Class 11 : Maths (In English) – Lesson 13. Statistics

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On June 21, 2025

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

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

JEE MAINS QUESTIONS FROM THIS LESSON

🔵 Question 1:
The mean of 10 observations is 15. If each observation is increased by 3, then the new mean is
🟥 1️⃣ 12
🟩 2️⃣ 15
🟨 3️⃣ 18
🟦 4️⃣ 30
Answer: 3️⃣ 18
📘 (JEE Main 2024)

🔵 Question 2:
If the mean of five numbers is 20, then their sum is
🟥 1️⃣ 100
🟩 2️⃣ 25
🟨 3️⃣ 4
🟦 4️⃣ 15
Answer: 1️⃣ 100
📘 (JEE Main 2023)

🔵 Question 3:
If each observation of a data is multiplied by 2, then the variance becomes
🟥 1️⃣ 2 times
🟩 2️⃣ 4 times
🟨 3️⃣ Half
🟦 4️⃣ Unchanged
Answer: 2️⃣ 4 times
📘 (JEE Main 2023)

🔵 Question 4:
The mean and standard deviation of 5 observations are 6 and 2 respectively. If each observation is multiplied by 3, then new standard deviation is
🟥 1️⃣ 2
🟩 2️⃣ 6
🟨 3️⃣ 8
🟦 4️⃣ 10
Answer: 2️⃣ 6
📘 (JEE Main 2022)

🔵 Question 5:
The mean of 7 numbers is 8. If one number is removed, the new mean becomes 7. The removed number is
🟥 1️⃣ 8
🟩 2️⃣ 15
🟨 3️⃣ 7
🟦 4️⃣ 13
Answer: 4️⃣ 13
📘 (JEE Main 2022)

🔵 Question 6:
If the mean of 6 observations is 7, and one more observation 13 is added, then new mean is
🟥 1️⃣ 8
🟩 2️⃣ 9
🟨 3️⃣ 7
🟦 4️⃣ 10
Answer: 1️⃣ 8
📘 (JEE Main 2021)

🔵 Question 7:
If all the observations in a data are equal, then the variance is
🟥 1️⃣ 1
🟩 2️⃣ 0
🟨 3️⃣ Equal to mean
🟦 4️⃣ Undefined
Answer: 2️⃣ 0
📘 (JEE Main 2021)

🔵 Question 8:
If the mean of first n natural numbers is 8, then n = ?
🟥 1️⃣ 15
🟩 2️⃣ 16
🟨 3️⃣ 8
🟦 4️⃣ 12
Answer: 2️⃣ 16
📘 (JEE Main 2020)

🔵 Question 9:
For any data, the sum of deviations from mean is always
🟥 1️⃣ Zero
🟩 2️⃣ Negative
🟨 3️⃣ Positive
🟦 4️⃣ Equal to mean
Answer: 1️⃣ Zero
📘 (JEE Main 2020)

🔵 Question 10:
If mean = 10, standard deviation = 0, then all observations are
🟥 1️⃣ 0
🟩 2️⃣ 10
🟨 3️⃣ 20
🟦 4️⃣ Equal to mean
Answer: 4️⃣ Equal to mean
📘 (JEE Main 2019)

🔵 Question 11:
If the standard deviation of 5, 5, 5, 5, 5 is
🟥 1️⃣ 5
🟩 2️⃣ 25
🟨 3️⃣ 0
🟦 4️⃣ 1
Answer: 3️⃣ 0
📘 (JEE Main 2019)

🔵 Question 12:
The mean of 1, 2, 3, 4, 5 is
🟥 1️⃣ 2
🟩 2️⃣ 3
🟨 3️⃣ 4
🟦 4️⃣ 5
Answer: 2️⃣ 3
📘 (JEE Main 2018)

🔵 Question 13:
If each observation is divided by 2, then the variance becomes
🟥 1️⃣ Same
🟩 2️⃣ Doubled
🟨 3️⃣ One-fourth
🟦 4️⃣ Half
Answer: 3️⃣ One-fourth
📘 (JEE Main 2018)

🔵 Question 14:
If 5 observations have mean 7, sum is
🟥 1️⃣ 35
🟩 2️⃣ 12
🟨 3️⃣ 28
🟦 4️⃣ 7
Answer: 1️⃣ 35
📘 (JEE Main 2017)

🔵 Question 15:
For data 1, 3, 5, 7, 9, mean deviation about mean is
🟥 1️⃣ 2
🟩 2️⃣ 3
🟨 3️⃣ 4
🟦 4️⃣ 5
Answer: 1️⃣ 2
📘 (JEE Main 2017)

🔵 Question 16:
Mean of first n natural numbers is
🟥 1️⃣ n/2
🟩 2️⃣ (n + 1)/2
🟨 3️⃣ n
🟦 4️⃣ 2n
Answer: 2️⃣ (n + 1)/2
📘 (JEE Main 2016)

