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Unit III Review
Unit III · Lecture 23

BDS113 — Introduction to Statistics

Unit III Review & Problem-Solving Session

Dispersion · Moments · Skewness · Kurtosis — Comprehensive Recap & Exam-Style Practice


Unit III · Lecture 23 of 40 · B.Sc. Data Science, 1st Year, Semester II

Unit III Review
Unit III · Lecture 23

🗺️ Session Roadmap


📋
Part A — Concept Recap (Quick Reference)
All key formulas and definitions from Lectures 17–22 in one place
⚠️
Part B — Common Mistakes & Exam Tips
Pitfalls to avoid, frequently confused concepts, marking traps
✏️
Part C — Solved Problems (Easy → Hard)
12 exam-style problems with step-by-step solutions
🧠
Part D — Comprehensive Quiz & Formula Sheet
20-question mixed quiz · Printable formula cheat-sheet
Unit III Review
Part A · Formula Sheet

📋 Quick Reference — All Unit III Formulas


📏 Mean Absolute Deviation

MAD = (1/n) Σ|xᵢ − x̄|
MAD (freq) = (1/N) Σfᵢ|xᵢ − x̄|
Coeff. of MAD = MAD / x̄

📊 Variance & Standard Deviation

σ² = (1/n)Σ(xᵢ − x̄)²
Shortcut: σ² = (Σxᵢ²/n) − x̄²
σ = √σ²
Sample: s² = Σ(xᵢ−x̄)² / (n−1)

📐 RMSD

RMSD(A) = √[(1/n)Σ(xᵢ−A)²]
RMSD(A) ≥ SD, equality iff A = x̄

📈 Coefficient of Variation

CV = (σ / x̄) × 100%
Lower CV → more consistent
Requires positive mean (ratio scale)

🔗 Moments

Raw: μᵣ' = (1/n)Σxᵢʳ
Central: μᵣ = (1/n)Σ(xᵢ−x̄)ʳ
μ₂ = μ₂' − (μ₁')²
μ₃ = μ₃' − 3μ₂'μ₁' + 2(μ₁')³
μ₄ = μ₄' − 4μ₃'μ₁' + 6μ₂'(μ₁')² − 3(μ₁')⁴

🔧 Sheppard's Correction

μ₂ᶜ = μ₂ − c²/12
μ₃ᶜ = μ₃ (no correction)
μ₄ᶜ = μ₄ − (c²/2)μ₂ + 7c⁴/240

Unit III Review
Part A · Formula Sheet

📋 Quick Reference — Skewness & Kurtosis


↔️ Skewness Measures

Karl Pearson (mode): SK = (Mean − Mode) / σ
Karl Pearson (median): SK = 3(Mean − Median) / σ
Bowley: SQ = (Q₃+Q₁−2Q₂) / (Q₃−Q₁)
Kelly: SKel = (P₉₀+P₁₀−2Median) / (P₉₀−P₁₀)
Moment: γ₁ = μ₃ / μ₂^(3/2)
β₁ = μ₃² / μ₂³ = γ₁²

Empirical: Mean − Mode ≈ 3(Mean − Median)

📐 Kurtosis Measures

β₂ = μ₄ / μ₂²
γ₂ = β₂ − 3 (excess kurtosis)

β₂ = 3 → Mesokurtic (normal)
β₂ > 3 → Leptokurtic (heavy tails)
β₂ < 3 → Platykurtic (light tails)

Percentile: K = (P₇₅−P₂₅) / (P₉₀−P₁₀)

🔑 Key Properties

• SD, CV, skewness, kurtosis are invariant under origin shift
• SD scales by |a| if y = ax+b; CV, γ₁, β₂ are fully invariant
• μ₁ = 0 always; μ₂ = variance
• Symmetric → γ₁ = 0, Mean = Median = Mode

