Introduction to Statistics

Unit II · Lecture 15

BDS113 — Introduction to Statistics

Unit II — Review & Problem-Solving

Consolidating Lectures 6–14 · Formula Sheet · Exam-Style Practice

Unit II · Lecture 15 of 40 · B.Sc. Data Science, 1st Year, Semester II

Introduction to Statistics

Unit II · Lecture 15

🗺️ Unit II Roadmap — What We Covered (Lectures 6–14)

📋

Lectures 6–7 — Data Organisation & Graphical Representation

📐

Lectures 8–10 — Arithmetic Mean (ungrouped, grouped, weighted, combined)

📊

Lectures 11–12 — Median, Mode & Empirical Relationship

📏

Lectures 13–14 — Quartiles, Deciles, Percentiles & Box Plot

💬 Today's Goal: We will review every formula, clarify common confusions, solve exam-style problems step by step, and build your exam strategy.

Introduction to Statistics

Unit II · Lecture 15

📖 Master Formula Sheet — All of Unit II

Ungrouped Mean

x̄ = Σxᵢ / n

Grouped Mean (Direct)

x̄ = Σfᵢxᵢ / Σfᵢ

xᵢ = class mark = (upper+lower)/2

Assumed Mean (Shortcut)

x̄ = A + Σfᵢdᵢ / Σfᵢ

dᵢ = xᵢ − A (A = assumed mean)

Step-Deviation

x̄ = A + (Σfᵢuᵢ / Σfᵢ) × h

uᵢ = (xᵢ − A)/h (h = class width)

Weighted Mean

x̄_w = Σwᵢxᵢ / Σwᵢ

Combined Mean

x̄₁₂ = (n₁x̄₁ + n₂x̄₂) / (n₁+n₂)

Property: Σ(xᵢ − x̄) = 0 always. The mean is the balancing point of data.

Ungrouped (Odd n)

M = value at position (n+1)/2

Ungrouped (Even n)

M = avg of values at n/2 & (n/2+1)

Grouped Data

Median = L + [(N/2 − CF) / f] × h

Median class = where CF first ≥ N/2. L = its lower limit, CF = CF before it, f = its frequency, h = class width

Key: Not affected by outliers. Can handle open-end classes. Q₂ = Median = P₅₀ = D₅.

Grouped Mode

Mode = L + [(f₁−f₀) / (2f₁−f₀−f₂)] × h

Modal class = highest frequency. f₁ = its freq, f₀ = freq before, f₂ = freq after, L = its lower limit, h = class width

Key: Works for qualitative data too. Can have no mode, one mode (unimodal), or multiple modes. Not unique for grouped data.

Karl Pearson's Empirical Formula

Mode = 3 Median − 2 Mean

Equivalently: Mean − Mode = 3(Mean − Median)
Rearranged: Median = (Mode + 2·Mean)/3 · Mean = (3·Median − Mode)/2

Distribution	Order
Symmetric	Mean = Median = Mode
Right-skewed (+ve skew)	Mean > Median > Mode
Left-skewed (−ve skew)	Mode > Median > Mean

Ungrouped Quartile

Qₖ = value at k(n+1)/4

k = 1, 2, 3

Grouped Quartile

Qₖ = L + [(kN/4−CF)/f] × h

IQR & Quartile Deviation

IQR = Q₃−Q₁ Q.D. = (Q₃−Q₁)/2

Coefficient of Q.D.

C.Q.D. = (Q₃−Q₁)/(Q₃+Q₁)

Deciles — Ungrouped

Dₖ = value at k(n+1)/10

k = 1, 2, ..., 9

Deciles — Grouped

Dₖ = L + [(kN/10−CF)/f] × h

Percentiles — Ungrouped

Pₖ = value at k(n+1)/100

k = 1, 2, ..., 99

Percentiles — Grouped

Pₖ = L + [(kN/100−CF)/f] × h

Master Equivalences: Q₁ = D₂ = P₂₅ | Q₂ = D₅ = P₅₀ = Median | Q₃ = D₈ = P₇₅

Introduction to Statistics

Unit II · Lecture 15

💡 The One Formula That Rules Them All

The Universal Partition Value Formula (Grouped Data)

Value = L + [(Position − CF) / f] × h

Median: Position = N/2
Quartile Qₖ: Position = kN/4
Decile Dₖ: Position = kN/10
Percentile Pₖ: Position = kN/100

In every case: Find the class where CF first reaches or exceeds the Position. Then apply the formula with that class's L, f, h and the CF of the class before it.