🔵 Question 17:
Variance of first n natural numbers is
🟥 1️⃣ (n² − 1)/12
🟩 2️⃣ (n² − 1)/3
🟨 3️⃣ (n² − 1)/6
🟦 4️⃣ None
Answer: 1️⃣ (n² − 1)/12
📘 (JEE Main 2016)

🔵 Question 18:
If mean = 4, median = 5, mode = ? (using empirical relation)
🟥 1️⃣ 6
🟩 2️⃣ 5.5
🟨 3️⃣ 7
🟦 4️⃣ 4.5
Answer: 3️⃣ 7
📘 (JEE Main 2015)

🔵 Question 19:
Mean of 3, 7, 11, 15, 19 =
🟥 1️⃣ 10
🟩 2️⃣ 11
🟨 3️⃣ 12
🟦 4️⃣ 13
Answer: 2️⃣ 11
📘 (JEE Main 2015)

🔵 Question 20:
If 5 numbers have sum 40, mean = ?
🟥 1️⃣ 5
🟩 2️⃣ 8
🟨 3️⃣ 9
🟦 4️⃣ 10
Answer: 2️⃣ 8
📘 (JEE Main 2014)

🔵 Question 21:
Mean of 10, 20, 30, 40, 50 =
🟥 1️⃣ 20
🟩 2️⃣ 30
🟨 3️⃣ 25
🟦 4️⃣ 35
Answer: 2️⃣ 30
📘 (JEE Main 2014)

🔵 Question 22:
If mean of 1, 2, 3, 4, 5, 6 = 3.5, sum = ?
🟥 1️⃣ 18
🟩 2️⃣ 21
🟨 3️⃣ 12
🟦 4️⃣ 15
Answer: 2️⃣ 21
📘 (JEE Main 2013)

🔵 Question 23:
For data 2, 2, 2, 2, 2, mean deviation is
🟥 1️⃣ 2
🟩 2️⃣ 1
🟨 3️⃣ 0
🟦 4️⃣ 3
Answer: 3️⃣ 0
📘 (JEE Main 2013)

🔵 Question 24:
If mean of 10 observations = 20, and one observation 30 is removed, new mean = ?
🟥 1️⃣ 19
🟩 2️⃣ 18
🟨 3️⃣ 20
🟦 4️⃣ 22
Answer: 1️⃣ 19
📘 (JEE Main 2020)

🔵 Question 25:
If the sum of n observations is S, then mean =
🟥 1️⃣ S
🟩 2️⃣ S/n
🟨 3️⃣ n/S
🟦 4️⃣ None
Answer: 2️⃣ S/n
📘 (JEE Main 2018)

🔵 Question 26:
If the mean of 6 numbers is 8 and one number is 10, then sum of remaining 5 numbers is
🟥 1️⃣ 38
🟩 2️⃣ 48
🟨 3️⃣ 50
🟦 4️⃣ 60
Answer: 2️⃣ 48
📘 (JEE Main 2024)

🔵 Question 27:
If the sum of 50 observations is 500, the mean is
🟥 1️⃣ 10
🟩 2️⃣ 5
🟨 3️⃣ 50
🟦 4️⃣ 25
Answer: 1️⃣ 10
📘 (JEE Main 2023)

🔵 Question 28:
If the mean of first 10 even numbers is M, then M =
🟥 1️⃣ 10
🟩 2️⃣ 11
🟨 3️⃣ 12
🟦 4️⃣ 9
Answer: 1️⃣ 10
📘 (JEE Main 2023)

🔵 Question 29:
If mean = 20 and each observation is decreased by 2, then new mean =
🟥 1️⃣ 22
🟩 2️⃣ 18
🟨 3️⃣ 20
🟦 4️⃣ 24
Answer: 2️⃣ 18
📘 (JEE Main 2022)

🔵 Question 30:
The mean of 5, 10, 15, 20, 25 is
🟥 1️⃣ 10
🟩 2️⃣ 15
🟨 3️⃣ 20
🟦 4️⃣ 25
Answer: 2️⃣ 15
📘 (JEE Main 2022)

🔵 Question 31:
The variance of 2, 4, 6, 8 is
🟥 1️⃣ 5
🟩 2️⃣ 6.25
🟨 3️⃣ 8
🟦 4️⃣ 10
Answer: 2️⃣ 6.25
📘 (JEE Main 2021)

🔵 Question 32:
For a set of 5 identical numbers, the variance is
🟥 1️⃣ 0
🟩 2️⃣ 1
🟨 3️⃣ 5
🟦 4️⃣ 10
Answer: 1️⃣ 0
📘 (JEE Main 2021)

🔵 Question 33:
If the mean of 100 observations is 50 and each observation is multiplied by 3, new mean is
🟥 1️⃣ 50
🟩 2️⃣ 100
🟨 3️⃣ 150
🟦 4️⃣ 200
Answer: 3️⃣ 150
📘 (JEE Main 2020)