Unit III Review
Part B · Common Mistakes

⚠️ Top 10 Mistakes Students Make in Unit III


1. Using n instead of (n−1) for sample variance. Population variance divides by N; sample variance divides by (n−1). Read the question carefully — does it say "population" or "sample"?
2. Forgetting to take the square root. Variance is in squared units. SD = √variance. If the question asks for SD and you report variance, you lose marks.
3. Computing |γ₁| instead of γ₁. The sign of γ₁ tells you the direction of skewness. Reporting only the magnitude loses critical information.
4. Applying Sheppard's correction to discrete data. Sheppard's correction is ONLY for continuous, equal-width grouped data. Never apply it to discrete frequency distributions.
5. Confusing β₁ with γ₁. β₁ = μ₃²/μ₂³ (always ≥ 0, no sign). γ₁ = μ₃/μ₂^(3/2) (signed). β₁ = γ₁².
6. Using CV when the mean is negative or zero. CV = (σ/x̄)×100 is undefined when x̄ = 0 and misleading when x̄ < 0.
7. Saying kurtosis measures "peakedness" only. Kurtosis measures tail heaviness. A sharp peak with thin tails can have LOW kurtosis.
8. Mixing up raw and central moments. μᵣ' (with prime) = raw moment about origin. μᵣ (no prime) = central moment about mean. Don't confuse the notation.
9. Not verifying Σ(xᵢ − x̄) = 0. As a check: the sum of deviations from the mean must always be zero. If it's not, your mean is wrong.
10. Applying Sheppard's correction to odd-order moments. Only even-order moments (μ₂, μ₄) need correction. μ₃ is unchanged.
Unit III Review
Part B · Exam Tips

🎯 Exam Strategy for Unit III


✅ Do This

1. Always show your table. Marks are awarded for the working table (xᵢ, fᵢ, fᵢxᵢ, deviations, etc.), not just the final answer.

2. State which formula you're using. Write "Using Karl Pearson's SK = (Mean−Mode)/σ" before substituting values.

3. Interpret every result. "SK = +0.6 means the distribution is moderately positively skewed" earns more marks than just "SK = 0.6."

4. Verify your work. Check: Σ(xᵢ−x̄) = 0? Variance ≥ 0? |SK| ≤ 3? These quick checks catch errors.

5. Use the shortcut formula for SD. σ² = (Σfᵢxᵢ²/N) − x̄² is faster and less error-prone than computing each deviation.

💡 Time-Saving Tricks

1. Assumed mean method: Choose A at the centre, compute dᵢ = (xᵢ−A)/c, then use σ² = c²[(Σfᵢdᵢ²/N) − (Σfᵢdᵢ/N)²].

2. Convert raw → central moments using the formulas rather than computing deviations individually.

3. For symmetric data: If frequencies are symmetric around the centre, immediately conclude γ₁ = 0 and μ₃ = 0 without computing.

4. For Sheppard's correction: Compute uncorrected moments first, then apply the correction as a final step.

5. Remember the empirical relation: Mean − Mode ≈ 3(Mean − Median). Use it to estimate the mode when it's not given.

Unit III Review
Part C · Problem 1

✏️ Problem 1 — Mean Absolute Deviation


📝 Problem (Easy)

Calculate the Mean Absolute Deviation of the following data: 6, 8, 10, 12, 14


Step-by-step Solution:

Step 1: Mean
x̄ = (6+8+10+12+14)/5 = 50/5 = 10

Step 2: Absolute Deviations

xᵢ|xᵢ − 10|
64
82
100
122
144
Sum12

Step 3: MAD = 12/5 = 2.4

Interpretation: On average, each value deviates by 2.4 units from the mean of 10.

Unit III Review
Part C · Problem 2

✏️ Problem 2 — Standard Deviation (Shortcut Method)


📝 Problem (Easy)

Find the variance and standard deviation of: 5, 10, 15, 20, 25 using the shortcut formula.


Step-by-step Solution:

Shortcut formula: σ² = (Σxᵢ²/n) − x̄²

n = 5, Σxᵢ = 75, x̄ = 15
Σxᵢ² = 25+100+225+400+625 = 1375

σ² = 1375/5 − 15² = 275 − 225 = 50
σ = √50 = 7.07

Verification: Σ(xᵢ−x̄) = (−10)+(−5)+0+5+10 = 0 ✓

Unit III Review
Part C · Problem 3

✏️ Problem 3 — SD of Frequency Distribution


📝 Problem (Medium)

Find the standard deviation and CV:

xᵢ246810
fᵢ35853

Step-by-step Solution:

N = 24, Σfᵢxᵢ = 6+20+48+40+30 = 144, x̄ = 144/24 = 6
Σfᵢxᵢ² = 12+80+288+320+300 = 1000

σ² = 1000/24 − 36 = 41.667 − 36 = 5.667
σ = √5.667 = 2.381

CV = (2.381/6) × 100 = 39.68%

Interpretation: The data has moderate variability (CV ≈ 40%). Each value deviates about 2.38 units from the mean on average.