Introduction to Statistics

Unit II · Lecture 15

⚠️ Top 10 Common Mistakes & Confusions

❌ Mistakes to Avoid

1. Using upper limit as xᵢ — Always use the class mark (midpoint), not the upper or lower limit.

2. Forgetting to order data for median/mode/quartiles — Data must be in ascending order first.

3. Wrong CF for grouped partition values — CF used in formula is of the class before the target class, not of the target class itself.

4. Confusing "N/2" with "n/2" — N = Σf for grouped data. n = number of observations for ungrouped data.

5. Assuming Mode = 3Median−2Mean is exact — It is an approximation for moderately skewed data only.

💡 Tricky Exam Scenarios

6. Even n median — Must average the two middle values, not pick one.

7. Position is fractional — For ungrouped quartiles/deciles/percentiles, interpolate between adjacent values.

8. Open-end classes — Mean cannot be calculated, but median/mode/quartiles can be.

9. Qualitative data — Only Mode works. Mean and Median (ordinal) have limited use.

10. Class width not uniform — The formula still works; just use the correct h for each class.

Introduction to Statistics

Unit II · Lecture 15

🧭 Decision Flowchart: Which Average to Use?

Introduction to Statistics

Unit II · Lecture 15

📝 Problem Set 1 — Mean

Problem 1.1 — Grouped Mean (Direct Method)

Find the mean marks from the following distribution:

Class	0–10	10–20	20–30	30–40	40–50
f	4	8	14	10	4

📝 Solution 1.1

Class marks (xᵢ): 5, 15, 25, 35, 45
fᵢxᵢ: 4×5=20, 8×15=120, 14×25=350, 10×35=350, 4×45=180
Σfᵢ = 40, Σfᵢxᵢ = 1020
x̄ = 1020/40 = 25.5 marks

Problem 1.2 — Step-Deviation Method

Using the same data as 1.1, find the mean using the step-deviation method with A = 25, h = 10.

📝 Solution 1.2

uᵢ = (xᵢ − 25)/10: u = −2, −1, 0, 1, 2
fᵢuᵢ: 4(−2)=−8, 8(−1)=−8, 14(0)=0, 10(1)=10, 4(2)=8
Σfᵢuᵢ = 2, Σfᵢ = 40
x̄ = 25 + (2/40) × 10 = 25 + 0.5 = 25.5 ✓ (Same answer!)

Problem 1.3 — Combined Mean

Section A has 30 students with mean marks 65. Section B has 20 students with mean marks 78. Find the combined mean.

📝 Solution 1.3

x̄₁₂ = (n₁x̄₁ + n₂x̄₂) / (n₁+n₂)
= (30×65 + 20×78) / (30+20)
= (1950 + 1560) / 50
= 3510/50 = 70.2 marks

Problem 1.4 — Weighted Mean

A student scores: Statistics 72 (weight 4), Maths 65 (weight 3), English 80 (weight 2), CS 85 (weight 1). Find the weighted mean.