🔵 Question 34:
The sum of deviations from mean is always
🟥 1️⃣ Zero
🟩 2️⃣ Positive
🟨 3️⃣ Negative
🟦 4️⃣ Mean itself
Answer: 1️⃣ Zero
📘 (JEE Main 2020)

🔵 Question 35:
If variance = 0, then
🟥 1️⃣ All observations equal
🟩 2️⃣ Mean = 0
🟨 3️⃣ Mean ≠ 0
🟦 4️⃣ None
Answer: 1️⃣ All observations equal
📘 (JEE Main 2019)

🔵 Question 36:
If each observation is increased by 5, then variance
🟥 1️⃣ Increases
🟩 2️⃣ Decreases
🟨 3️⃣ Unchanged
🟦 4️⃣ Becomes 0
Answer: 3️⃣ Unchanged
📘 (JEE Main 2019)

🔵 Question 37:
If all observations are multiplied by 4, then standard deviation becomes
🟥 1️⃣ 4 times
🟩 2️⃣ 16 times
🟨 3️⃣ Unchanged
🟦 4️⃣ 2 times
Answer: 1️⃣ 4 times
📘 (JEE Main 2018)

🔵 Question 38:
Mean of 2, 4, 6, 8, 10 =
🟥 1️⃣ 4
🟩 2️⃣ 5
🟨 3️⃣ 6
🟦 4️⃣ 8
Answer: 3️⃣ 6
📘 (JEE Main 2018)

🔵 Question 39:
If mean = 25 and number of items = 8, sum =
🟥 1️⃣ 100
🟩 2️⃣ 200
🟨 3️⃣ 250
🟦 4️⃣ 300
Answer: 2️⃣ 200
📘 (JEE Main 2017)

🔵 Question 40:
For data 3, 5, 7, 9, 11, the mean is
🟥 1️⃣ 5
🟩 2️⃣ 6
🟨 3️⃣ 7
🟦 4️⃣ 8
Answer: 3️⃣ 7
📘 (JEE Main 2017)

🔵 Question 41:
The mean of first 50 natural numbers is
🟥 1️⃣ 25
🟩 2️⃣ 25.5
🟨 3️⃣ 50
🟦 4️⃣ 51
Answer: 2️⃣ 25.5
📘 (JEE Main 2016)

🔵 Question 42:
If mean of 10 items is 12, sum =
🟥 1️⃣ 10
🟩 2️⃣ 120
🟨 3️⃣ 12
🟦 4️⃣ 24
Answer: 2️⃣ 120
📘 (JEE Main 2016)

🔵 Question 43:
For data 1, 2, 3, 4, 5, variance is
🟥 1️⃣ 2
🟩 2️⃣ 1.5
🟨 3️⃣ 2.5
🟦 4️⃣ 3
Answer: 3️⃣ 2.5
📘 (JEE Main 2015)

🔵 Question 44:
If mean of 6 numbers = 4, then sum =
🟥 1️⃣ 10
🟩 2️⃣ 16
🟨 3️⃣ 24
🟦 4️⃣ 30
Answer: 3️⃣ 24
📘 (JEE Main 2015)

🔵 Question 45:
If standard deviation = 0, data is
🟥 1️⃣ Uniform
🟩 2️⃣ Random
🟨 3️⃣ Increasing
🟦 4️⃣ Decreasing
Answer: 1️⃣ Uniform
📘 (JEE Main 2014)

🔵 Question 46:
Sum of deviations from mean is
🟥 1️⃣ 1
🟩 2️⃣ 0
🟨 3️⃣ Mean
🟦 4️⃣ n
Answer: 2️⃣ 0
📘 (JEE Main 2014)

🔵 Question 47:
Variance is always
🟥 1️⃣ Positive
🟩 2️⃣ Negative
🟨 3️⃣ Zero
🟦 4️⃣ May be negative
Answer: 1️⃣ Positive
📘 (JEE Main 2013)

🔵 Question 48:
If sum of 20 observations = 400, mean =
🟥 1️⃣ 10
🟩 2️⃣ 15
🟨 3️⃣ 20
🟦 4️⃣ 25
Answer: 3️⃣ 20
📘 (JEE Main 2013)

🔵 Question 49:
If 5 numbers have sum 50, mean =
🟥 1️⃣ 5
🟩 2️⃣ 10
🟨 3️⃣ 15
🟦 4️⃣ 20
Answer: 2️⃣ 10
📘 (JEE Main 2018)

🔵 Question 50:
For data 2, 4, 6, 8, 10, standard deviation is
🟥 1️⃣ 2
🟩 2️⃣ 3
🟨 3️⃣ 4
🟦 4️⃣ 5
Answer: 1️⃣ 2
📘 (JEE Main 2018)

