Σ fᵢuᵢ = 6
➡️ Σ fᵢuᵢ² = 254
✳️ Step 5 ➤ Mean
➡️ x̄ = A + (Σfᵢuᵢ / Σfᵢ) · h = 92.5 + (6/60)·5 = 92.5 + 0.5 = 93.0 cm
✳️ Step 6 ➤ Variance
➡️ σ² = [Σfᵢuᵢ² / Σfᵢ − (Σfᵢuᵢ / Σfᵢ)²] · h²
➡️ = [254/60 − (6/60)²] · 5²
➡️ = (4.2333 − 0.01) · 25 = 105.5833
✳️ Step 7 ➤ Standard deviation
➡️ σ = √σ² = √105.5833 = 10.275… ≈ 10.28 cm
✔️ Final (Q9): Mean = 93.0 cm, Variance ≈ 105.58, Standard Deviation ≈ 10.28 cm
🔵 Question 10
Diameters of circles (mm) in a design:
Classes: 33–36, 37–40, 41–44, 45–48, 49–52
No. of circles: 15, 17, 21, 22, 25
Hint: Make classes continuous → 32.5–36.5, 36.5–40.5, 40.5–44.5, 44.5–48.5, 48.5–52.5
🟢 Answer (short-cut method):
✳️ Step 1 ➤ Midpoints xᵢ = 34.5, 38.5, 42.5, 46.5, 50.5
✳️ Step 2 ➤ Choose A = 42.5, width h = 4, total Σfᵢ = 100
✳️ Step 3 ➤ uᵢ = (xᵢ − A)/h = −2, −1, 0, 1, 2
✳️ Step 4 ➤ Sums with frequency
➡️ Σ fᵢuᵢ = 25
➡️ Σ fᵢuᵢ² = 199
✳️ Step 5 ➤ Mean diameter
➡️ x̄ = A + (Σfᵢuᵢ / Σfᵢ) · h = 42.5 + (25/100)·4 = 42.5 + 1 = 43.5 mm
✳️ Step 6 ➤ Variance
➡️ σ² = [Σfᵢuᵢ² / Σfᵢ − (Σfᵢuᵢ / Σfᵢ)²] · h²
➡️ = [199/100 − (25/100)²] · 4²
➡️ = (1.99 − 0.0625) · 16 = 30.84
✳️ Step 7 ➤ Standard deviation
➡️ σ = √30.84 = 5.553… ≈ 5.55 mm
✔️ Final (Q10): Mean diameter = 43.5 mm, Variance ≈ 30.84, Standard Deviation ≈ 5.55 mm
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OTHER IMPORTANT QUESTIONS FOR EXAMS
CBSE STYLE MODEL PAPER
ESPECIALLY FROM THIS CHAPTER ONLY
🧭 Section A – Very Short / Objective Type (1 mark each)
🔵 Question 1:
The range of the data 5, 12, 18, 25, 30 is
🔵 (A) 25
🟢 (B) 30
🟠 (C) 20
🔴 (D) 35
🟢 Answer: (C) 20
➡️ Range = Maximum − Minimum = 30 − 10 = 20
🔵 Question 2:
The mean deviation of 2, 4, 6 about mean is
🔵 (A) 0
🟢 (B) 2
🟠 (C) 4
🔴 (D) 1.33
🟢 Answer: (D) 1.33
🔵 Question 3:
Which of the following is independent of change of origin?