Unit III Review
Part C · Problem 4

✏️ Problem 4 — SD of Grouped Data (Assumed Mean)


📝 Problem (Medium)

Using the assumed mean method, find the SD of:

Class0–1010–2020–3030–4040–50
fᵢ4812106

Step-by-step Solution:

Mid-points: 5, 15, 25, 35, 45. Let A = 25, c = 10.
dᵢ = (xᵢ−25)/10: d = −2, −1, 0, 1, 2

fᵢdᵢfᵢdᵢfᵢdᵢ²
4−2−816
8−1−88
12000
1011010
621224
40658

σ²(d) = (58/40) − (6/40)² = 1.45 − 0.0225 = 1.4275
σ² = 1.4275 × 100 = 142.75
σ = √142.75 = 11.95

x̄ = 25 + (6/40)×10 = 25 + 1.5 = 26.5

Unit III Review
Part C · Problem 5

✏️ Problem 5 — Sheppard's Correction


📝 Problem (Medium)

For a continuous grouped dataset with class width c = 10, the uncorrected central moments are: μ₂ = 85, μ₃ = 40, μ₄ = 12000. Apply Sheppard's correction and find the corrected SD and kurtosis.


Step-by-step Solution:

Sheppard's Corrections (c = 10, c² = 100):

μ₂ᶜ = 85 − 100/12 = 85 − 8.333 = 76.667
μ₃ᶜ = μ₃ = 40 (no correction for odd moments)
μ₄ᶜ = 12000 − (100/2)(85) + (7×10000/240)
    = 12000 − 4250 + 291.667 = 8041.667

Corrected SD: σᶜ = √76.667 = 8.756

Corrected Kurtosis:
β₂ᶜ = 8041.667 / (76.667)² = 8041.667 / 5877.83 = 1.368
γ₂ᶜ = 1.368 − 3 = −1.632 (Platykurtic)

Corrected Skewness:
γ₁ᶜ = 40 / (76.667)^1.5 = 40 / 670.82 = 0.0596 (approximately symmetric)

Unit III Review
Part C · Problem 6

✏️ Problem 6 — Moments & Shape Measures


📝 Problem (Medium)

Given raw moments: μ₁' = 5, μ₂' = 30, μ₃' = 200, μ₄' = 1500. Find all central moments, skewness (γ₁), and kurtosis (β₂).


Step-by-step Solution:

Central Moments:

μ₂ = 30 − 25 = 5
μ₃ = 200 − 3(30)(5) + 2(125) = 200 − 450 + 250 = 0
μ₄ = 1500 − 4(200)(5) + 6(30)(25) − 3(625)
    = 1500 − 4000 + 4500 − 1875 = 125

Shape Measures:

γ₁ = 0 / 5^(3/2) = 0 → Symmetric
β₂ = 125 / 25 = 5.0 → Leptokurtic
γ₂ = 5.0 − 3 = +2.0

Interpretation: The distribution is symmetric but leptokurtic — it has a sharp peak and heavy tails, similar to a Laplace distribution.

Unit III Review
Part C · Problem 7

✏️ Problem 7 — Comparing Two Distributions


📝 Problem (Hard)

Two factories produce bolts. Factory A: Mean diameter = 5 cm, SD = 0.05 cm. Factory B: Mean diameter = 2 cm, SD = 0.04 cm. Which factory is more consistent? Justify your answer with appropriate measures.


Step-by-step Solution:

Why SD alone is insufficient:
Factory A has SD = 0.05 and Factory B has SD = 0.04. By SD alone, B seems more consistent. But their means are very different (5 vs 2 cm), so we need a relative measure.

CV Comparison:

CV_A = (0.05/5) × 100 = 1.0%
CV_B = (0.04/2) × 100 = 2.0%

Conclusion: Factory A (CV = 1%) is more consistent than Factory B (CV = 2%). Despite having a larger absolute SD, Factory A's variability relative to its mean is smaller.

Key lesson: When means differ significantly, always use CV, not SD, for comparison.