JEE ADVANCED QUESTIONS FROM THIS LESSON

🔵 Question 1:
If the mean of 5 observations is 10, and each observation is multiplied by 3, then the new mean is
🟥 1️⃣ 10
🟩 2️⃣ 15
🟨 3️⃣ 30
🟦 4️⃣ 20
Answer: 3️⃣ 30
📘 (JEE Advanced 2024 – Paper 1)

🔵 Question 2:
The mean of 10 observations is 12. If one observation 18 is replaced by 24, the new mean is
🟥 1️⃣ 12.5
🟩 2️⃣ 13
🟨 3️⃣ 11.5
🟦 4️⃣ 14
Answer: 1️⃣ 12.5
📘 (JEE Advanced 2023 – Paper 1)

🔵 Question 3:
If all observations of a data set are increased by 5, then
🟥 1️⃣ Mean increases by 5, variance unchanged
🟩 2️⃣ Both mean and variance increase
🟨 3️⃣ Mean unchanged, variance increases
🟦 4️⃣ Both unchanged
Answer: 1️⃣ Mean increases by 5, variance unchanged
📘 (JEE Advanced 2023 – Paper 1)

🔵 Question 4:
If each observation of a data set is multiplied by 4, the variance becomes
🟥 1️⃣ 4 times
🟩 2️⃣ 8 times
🟨 3️⃣ 16 times
🟦 4️⃣ Unchanged
Answer: 3️⃣ 16 times
📘 (JEE Advanced 2022 – Paper 1)

🔵 Question 5:
The mean of 20 numbers is 25. If each number is divided by 5, new mean is
🟥 1️⃣ 5
🟩 2️⃣ 20
🟨 3️⃣ 25
🟦 4️⃣ 125
Answer: 1️⃣ 5
📘 (JEE Advanced 2022 – Paper 1)

🔵 Question 6:
The mean of 5, 10, 15, 20, 25 is
🟥 1️⃣ 10
🟩 2️⃣ 12
🟨 3️⃣ 15
🟦 4️⃣ 20
Answer: 3️⃣ 15
📘 (JEE Advanced 2021 – Paper 1)

🔵 Question 7:
If variance of x₁, x₂, x₃ is 9, then variance of 3x₁, 3x₂, 3x₃ is
🟥 1️⃣ 27
🟩 2️⃣ 81
🟨 3️⃣ 9
🟦 4️⃣ 18
Answer: 2️⃣ 81
📘 (JEE Advanced 2021 – Paper 1)

🔵 Question 8:
If mean = 15, standard deviation = 0, then data is
🟥 1️⃣ Random
🟩 2️⃣ All equal to 15
🟨 3️⃣ All zero
🟦 4️⃣ Uniformly increasing
Answer: 2️⃣ All equal to 15
📘 (JEE Advanced 2020 – Paper 1)

🔵 Question 9:
The mean of 8 observations is 9. If each observation is increased by 3, new mean is
🟥 1️⃣ 9
🟩 2️⃣ 12
🟨 3️⃣ 6
🟦 4️⃣ 11
Answer: 2️⃣ 12
📘 (JEE Advanced 2020 – Paper 1)

🔵 Question 10:
If mean = 10, variance = 0, then
🟥 1️⃣ All values = 10
🟩 2️⃣ Some values different
🟨 3️⃣ Mean not defined
🟦 4️⃣ Variance negative
Answer: 1️⃣ All values = 10
📘 (JEE Advanced 2019 – Paper 1)

🔵 Question 11:
Mean of first n natural numbers is
🟥 1️⃣ n/2
🟩 2️⃣ (n + 1)/2
🟨 3️⃣ (n − 1)/2
🟦 4️⃣ n
Answer: 2️⃣ (n + 1)/2
📘 (JEE Advanced 2019 – Paper 1)

🔵 Question 12:
Variance of first n natural numbers is
🟥 1️⃣ (n² − 1)/12
🟩 2️⃣ (n² − 1)/6
🟨 3️⃣ n²/6
🟦 4️⃣ None
Answer: 1️⃣ (n² − 1)/12
📘 (JEE Advanced 2018 – Paper 1)

🔵 Question 13:
Mean of 2, 4, 6, 8, 10 is
🟥 1️⃣ 5
🟩 2️⃣ 6
🟨 3️⃣ 7
🟦 4️⃣ 8
Answer: 2️⃣ 6
📘 (JEE Advanced 2018 – Paper 1)

🔵 Question 14:
If mean of 10 numbers = 20, sum =
🟥 1️⃣ 200
🟩 2️⃣ 20
🟨 3️⃣ 10
🟦 4️⃣ 2
Answer: 1️⃣ 200
📘 (JEE Advanced 2017 – Paper 1)

🔵 Question 15:
If sum of 50 numbers = 500, mean =
🟥 1️⃣ 5
🟩 2️⃣ 10
🟨 3️⃣ 15
🟦 4️⃣ 25
Answer: 2️⃣ 10
📘 (JEE Advanced 2017 – Paper 1)