🔵 (A) Range
🟢 (B) Mean deviation
🟠 (C) Standard deviation
🔴 (D) Variance
🟢 Answer: (C) Standard deviation
🔵 Question 4:
If all observations of a data set are equal, the standard deviation is
🔵 (A) 1
🟢 (B) Mean
🟠 (C) 0
🔴 (D) Undefined
🟢 Answer: (C) 0
🔵 Question 5:
The formula for coefficient of variation is
🔵 (A) Mean/σ × 100
🟢 (B) σ/Mean × 100
🟠 (C) σ × Mean
🔴 (D) σ − Mean
🟢 Answer: (B) σ/Mean × 100
🔵 Question 6:
Mean deviation is based on
🔵 (A) Absolute deviations
🟢 (B) Squares of deviations
🟠 (C) Simple deviations
🔴 (D) None
🟢 Answer: (A) Absolute deviations
🔵 Question 7:
Variance is
🔵 (A) Square of mean
🟢 (B) Mean of squares
🟠 (C) Mean of squared deviations from mean
🔴 (D) Square of median
🟢 Answer: (C) Mean of squared deviations from mean
🔵 Question 8:
The step-deviation method is used to
🔵 (A) Decrease the mean
🟢 (B) Simplify calculations
🟠 (C) Increase the mean
🔴 (D) Change standard deviation
🟢 Answer: (B) Simplify calculations
🔵 Question 9:
If σ = 0, then
🔵 (A) All observations are equal
🟢 (B) Mean is 0
🟠 (C) Mean = Median
🔴 (D) None
🟢 Answer: (A) All observations are equal
🔵 Question 10:
The most commonly used measure of dispersion is
🔵 (A) Range
🟢 (B) Mean deviation
🟠 (C) Standard deviation
🔴 (D) Variance
🟢 Answer: (C) Standard deviation
🔵 Question 11:
If each observation is multiplied by 5, the new standard deviation is
🔵 (A) 5 times old
🟢 (B) 1/5 times old
🟠 (C) Same
🔴 (D) Zero
🟢 Answer: (A) 5 times old
🔵 Question 12:
The mean deviation about mean of 4, 8, 12 is
🔵 (A) 2.66
🟢 (B) 3
🟠 (C) 4
🔴 (D) 6
🟢 Answer: (B) 3
🔵 Question 13:
Which measure of dispersion is affected most by extreme values?
🔵 (A) Range
🟢 (B) Standard deviation
🟠 (C) Mean deviation
🔴 (D) All
🟢 Answer: (A) Range
🔵 Question 14:
The standard deviation is always
🔵 (A) Negative
🟢 (B) Positive or zero
🟠 (C) One
🔴 (D) Undefined
🟢 Answer: (B) Positive or zero
🔵 Question 15:
Coefficient of variation is used to compare
🔵 (A) Two means
🟢 (B) Two dispersions
🟠 (C) Two data sets
🔴 (D) Two medians
🟢 Answer: (C) Two data sets
🔵 Question 16:
If all data values are multiplied by a constant, mean deviation is
🔵 (A) Multiplied by same constant
🟢 (B) Divided by constant
🟠 (C) Unchanged
🔴 (D) Zero
🟢 Answer: (A) Multiplied by same constant
🔵 Question 17:
Which is the simplest measure of dispersion?
🔵 (A) Range
🟢 (B) Mean deviation
🟠 (C) Variance
🔴 (D) Standard deviation
🟢 Answer: (A) Range
🔵 Question 18:
If mean = 40, σ = 4, then C.V. =
🔵 (A) 1%
🟢 (B) 4%
🟠 (C) 10%
🔴 (D) 40%
🟢 Answer: (C) 10%
🔵 Question 19:
Find the mean deviation about the mean for the data: 5, 10, 15, 20, 25
🟢 Answer:
➡️ Step 1: Mean (x̄) = (5 + 10 + 15 + 20 + 25) / 5 = 15
➡️ Step 2: Deviations |xᵢ − x̄| = 10, 5, 0, 5, 10
➡️ Step 3: Sum = 30
➡️ Step 4: Mean Deviation = 30 / 5 = 6
✔️ Final Answer: Mean Deviation = 6
🔵 Question 20:
Find the variance and standard deviation for the data: 4, 8, 12, 16, 20
🟢 Answer:
➡️ Step 1: Mean (x̄) = (4 + 8 + 12 + 16 + 20) / 5 = 12
➡️ Step 2: Deviations (xᵢ − x̄) = −8, −4, 0, 4, 8
➡️ Step 3: Squares = 64, 16, 0, 16, 64 → Sum = 160
➡️ Step 4: Variance = 160 / 5 = 32
➡️ Step 5: Standard Deviation = √32 = 5.66
✔️ Final Answer: Variance = 32, Standard Deviation = 5.66
🔵 Question 21:
Find the standard deviation of 3, 7, 11, 15, 19 using the step-deviation method
🟢 Answer:
➡️ Step 1: Let assumed mean (A) = 11, h = 4
➡️ Step 2: xᵢ = 3, 7, 11, 15, 19
➡️ Step 3: dᵢ = (xᵢ − 11) / 4 = −2, −1, 0, 1, 2
➡️ Step 4: Σdᵢ = 0, Σdᵢ² = 10
➡️ Step 5: S.D. = h × √(Σdᵢ² / N − (Σdᵢ / N)²)
➡️ Step 6: S.D. = 4 × √(10 / 5) = 4 × √2 = 5.66
✔️ Final Answer: Standard Deviation = 5.66
🔵 Question 22:
Find the mean, variance, and standard deviation of the following frequency distribution:
xᵢ = 10, 20, 30, 40, 50
fᵢ = 3, 5, 7, 3, 2
🟢 Answer:
➡️ Step 1: Σfᵢ = 20
➡️ Step 2: Σfᵢxᵢ = 10×3 + 20×5 + 30×7 + 40×3 + 50×2 = 600
➡️ Step 3: Mean (x̄) = 600 / 20 = 30
➡️ Step 4: (xᵢ − x̄)² = 400, 100, 0, 100, 400
➡️ Step 5: Σfᵢ(xᵢ − x̄)² = 3×400 + 5×100 + 7×0 + 3×100 + 2×400 = 3200
➡️ Step 6: Variance = 3200 / 20 = 160
➡️ Step 7: Standard Deviation = √160 = 12.65
✔️ Final Answer: Mean = 30, Variance = 160, Standard Deviation = 12.65
🔵 Question 23:
Find the coefficient of variation for the data: 10, 20, 30, 40, 50
🟢 Answer:
➡️ Step 1: Mean (x̄) = 30
➡️ Step 2: Variance = 200, Standard Deviation = 14.14
➡️ Step 3: C.V. = (S.D. / Mean) × 100 = (14.14 / 30) × 100 = 47.13%
✔️ Final Answer: Coefficient of Variation = 47.13%
🔵 Question 24:
Which data is more consistent?
Set A: Mean = 40, S.D. = 4
Set B: Mean = 80, S.D. = 10
🟢 Answer:
➡️ C.V.(A) = (4 / 40) × 100 = 10%
➡️ C.V.(B) = (10 / 80) × 100 = 12.5%
✔️ Final Answer: Set A is more consistent (smaller C.V.)
🔵 Question 25:
Show that standard deviation is not affected by origin but affected by scale
🟢 Answer:
➡️ If each observation xᵢ is replaced by a + bxᵢ,
➡️ New Standard Deviation = |b| × Old Standard Deviation
✔️ Hence, unaffected by origin (a), affected by scale (b)
🔵 Question 26:
For what type of data is range a suitable measure of dispersion?
🟢 Answer:
✔️ Range is suitable for small data sets where extreme values are important,
such as daily temperature or stock prices
🔵 Question 27:
If S.D. = 0, what can you say about the data?
🟢 Answer:
✔️ All observations are equal and there is no variation in the data
🔵 Question 28:
Calculate the mean deviation about the mean for the following data:
x: 10, 20, 30, 40, 50
f: 3, 5, 8, 4, 2
🟢 Answer:
➡️ Step 1: Σf = 3 + 5 + 8 + 4 + 2 = 22
➡️ Step 2: Σfx = (10×3) + (20×5) + (30×8) + (40×4) + (50×2) = 660
➡️ Step 3: Mean (x̄) = 660 ÷ 22 = 30
➡️ Step 4: |xᵢ − x̄| = 20, 10, 0, 10, 20
➡️ Step 5: f × |xᵢ − x̄| = (3×20) + (5×10) + (8×0) + (4×10) + (2×20) = 200
➡️ Step 6: Mean Deviation = 200 ÷ 22 = 9.09
✔️ Final Answer: Mean Deviation = 9.09
🔵 Question 29:
Find the variance and standard deviation using the step-deviation method for the following data:
Class: 0–10, 10–20, 20–30, 30–40, 40–50
Frequency (f): 5, 8, 15, 16, 6
🟢 Answer:
➡️ Step 1: Class mark (xᵢ): 5, 15, 25, 35, 45
➡️ Step 2: Assumed mean (A) = 25, class width (h) = 10
➡️ Step 3: dᵢ = (xᵢ − A)/h = −2, −1, 0, 1, 2
➡️ Step 4: f × dᵢ = (5×−2) + (8×−1) + (15×0) + (16×1) + (6×2) = 10
➡️ Step 5: f × dᵢ² = (5×4) + (8×1) + (15×0) + (16×1) + (6×4) = 68
➡️ Step 6: Σf = 50
➡️ Step 7: Variance = h² × [ (Σf dᵢ² / Σf) − (Σf dᵢ / Σf)² ]
Variance = 10² × [ (68 / 50) − (10 / 50)² ]
Variance = 100 × [ 1.36 − 0.04 ] = 132
➡️ Step 8: Standard Deviation = √132 = 11.49
✔️ Final Answer: Variance = 132, Standard Deviation = 11.49
🔵 Question 30:
For the following data, find mean, variance, and standard deviation:
x: 5, 10, 15, 20, 25
f: 6, 8, 10, 7, 9
🟢 Answer:
➡️ Step 1: Σf = 6 + 8 + 10 + 7 + 9 = 40
➡️ Step 2: Σfx = (5×6) + (10×8) + (15×10) + (20×7) + (25×9) = 625
➡️ Step 3: Mean (x̄) = 625 ÷ 40 = 15.625
➡️ Step 4: (xᵢ − x̄) = −10.625, −5.625, −0.625, 4.375, 9.375
➡️ Step 5: (xᵢ − x̄)² = 112.89, 31.64, 0.39, 19.14, 87.89
➡️ Step 6: f × (xᵢ − x̄)² = (6×112.89) + (8×31.64) + (10×0.39) + (7×19.14) + (9×87.89) = 1859.32
➡️ Step 7: Variance = 1859.32 ÷ 40 = 46.48
➡️ Step 8: Standard Deviation = √46.48 = 6.82
✔️ Final Answer: Mean = 15.63, Variance = 46.48, Standard Deviation = 6.82
🔵 Question 31:
Compare the consistency of two data sets:
Set A: Mean = 60, S.D. = 10
Set B: Mean = 80, S.D. = 15
🟢 Answer:
➡️ Coefficient of Variation (C.V.) = (S.D. / Mean) × 100
➡️ C.V.(A) = (10 / 60) × 100 = 16.67%
➡️ C.V.(B) = (15 / 80) × 100 = 18.75%
✔️ Final Answer: Set A is more consistent (smaller C.V.)
🔵 Question 32 (Case Study):
A factory produces bulbs with lifetimes (in hours) as follows:
Lifetime: 1000–1200, 1200–1400, 1400–1600, 1600–1800, 1800–2000
Frequency: 5, 15, 20, 8, 2
🟢 Answer:
➡️ Step 1: Class marks (xᵢ): 1100, 1300, 1500, 1700, 1900
➡️ Step 2: Σf = 5 + 15 + 20 + 8 + 2 = 50
➡️ Step 3: Σfx = (5×1100) + (15×1300) + (20×1500) + (8×1700) + (2×1900) = 73,800
➡️ Step 4: Mean (x̄) = 73,800 ÷ 50 = 1476
➡️ Step 5: (xᵢ − x̄) = −376, −176, 24, 224, 424
➡️ Step 6: (xᵢ − x̄)² = 141376, 30976, 576, 50176, 179776
➡️ Step 7: f × (xᵢ − x̄)² = (5×141376) + (15×30976) + (20×576) + (8×50176) + (2×179776) = 1,943,000
➡️ Step 8: Variance = 1,943,000 ÷ 50 = 38,860
➡️ Step 9: Standard Deviation = √38,860 = 197.13
✔️ Final Answer: Mean = 1476, Standard Deviation = 197.13
🔵 Question 33 (Application):
The average daily income of 10 workers is ₹200 with a standard deviation of ₹40. Each worker is given a raise of ₹20 and the new currency is ₹1 = old ₹2. Find the new mean and new standard deviation.
🟢 Answer:
➡️ Old Mean = 200, Old S.D. = 40
➡️ New Mean = (200 + 20) ÷ 2 = 110
➡️ New S.D. = 40 ÷ 2 = 20
✔️ Final Answer: New Mean = ₹110, New Standard Deviation = ₹20
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