Unit III Review
Part C · Problem 8

✏️ Problem 8 — Complete Shape Analysis


📝 Problem (Hard)

For a distribution: Mean = 45, Median = 48, Mode = 52, SD = 12, Q₁ = 35, Q₃ = 58. Calculate ALL available skewness measures and comment on consistency.


Step-by-step Solution:

1. Karl Pearson's SK (using mode):
SK = (45 − 52) / 12 = −7/12 = −0.583

2. Karl Pearson's SK (using median):
SK = 3(45 − 48) / 12 = −9/12 = −0.75

3. Bowley's SQ:
SQ = (58 + 35 − 2×48) / (58 − 35) = (93 − 96) / 23 = −3/23 = −0.130

4. Empirical mode estimate:
Mode ≈ 3(45) − 2(48) = 135 − 96 = 39. But actual mode = 52. The empirical relation doesn't hold well here — the distribution may be significantly skewed.

Interpretation: All three measures are negative, confirming left (negative) skewness. The Mean < Median < Mode pattern is classic left-skew. Karl Pearson's measures indicate moderate-to-strong skewness; Bowley's indicates mild skewness (quartiles are less sensitive to asymmetry in the tails).

Unit III Review
Part C · Problem 9

✏️ Problem 9 — Full Distribution Analysis


📝 Problem (Hard)

For a grouped frequency distribution (class width c = 5), the first four raw moments about origin are: μ₁' = 25, μ₂' = 650, μ₃' = 17500, μ₄' = 500000. Find: (a) Central moments, (b) SD, (c) γ₁, (d) β₂, (e) Apply Sheppard's correction to all applicable moments.


Step-by-step Solution:

(a) Central Moments:

μ₂ = 650 − 625 = 25
μ₃ = 17500 − 3(650)(25) + 2(15625) = 17500 − 48750 + 31250 = 0
μ₄ = 500000 − 4(17500)(25) + 6(650)(625) − 3(390625)
    = 500000 − 1750000 + 2437500 − 1171875 = 15625

(b) SD: σ = √25 = 5

(c) Skewness: γ₁ = 0/125 = 0 (Symmetric)

(d) Kurtosis: β₂ = 15625/625 = 25.0, γ₂ = +22.0 (Extremely leptokurtic)

(e) Sheppard's Correction (c = 5, c² = 25):

μ₂ᶜ = 25 − 25/12 = 25 − 2.083 = 22.917
μ₃ᶜ = 0 (unchanged)
μ₄ᶜ = 15625 − (25/2)(25) + (7×625/240)
    = 15625 − 312.5 + 18.229 = 15330.729
β₂ᶜ = 15330.729 / (22.917)² = 15330.729 / 525.19 = 29.19

Note: The extremely high kurtosis (β₂ = 25 before, 29 after correction) suggests very heavy tails — possibly outliers or a non-standard distribution shape.

Unit III Review
Part C · Problem 10

✏️ Problem 10 — RMSD and Minimum Property


📝 Problem (Hard)

For the data 3, 7, 8, 10, 12, compute RMSD about A = 5, RMSD about A = 8, and the SD. Verify the minimum RMSD property.


Step-by-step Solution:

Mean = (3+7+8+10+12)/5 = 40/5 = 8

RMSD(A=5):
Σ(xᵢ−5)² = 4+9+9+25+49 = 96
RMSD(5) = √(96/5) = √19.2 = 4.382

RMSD(A=8):
Σ(xᵢ−8)² = 25+1+0+4+16 = 46
RMSD(8) = √(46/5) = √9.2 = 3.033

SD (RMSD about mean = 8):
SD = √9.2 = 3.033

Verification:
RMSD(5) = 4.382 > SD = 3.033 ✓
RMSD(8) = 3.033 = SD ✓ (because A = 8 = mean)
The SD is indeed the minimum RMSD.

Unit III Review
Part C · Problem 11

✏️ Problem 11 — CV and Investment Decision


📝 Problem (Hard)

An investor records annual returns (%) over 6 years:
Fund X: 10, 12, 14, 16, 18, 20
Fund Y: −5, 5, 15, 15, 25, 35
Compare their risk using SD and CV. Which fund is better for a risk-averse investor?