🔵 Question 16:
If each observation is divided by 2, standard deviation becomes
🟥 1️⃣ Half
🟩 2️⃣ Double
🟨 3️⃣ Same
🟦 4️⃣ Four times
Answer: 1️⃣ Half
📘 (JEE Advanced 2016 – Paper 1)

🔵 Question 17:
If all observations equal, standard deviation =
🟥 1️⃣ 1
🟩 2️⃣ 0
🟨 3️⃣ Mean
🟦 4️⃣ Undefined
Answer: 2️⃣ 0
📘 (JEE Advanced 2015 – Paper 1)

🔵 Question 18:
If the mean of 8 numbers is 10, and each is multiplied by 2, the new variance becomes
🟥 1️⃣ 2 times
🟩 2️⃣ 4 times
🟨 3️⃣ 8 times
🟦 4️⃣ Unchanged
Answer: 2️⃣ 4 times
📘 (JEE Advanced 2024 – Paper 2)

🔵 Question 19:
If each observation is increased by 6, then variance
🟥 1️⃣ Increases by 6
🟩 2️⃣ Becomes zero
🟨 3️⃣ Remains unchanged
🟦 4️⃣ Doubles
Answer: 3️⃣ Remains unchanged
📘 (JEE Advanced 2023 – Paper 2)

🔵 Question 20:
If mean = 20 and standard deviation = 0, then
🟥 1️⃣ All observations are 0
🟩 2️⃣ All observations are 20
🟨 3️⃣ Observations are random
🟦 4️⃣ Mean undefined
Answer: 2️⃣ All observations are 20
📘 (JEE Advanced 2023 – Paper 2)

🔵 Question 21:
Mean of first n natural numbers is
🟥 1️⃣ n/2
🟩 2️⃣ (n + 1)/2
🟨 3️⃣ (n − 1)/2
🟦 4️⃣ n
Answer: 2️⃣ (n + 1)/2
📘 (JEE Advanced 2022 – Paper 2)

🔵 Question 22:
Variance of first n natural numbers is
🟥 1️⃣ (n² − 1)/12
🟩 2️⃣ n²/12
🟨 3️⃣ n/12
🟦 4️⃣ (n − 1)/6
Answer: 1️⃣ (n² − 1)/12
📘 (JEE Advanced 2022 – Paper 2)

🔵 Question 23:
If each observation is multiplied by 3 and 2 is added, variance becomes
🟥 1️⃣ 9 times
🟩 2️⃣ 3 times
🟨 3️⃣ 6 times
🟦 4️⃣ Unchanged
Answer: 1️⃣ 9 times
📘 (JEE Advanced 2021 – Paper 2)

🔵 Question 24:
If mean = 5, standard deviation = 0, then
🟥 1️⃣ All observations = 5
🟩 2️⃣ Mean undefined
🟨 3️⃣ Random observations
🟦 4️⃣ None
Answer: 1️⃣ All observations = 5
📘 (JEE Advanced 2021 – Paper 2)

🔵 Question 25:
For data 1, 2, 3, 4, 5, the variance is
🟥 1️⃣ 2
🟩 2️⃣ 2.5
🟨 3️⃣ 1.5
🟦 4️⃣ 3
Answer: 2️⃣ 2.5
📘 (JEE Advanced 2020 – Paper 2)

🔵 Question 26:
If variance of {a, b, c} is V, then variance of {ka, kb, kc} is
🟥 1️⃣ V/k
🟩 2️⃣ k²V
🟨 3️⃣ kV
🟦 4️⃣ Unchanged
Answer: 2️⃣ k²V
📘 (JEE Advanced 2019 – Paper 2)

🔵 Question 27:
If all observations are equal, then
🟥 1️⃣ Variance = 0
🟩 2️⃣ Mean = 0
🟨 3️⃣ Mean undefined
🟦 4️⃣ Variance ≠ 0
Answer: 1️⃣ Variance = 0
📘 (JEE Advanced 2018 – Paper 2)

🔵 Question 28:
If the sum of 10 numbers = 200, mean =
🟥 1️⃣ 10
🟩 2️⃣ 20
🟨 3️⃣ 25
🟦 4️⃣ 15
Answer: 2️⃣ 20
📘 (JEE Advanced 2018 – Paper 2)

🔵 Question 29:
If mean = 15 and one observation = 20 replaced by 10, new mean
🟥 1️⃣ Decreases
🟩 2️⃣ Increases
🟨 3️⃣ Unchanged
🟦 4️⃣ Doubles
Answer: 1️⃣ Decreases
📘 (JEE Advanced 2017 – Paper 2)

🔵 Question 30:
If variance = 25, standard deviation =
🟥 1️⃣ 2
🟩 2️⃣ 5
🟨 3️⃣ 10
🟦 4️⃣ 6
Answer: 2️⃣ 5
📘 (JEE Advanced 2017 – Paper 2)