📝 Solution 1.4

Σwᵢxᵢ = 4(72)+3(65)+2(80)+1(85) = 288+195+160+85 = 728
Σwᵢ = 4+3+2+1 = 10
x̄_w = 728/10 = 72.8

Introduction to Statistics

Unit II · Lecture 15

📝 Problem Set 2 — Median & Mode

Problem 2.1 — Grouped Median

Find the median:

Class	0–20	20–40	40–60	60–80	80–100
f	6	10	18	12	4

📝 Solution 2.1

N = 50. N/2 = 25. CF: 6, 16, 34, 46, 50.
CF first ≥ 25 at 40–60 (CF=34). Median class = 40–60.
L=40, CF(before)=16, f=18, h=20
Median = 40 + [(25−16)/18] × 20 = 40 + [9/18]×20 = 40 + 10 = 50

Problem 2.2 — Grouped Mode

Find the mode for the same data as 2.1.

📝 Solution 2.2

Highest frequency = 18 at class 40–60 → Modal class = 40–60
L=40, f₁=18, f₀=10(freq before), f₂=12(freq after), h=20
Mode = 40 + [(18−10)/(2×18−10−12)] × 20
= 40 + [8/(36−22)] × 20 = 40 + [8/14] × 20 = 40 + 11.43 = 51.43

Problem 2.3 — Empirical Relationship

For the data in 2.1, the Mean = 49.6, Median = 50, Mode = 51.43. Verify the empirical formula and comment on skewness.

📝 Solution 2.3

Empirical: Mode = 3·Median − 2·Mean = 3(50) − 2(49.6) = 150 − 99.2 = 50.8
Actual Mode = 51.43. Difference = 0.63 (small — formula is approximately correct).

Since Mean(49.6) ≈ Median(50) ≈ Mode(51.43), the distribution is approximately symmetric (very slight left skew).

Introduction to Statistics

Unit II · Lecture 15

📝 Problem Set 3 — Quartiles & Partition Values

Problem 3.1 — Q₁, Q₃, IQR, Q.D.

Find Q₁, Q₃, IQR, and Quartile Deviation from the data in Problem 2.1.

📝 Solution 3.1

N=50, CF: 6, 16, 34, 46, 50.

Q₁: N/4=12.5. CF first ≥ 12.5 at 20–40 (CF=16). L=20, CF(before)=6, f=10, h=20.
Q₁ = 20 + [(12.5−6)/10]×20 = 20 + 13 = 33

Q₃: 3N/4=37.5. CF first ≥ 37.5 at 60–80 (CF=46). L=60, CF(before)=34, f=12, h=20.
Q₃ = 60 + [(37.5−34)/12]×20 = 60 + 5.83 = 65.83

IQR = Q₃ − Q₁ = 65.83 − 33 = 32.83
Q.D. = 32.83/2 = 16.42
C.Q.D. = 32.83 / (65.83+33) = 32.83/98.83 = 0.332

Problem 3.2 — Percentiles

Find P₂₅, P₅₀, P₇₅ from the same data. Verify P₂₅ = Q₁.

📝 Solution 3.2

P₂₅: kN/100 = 25×50/100 = 12.5 → same position as Q₁ → P₂₅ = 33 = Q₁ ✓
P₅₀: kN/100 = 50×50/100 = 25 → same position as Median → P₅₀ = 50 = Median ✓
P₇₅: kN/100 = 75×50/100 = 37.5 → same position as Q₃ → P₇₅ = 65.83 = Q₃ ✓

Problem 3.3 — Ungrouped Quartiles

Find Q₁, Q₂, Q₃ for: 15, 22, 28, 33, 38, 42, 48, 55, 62, 70, 78

📝 Solution 3.3

n=11, data is ordered.
Q₁ = pos (12)/4 = 3rd value = 28
Q₂ = pos (12)/2 = 6th value = 42 (= Median)
Q₃ = pos 3(12)/4 = 9th value = 62
IQR = 62 − 28 = 34

Introduction to Statistics

Unit II · Lecture 15

📝 Problem Set 4 — Mixed & Exam-Style Problems

Problem 4.1 — Finding Missing Frequency

The mean of the following distribution is 35.8. Find the missing frequency a:

Class	10–20	20–30	30–40	40–50	50–60
f	5	a	15	10	5

📝 Solution 4.1

xᵢ: 15, 25, 35, 45, 55
fᵢxᵢ: 75, 25a, 525, 450, 275
Σfᵢ = 35+a, Σfᵢxᵢ = 1325+25a

x̄ = (1325+25a)/(35+a) = 35.8
1325+25a = 35.8(35+a) = 1253+35.8a
1325−1253 = 35.8a−25a
72 = 10.8a
a = 72/10.8 ≈ 6.67 → round to 7
(If exact answer needed: a = 72/10.8 = 20/3. Check data — likely a = 5 or integer; recheck: with a=5, mean = (1325+125)/40 = 1450/40 = 36.25. With the exam given 35.8, a is approximately 7.)

Problem 4.2 — Mean from One Missing Value

The mean of 8 observations is 15. Six of them are: 11, 13, 15, 17, 19, 21. One value is known to be less than 10. Find the two missing values if the 7th value is twice the 8th value.

📝 Solution 4.2

Σxᵢ = 8×15 = 120. Sum of 6 known = 11+13+15+17+19+21 = 96.
Let 8th value = k, then 7th value = 2k.
96 + 2k + k = 120 → 3k = 24 → k = 8, 2k = 16
Check: value less than 10 → 8 ✓. Missing values are 8 and 16.

Problem 4.3 — Complete Analysis

For the data below, calculate ALL measures: Mean, Median, Mode, Q₁, Q₃, IQR, Q.D., D₅, P₇₅. Verify the empirical relationship.

Class	10–20	20–30	30–40	40–50	50–60	60–70
f	4	8	12	16	10	5

📝 Solution 4.3 — Complete Analysis

N = 55. xᵢ: 15, 25, 35, 45, 55, 65. CF: 4, 12, 24, 40, 50, 55.

Mean (step-deviation, A=35, h=10):
d: −20,−10,0,10,20,30 → u: −2,−1,0,1,2,3
fu: −8,−8,0,16,20,15 → Σfu = 35
x̄ = 35 + (35/55)×10 = 35 + 6.36 = 41.36

Median: N/2 = 27.5. CF first ≥ 27.5 at 40–50 (CF=40).
M = 40 + [(27.5−24)/16]×10 = 40 + 2.19 = 42.19

Mode: Highest freq = 16 at 40–50. f₁=16, f₀=12, f₂=10.
Z = 40 + [(16−12)/(32−12−10)]×10 = 40 + [4/10]×10 = 44

Empirical check: 3M−2x̄ = 3(42.19)−2(41.36) = 126.57−82.72 = 43.85 ≈ Mode(44) ✓

Q₁: N/4=13.75. CF first ≥ 13.75 at 30–40 (CF=24).
Q₁ = 30 + [(13.75−12)/12]×10 = 30 + 1.46 = 31.46

Q₃: 3N/4=41.25. CF first ≥ 41.25 at 50–60 (CF=50).
Q₃ = 50 + [(41.25−40)/10]×10 = 50 + 1.25 = 51.25

IQR = 51.25−31.46 = 19.79 · Q.D. = 9.90
D₅: 5N/10=27.5 → same as median → D₅ = 42.19 ✓
P₇₅: 75N/100=41.25 → same as Q₃ → P₇₅ = 51.25 ✓

Introduction to Statistics

Unit II · Lecture 15

🎯 Exam Strategy — How to Score Maximum

Before the Exam

1. Memorize the one master formula for grouped partition values.

2. Practice the step-deviation method — it's the fastest for calculations.

3. Know the equivalences: Q₁=D₂=P₂₅ etc.

4. Draw a CF table before solving any grouped problem.

During the Exam

1. Read the question fully — identify if data is grouped or ungrouped.

2. Always arrange data in ascending order for ungrouped problems.

3. Show all steps — CF table, identifying the class, substituting values.

4. Verify using the empirical formula if asked.

Common Exam Questions

1. Calculate Mean, Median, Mode for grouped data.

2. Find missing frequency given the mean.

3. Verify Karl Pearson's empirical relationship.

4. Calculate Q₁, Q₃, IQR, Quartile Deviation.

5. Find specific deciles or percentiles.

Introduction to Statistics

Unit II · Lecture 15

🧠 Comprehensive Quiz — Unit II

Q1: Σ(xᵢ − x̄) for any dataset always equals:

A) A positive number

B) Zero

C) Σxᵢ

D) n

Q2: The class mark of 20–30 is:

A) 20

B) 30

C) 25

D) 10

Q3: If every observation is multiplied by 3, the new mean is:

A) x̄ + 3

B) 3x̄

C) x̄/3

D) x̄

Q4: The median class is identified by:

A) Highest frequency

B) CF first ≥ N/2

C) Middle class in table

D) CF = N

Q5: In a left-skewed distribution:

A) Mean > Median > Mode

B) Mode > Median > Mean

C) Mean = Median = Mode

D) Median > Mode > Mean

Q6: Which can be used for nominal (qualitative) data?

A) Mean

B) Median

C) Mode

D) All three

Q7: P₇₅ is equivalent to:

A) Q₁ and D₂

B) Q₂ and D₅

C) Q₃ and D₈

D) Median

Q8: The step-deviation method formula is:

A) x̄ = A + Σfᵢdᵢ/N

B) x̄ = A + (Σfᵢuᵢ/N) × h

C) x̄ = Σfᵢxᵢ/N

D) x̄ = A + Σfᵢuᵢ

Q9: If Mean=50, Median=45, the Mode (empirical) is approximately:

A) 40

B) 35

C) 55

D) 48

Q10: The IQR of a dataset is 24. The Quartile Deviation is:

A) 24

B) 48

C) 12

D) 6

Q11: For grouped data with N=60, Q₃ class is found where CF first reaches:

A) 15

B) 30

C) 45

D) 60

Q12: The weighted mean is preferred when:

A) All items are equally important

B) Items have different levels of importance

C) Data is qualitative

D) The distribution is skewed

Q13: A box plot displays:

A) Mean and Mode

B) Min, Q₁, Median, Q₃, Max

C) All percentiles

D) Frequency distribution

Q14: For n=30, the position of D₄ is:

A) 12.0

B) 12.4

C) 13.0

D) 4.0

Q15: The mode formula for grouped data uses:

A) All class frequencies

B) Modal class freq + freq of classes before & after it

C) Only the highest frequency

D) Cumulative frequencies

Introduction to Statistics

Unit II · Lecture 15

🔑 Unit II — Final Summary

Central Tendency summarizes data with a single representative value. Three main measures: Mean (mathematical centre), Median (positional centre), Mode (most frequent value).
Mean uses all observations and is algebraically tractable, but is sensitive to outliers. Use for symmetric data. Shortcut methods: assumed mean and step-deviation.
Median is robust to outliers, works for open-end and ordinal data, but ignores extremes. Use for skewed data.
Mode works for qualitative data and finds the most typical value, but may not be unique. Use for nominal data or to find the peak.
Empirical formula: Mode ≈ 3 Median − 2 Mean (moderate skew only). Symmetric: all three equal.
Partition values (Quartiles, Deciles, Percentiles) all follow one unified formula: Value = L + [(Position−CF)/f] × h.
Key equivalences: Q₁=D₂=P₂₅, Q₂=D₅=P₅₀=Median, Q₃=D₈=P₇₅.
IQR = Q₃−Q₁ measures the spread of the middle 50% — robust to outliers. Box plot visualises this beautifully.
Always draw a CF table for grouped data problems. Always order data for ungrouped problems.

Introduction to Statistics

Unit II · Lecture 15

Unit II Complete!

You now have a solid foundation in measures of central tendency and partition values.

Coming Up: Unit III

Measures of Dispersion — Range, Quartile Deviation, Mean Deviation, Standard Deviation, Variance

Skewness & Kurtosis — Shape of distributions beyond the centre

Unit II · Lecture 15 · BDS113 Introduction to Statistics · B.Sc. Data Science, Semester II