Step-by-step Solution:

Fund X: x̄ = 90/6 = 15, Σxᵢ² = 100+144+196+256+324+400 = 1420
σ² = 1420/6 − 225 = 236.67 − 225 = 11.67, σ = 3.42
CV_X = (3.42/15)×100 = 22.8%

Fund Y: x̄ = 90/6 = 15, Σxᵢ² = 25+25+225+225+625+1225 = 2350
σ² = 2350/6 − 225 = 391.67 − 225 = 166.67, σ = 12.91
CV_Y = (12.91/15)×100 = 86.1%

Comparison:

Fund XFund Y
Mean15%15%
SD3.42%12.91%
CV22.8%86.1%

Conclusion: Both funds have the same average return (15%), but Fund X is far more consistent (CV = 22.8% vs 86.1%). A risk-averse investor should choose Fund X.

Unit III Review
Part C · Problem 12

✏️ Problem 12 — Complete Unit III Challenge


📝 Problem (Hard — Exam-Style)

The marks of 50 students are given below. Find: (a) Mean, (b) SD, (c) CV, (d) Apply Sheppard's correction, (e) If the raw third and fourth central moments are μ₃ = 120 and μ₄ = 8500, find γ₁ and β₂ (corrected).

Marks20–3030–4040–5050–6060–7070–80
fᵢ48141284

Step-by-step Solution:

Mid-points: 25, 35, 45, 55, 65, 75. A = 50, c = 10, dᵢ = (xᵢ−50)/10.

fᵢdᵢfᵢdᵢfᵢdᵢ²
4−2.5−1025
8−1.5−1218
14−0.5−73.5
120.563
81.51218
42.51025
50−192.5

(a) Mean: x̄ = 50 + (−1/50)×10 = 50 − 0.2 = 49.8

(b) SD: σ²(d) = 92.5/50 − (−1/50)² = 1.85 − 0.0004 = 1.8496
σ² = 1.8496 × 100 = 184.96, σ = 13.60

(c) CV: (13.60/49.8)×100 = 27.31%

(d) Sheppard's correction (c = 10):
μ₂ᶜ = 184.96 − 100/12 = 184.96 − 8.33 = 176.63
σᶜ = √176.63 = 13.29

(e) Shape measures (corrected):
μ₃ᶜ = 120 (unchanged)
μ₄ᶜ = 8500 − (100/2)(184.96) + (7×10000/240)
    = 8500 − 9248 + 291.67 = −456.33

Note: Negative μ₄ᶜ suggests Sheppard's correction may not be fully appropriate here (the distribution may not be smooth enough). In practice, report uncorrected μ₄ if corrected value is implausible.

Using uncorrected moments:
γ₁ = 120 / (184.96)^1.5 = 120 / 2513.5 = 0.0478 (approximately symmetric)
β₂ = 8500 / (184.96)² = 8500 / 34210.2 = 0.248

Interpretation: The distribution is approximately symmetric (γ₁ ≈ 0) with light tails (β₂ < 3). The symmetric frequencies (4, 8, 14, 12, 8, 4) confirm this.

Unit III Review
Part C · Interactive Tool

🎬 Interactive: Full Distribution Analyzer


Enter any dataset to compute ALL Unit III measures at once — MAD, SD, CV, Moments, Skewness, and Kurtosis:



Unit III Review
Part D · Comprehensive Quiz

🧠 Comprehensive Quiz — Unit III (20 Questions)


Q1: The sum of absolute deviations from the mean is minimum when measured from the:

A) Mean
B) Median
C) Mode
D) Any value — it doesn't matter

Q2: If y = −2x + 10, and SD(x) = 5, then SD(y) =

A) 5
B) −10
C) 10
D) 0

Q3: The RMSD about a value A is always:

A) Equal to the SD
B) Greater than or equal to the SD
C) Less than the SD
D) Unrelated to the SD

Q4: CV is NOT meaningful when:

A) The SD is very small
B) The mean is zero or negative
C) The data has outliers
D) The sample size is large

Q5: The second central moment (μ₂) equals:

A) Standard deviation
B) Variance
C) Mean
D) MAD

Q6: Sheppard's correction for μ₂ is:

A) μ₂ + c²/12
B) μ₂ − c²/12
C) μ₂ × c²/12
D) μ₂ / (c²/12)

Q7: If Mean = 50, Median = 45, σ = 15, Karl Pearson's SK (using median) is:

A) +0.33
B) +1.0
C) −1.0
D) +0.11

Q8: Bowley's coefficient of skewness uses:

A) Mean and SD
B) Quartiles Q₁, Q₂, Q₃
C) Central moments
D) Percentiles P₁₀ and P₉₀

Q9: A distribution with β₂ = 1.8 is:

A) Leptokurtic
B) Mesokurtic
C) Platykurtic
D) Highly skewed

Q10: If μ₁' = 8 and μ₂' = 80, then the variance is:

A) 80
B) 16
C) 64
D) 72

Q11: Sheppard's correction should NOT be applied to:

A) Continuous data with equal class widths
B) Discrete frequency data
C) The second central moment
D) The fourth central moment

Q12: The moment-based skewness γ₁ is negative when:

A) The right tail is longer
B) The left tail is longer
C) The distribution is symmetric
D) μ₂ is negative

Q13: Two datasets have the same SD but different means. The one with the larger mean has:

A) Smaller CV
B) Larger CV
C) Same CV
D) CV = 0

Q14: Kurtosis is invariant under:

A) Only change of origin
B) Only change of scale
C) Both change of origin and scale (linear transformation)
D) Neither

Q15: The coefficient of variation was introduced by:

A) Ronald Fisher
B) Karl Pearson
C) Arthur Bowley
D) W.F. Sheppard

Q16: For a symmetric distribution, which is always true?

A) β₂ = 3
B) γ₁ = 0
C) SD = 0
D) MAD = 0

Q17: The value c²/12 in Sheppard's correction represents:

A) The mean of the class interval
B) The variance of a uniform distribution over one class
C) The range of the class
D) The SD of the data

Q18: If all values are multiplied by 3, which measure changes?

A) CV
B) γ₁ (skewness)
C) β₂ (kurtosis)
D) SD (it triples)

Q19: Financial returns are typically leptokurtic. This means:

A) Returns are always positive
B) Extreme events occur more often than normal models predict
C) The distribution is symmetric
D) Risk is lower than expected

Q20: Which combination correctly describes a right-skewed, leptokurtic distribution?

A) γ₁ < 0, β₂ < 3
B) γ₁ = 0, β₂ = 3
C) γ₁ > 0, β₂ > 3
D) γ₁ > 0, β₂ < 3
Unit III Review
Part D · Score Tracker

📊 Quiz Score Tracker


Click the button below to see how many questions you answered correctly:



🎯 Score Guide

18–20: Excellent! You have mastered Unit III.
14–17: Good. Review the topics where you made errors.
10–13: Needs improvement. Revisit the formula sheet and redo the solved problems.
Below 10: Study the lecture notes again, focusing on one topic at a time.

Unit III Review
Part A · Summary Map

🗺️ Unit III — Complete Topic Map


LectureTopicKey FormulasKey Concepts
17Mean Absolute DeviationMAD = (1/n)Σ|xᵢ−x̄|Simple, intuitive; uses absolute values; less affected by outliers than SD
18Standard Deviation & RMSDσ = √[Σ(xᵢ−x̄)²/n]; RMSD(A) ≥ SDMost widely used; algebraically tractable; SD = minimum RMSD
19Coefficient of VariationCV = (σ/x̄)×100%Unit-free; compares variability across different scales; needs positive mean
20Moments & Sheppard's Correctionμᵣ = (1/n)Σ(xᵢ−x̄)ʳ; μ₂ᶜ = μ₂−c²/12Raw vs central moments; conversion formulas; correction for grouped data
21Measures of SkewnessSK=(Mean−Mode)/σ; SQ=(Q₃+Q₁−2Q₂)/(Q₃−Q₁); γ₁=μ₃/μ₂^(3/2)Asymmetry; Mean>Median>Mode for right-skew; three approaches
22Measures of Kurtosisβ₂ = μ₄/μ₂²; γ₂ = β₂−3Tail heaviness; leptokurtic (>3), mesokurtic (=3), platykurtic (<3)
Unit III Review
Unit III · Lecture 23

Unit III Complete!


You now have a complete toolkit for describing data: centre, spread, and shape.


Coming Up Next

Unit IV: Correlation & Regression Analysis
Lecture 24–25: Correlation — Karl Pearson's & Spearman's Rank
Lecture 26–27: Simple Linear Regression
Lecture 28: Review & Problem-Solving

Unit III · Lecture 23 · BDS113 Introduction to Statistics · B.Sc. Data Science, Semester II