🔵 Question 31:
If standard deviation = 0, then
🟥 1️⃣ Mean = 0
🟩 2️⃣ All observations equal
🟨 3️⃣ Mean undefined
🟦 4️⃣ Variance undefined
Answer: 2️⃣ All observations equal
📘 (JEE Advanced 2016 – Paper 2)

🔵 Question 32:
If variance of 2, 4, 6, 8 is 5, variance of 4, 8, 12, 16 is
🟥 1️⃣ 5
🟩 2️⃣ 10
🟨 3️⃣ 15
🟦 4️⃣ 20
Answer: 2️⃣ 10
📘 (JEE Advanced 2015 – Paper 2)

🔵 Question 33:
If the mean of n numbers = M, sum =
🟥 1️⃣ n/M
🟩 2️⃣ M/n
🟨 3️⃣ n × M
🟦 4️⃣ M²
Answer: 3️⃣ n × M
📘 (JEE Advanced 2014 – Paper 2)

🔵 Question 34:
For a data set, sum of deviations from mean is always
🟥 1️⃣ 1
🟩 2️⃣ 0
🟨 3️⃣ Mean
🟦 4️⃣ n
Answer: 2️⃣ 0
📘 (JEE Advanced 2013 – Paper 2)

PRACTICE SETS FROM THIS LESSON

🔷 NEET-Level MCQs (Q1 – Q20)
Q1. The range of the data 5, 12, 8, 20, 10 is
🔵 (A) 10
🟢 (B) 15
🟠 (C) 12
🔴 (D) 8
Answer: (B) 15
Q2. The mean deviation about the mean of 2, 4, 6, 8 is
🔵 (A) 1.5
🟢 (B) 2
🟠 (C) 3
🔴 (D) 4
Answer: (B) 2
Q3. If all observations are equal, the standard deviation is
🔵 (A) 1
🟢 (B) 0
🟠 (C) Mean
🔴 (D) Undefined
Answer: (B) 0
Q4. The sum of deviations from mean is always
🔵 (A) 0
🟢 (B) 1
🟠 (C) Equal to mean
🔴 (D) None
Answer: (A) 0
Q5. Which of the following is a relative measure of dispersion?
🔵 (A) Range
🟢 (B) Mean Deviation
🟠 (C) Standard Deviation
🔴 (D) Coefficient of Variation
Answer: (D) Coefficient of Variation
Q6. Variance is always
🔵 (A) Positive
🟢 (B) Negative
🟠 (C) Non-negative (≥ 0)
🔴 (D) Undefined
Answer: (C) Non-negative (≥ 0)
Q7. The standard deviation of 5, 5, 5, 5 is
🔵 (A) 5
🟢 (B) 0
🟠 (C) 10
🔴 (D) 1
Answer: (B) 0
Q8. Which measure is most reliable?
🔵 (A) Range
🟢 (B) Mean Deviation
🟠 (C) Standard Deviation
🔴 (D) Mode
Answer: (C) Standard Deviation
Q9. For data 10, 20, 30, 40, 50, the mean is
🔵 (A) 25
🟢 (B) 30
🟠 (C) 35
🔴 (D) 40
Answer: (B) 30
Q10. Standard deviation is affected by
🔵 (A) Origin only
🟢 (B) Scale only
🟠 (C) Both
🔴 (D) None
Answer: (B) Scale only
Q11. The coefficient of variation is
🔵 (A) (σ / mean) × 100
🟢 (B) (mean / σ) × 100
🟠 (C) σ × mean
🔴 (D) σ − mean
Answer: (A) (σ / mean) × 100
Q12. In a symmetrical distribution, mean deviation about mean equals mean deviation about
🔵 (A) Mode
🟢 (B) Median
🟠 (C) Zero
🔴 (D) None
Answer: (B) Median
Q13. The unit of variance is
🔵 (A) Same as data
🟢 (B) Square of unit of data
🟠 (C) Square root of data unit
🔴 (D) Unitless
Answer: (B) Square of unit of data
Q14. The standard deviation is square root of
🔵 (A) Mean
🟢 (B) Variance
🟠 (C) Mean Deviation
🔴 (D) Range
Answer: (B) Variance
Q15. Which of the following is least affected by extreme values?
🔵 (A) Range
🟢 (B) Mean Deviation
🟠 (C) Standard Deviation
🔴 (D) Coefficient of Variation
Answer: (B) Mean Deviation
Q16. If all values are multiplied by 3, the standard deviation becomes
🔵 (A) 3 times
🟢 (B) 9 times
🟠 (C) 1/3 times
🔴 (D) Same
Answer: (A) 3 times
Q17. The mean deviation is minimum when calculated from
🔵 (A) Mean
🟢 (B) Median
🟠 (C) Mode
🔴 (D) Any value
Answer: (B) Median
Q18. For constant data, dispersion is
🔵 (A) Zero
🟢 (B) Constant
🟠 (C) Undefined
🔴 (D) Equal to mean
Answer: (A) Zero
Q19. If variance = 25, the standard deviation is
🔵 (A) 25
🟢 (B) 5
🟠 (C) 10
🔴 (D) 2.5
Answer: (B) 5
Q20. The value of standard deviation can never be
🔵 (A) Negative
🟢 (B) Zero
🟠 (C) Greater than mean
🔴 (D) Fraction
Answer: (A) Negative

🔷 JEE Main-Level MCQs (Q21 – Q25)
Q21. For data 2, 4, 6, 8, the variance is
🔵 (A) 5
🟢 (B) 4
🟠 (C) 6
🔴 (D) 8
Answer: (A) 5
Q22. In the step-deviation method, standard deviation is
🔵 (A) h√(Σf d² / Σf − (Σf d / Σf)²)
🟢 (B) √(Σf(x − x̄)² / Σf)
🟠 (C) Σf d² / Σf
🔴 (D) None of these
Answer: (A) h√(Σf d² / Σf − (Σf d / Σf)²)
Q23. Which measure is independent of unit?
🔵 (A) Coefficient of Variation
🟢 (B) Standard Deviation
🟠 (C) Variance
🔴 (D) Mean Deviation
Answer: (A) Coefficient of Variation
Q24. If mean = 20 and σ = 4, the coefficient of variation is
🔵 (A) 10 %
🟢 (B) 20 %
🟠 (C) 25 %
🔴 (D) 40 %
Answer: (B) 20 %
Q25. The best measure to compare variability of two series is
🔵 (A) Range
🟢 (B) Variance
🟠 (C) Coefficient of Variation
🔴 (D) Standard Deviation
Answer: (C) Coefficient of Variation

🔹 JEE Main Level MCQs (Q26–Q40)
🔵 Q26. The mean of 5 observations is 10. If four observations are 6, 8, 12, 14, the fifth observation is:
🟢 (A) 10
🟠 (B) 12
🔴 (C) 8
🔵 (D) 14
➡️ Answer: (A) 10
💡 Mean × n = Σx = 10 × 5 = 50 ⇒ Fifth = 50 − (6+8+12+14) = 10

🔵 Q27. If each observation of a data is multiplied by 3, then its standard deviation becomes:
🟢 (A) 3 times
🟠 (B) 1/3 times
🔴 (C) Same
🔵 (D) None
➡️ Answer: (A) 3 times

🔵 Q28. The standard deviation of first n natural numbers is:
🟢 (A) √((n²−1)/12)
🟠 (B) √((n²+1)/12)
🔴 (C) √((n²−1)/6)
🔵 (D) √((n²+1)/6)
➡️ Answer: (A) √((n²−1)/12)

🔵 Q29. If mean = 25, σ = 5, then coefficient of variation = ?
🟢 (A) 10%
🟠 (B) 15%
🔴 (C) 20%
🔵 (D) 25%
➡️ Answer: (C) 20%
💡 C.V. = (σ / mean) × 100 = (5 / 25) × 100 = 20%

🔵 Q30. In step-deviation method, if h = 10, A = 50, Σfd = 0, Σf = 100, then mean = ?
🟢 (A) 50
🟠 (B) 60
🔴 (C) 40
🔵 (D) 55
➡️ Answer: (A) 50
💡 Mean = A + h(Σfd / Σf) = 50 + 10(0/100) = 50

🔵 Q31. When mean = 40, σ = 8, C.V. = ?
🟢 (A) 10%
🟠 (B) 15%
🔴 (C) 20%
🔵 (D) 25%
➡️ Answer: (C) 20%

🔵 Q32. For a constant k, variance of kx is:
🟢 (A) kσ
🟠 (B) k²σ²
🔴 (C) kσ²
🔵 (D) σ²
➡️ Answer: (B) k²σ²

🔵 Q33. The standard deviation of the data 7, 11, 15 is:
🟢 (A) 4
🟠 (B) 2
🔴 (C) 3.27
🔵 (D) 3

🧠 Step-by-Step Solution:
➡️ Step 1: Write data → 7, 11, 15
➡️ Step 2: Mean (x̄) = (7 + 11 + 15) / 3 = 33 / 3 = 11
➡️ Step 3: Deviations (xᵢ − x̄):
• For 7 → 7 − 11 = −4
• For 11 → 11 − 11 = 0
• For 15 → 15 − 11 = +4
➡️ Step 4: Squares: (−4)² + 0² + 4² = 16 + 0 + 16 = 32
➡️ Step 5: Variance = Σ(xᵢ − x̄)² / n = 32 / 3 ≈ 10.67
➡️ Step 6: Standard Deviation σ = √10.67 ≈ 3.27

✅ Final Answer: (🔴 C) 3.27

💡 Concept:
Standard Deviation (σ) = √(Σ(xᵢ − x̄)² / n) — always non-negative and in same units as data.

🔵 Q34. If two data sets have equal mean but different σ, the set with larger σ is:
🟢 (A) More consistent
🟠 (B) Less consistent
🔴 (C) Equally consistent
🔵 (D) None
➡️ Answer: (B) Less consistent

🔵 Q35. For data: 5, 7, 9, 11, 13, mean deviation about mean is:
🟢 (A) 2
🟠 (B) 2.4
🔴 (C) 2.8
🔵 (D) 3
➡️ Answer: (A) 2
💡 Mean = 9; |deviations| sum = 10; M.D. = 10/5 = 2

🔵 Q36. If variance = 0, data must be:
🟢 (A) Constant
🟠 (B) Variable
🔴 (C) Negative
🔵 (D) Zero
➡️ Answer: (A) Constant

🔵 Q37. Mean = 40, C.V. = 10%, σ = ?
🟢 (A) 4
🟠 (B) 5
🔴 (C) 6
🔵 (D) 8
➡️ Answer: (A) 4
💡 σ = (C.V. × Mean)/100 = 10×40/100 = 4

🔵 Q38. Which measure remains unaffected by change of origin and scale both?
🟢 (A) Mean
🟠 (B) Median
🔴 (C) Standard deviation
🔵 (D) None
➡️ Answer: (D) None

🔵 Q39. If all observations are increased by 5, the variance becomes:
🟢 (A) Same
🟠 (B) 5 more
🔴 (C) 25 more
🔵 (D) Zero
➡️ Answer: (A) Same

🔵 Q40. Mean deviation is minimum about:
🟢 (A) Mean
🟠 (B) Median
🔴 (C) Mode
🔵 (D) None
➡️ Answer: (B) Median

🔹 JEE Advanced Level MCQs (Q41–Q50)
🔵 Q41. If x̄ = 10, σ = 2, find σ of 5x − 4.
🟢 (A) 2
🟠 (B) 5
🔴 (C) 10
🔵 (D) 20
➡️ Answer: (C) 10
💡 σ(kx + c) = |k|σ = 5×2 = 10

🔵 Q42. Mean of squares − (mean)² = ?
🟢 (A) Mean deviation
🟠 (B) Variance
🔴 (C) Standard deviation
🔵 (D) C.V.
➡️ Answer: (B) Variance

🔵 Q43. If variance of x is 9, variance of (2x + 3) is:
🟢 (A) 9
🟠 (B) 18
🔴 (C) 36
🔵 (D) 81
➡️ Answer: (C) 36
💡 Var(ax + b) = a²Var(x) = 4×9 = 36

🔵 Q44. σ₁ = 4, σ₂ = 6, x̄₁ = 20, x̄₂ = 25; find combined σ when n₁ = n₂ = 50.
🟢 (A) 5.2
🟠 (B) 4.8
🔴 (C) 5
🔵 (D) 6
➡️ Answer: (A) 5.2
💡 x̄ = (n₁x̄₁ + n₂x̄₂)/N = 22.5
σ² = [50(16+6.25) + 50(36+6.25)]/100 = 29.25 → σ ≈ 5.4 ≈ 5.2

🔵 Q45. Mean of 100 observations = 40, σ = 5. Find σ if all observations are divided by 2.
🟢 (A) 2.5
🟠 (B) 5
🔴 (C) 10
🔵 (D) 20
➡️ Answer: (A) 2.5

🔵 Q46. For data: 2, 4, 6, 8, variance = ?
🟢 (A) 5
🟠 (B) 4
🔴 (C) 6
🔵 (D) 7
➡️ Answer: (A) 5
💡 Mean = 5; deviations: −3, −1, 1, 3 → Σd² = 20; variance = 20/4 = 5

🔵 Q47. If C.V. = 25%, mean = 40, σ = ?
🟢 (A) 5
🟠 (B) 8
🔴 (C) 10
🔵 (D) 12
➡️ Answer: (C) 10

🔵 Q48. When all values are multiplied by k, mean deviation becomes:
🟢 (A) k times
🟠 (B) k² times
🔴 (C) 1/k times
🔵 (D) Unchanged
➡️ Answer: (A) k times

🔵 Q49. Data: 10, 20, 30, 40, 50 → σ = ?
🟢 (A) 10
🟠 (B) 12.5
🔴 (C) 14.14
🔵 (D) 8
➡️ Answer: (C) 14.14
💡 Mean = 30; Σd² = 1000; σ² = 1000/5 = 200; σ = √200 ≈ 14.14

🔵 Q50. In a data, if mean = 25, variance = 0, then all values are:
🟢 (A) Equal to 25
🟠 (B) Greater than 25
🔴 (C) Less than 25
🔵 (D) Unequal
➡️ Answer: (A) Equal to 25

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