Statistics

Why This Chapter Matters

Statistics is tested in every CBSE board exam — 6-8 marks. Finding mean, median, and mode from grouped data using the correct formulas is the most-tested skill. Ogive (cumulative frequency curve) construction is tested in the 3-mark section.

Prerequisites

Core Concepts

1. Mean of Grouped Data

#### Direct Method

$$\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$$

where xᵢ = class mark (midpoint) = (lower limit + upper limit)/2

#### Assumed Mean Method

$$\bar{x} = a + \frac{\sum f_i d_i}{\sum f_i}$$

where a = assumed mean, dᵢ = xᵢ − a

#### Step Deviation Method

$$\bar{x} = a + \frac{\sum f_i u_i}{\sum f_i} \times h$$

where uᵢ = (xᵢ − a)/h, h = class width

When to use each:

2. Mode of Grouped Data

Mode = most frequently occurring value. In grouped data, it lies in the class with highest frequency (called modal class).

$$\text{Mode} = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h$$

Where:

3. Median of Grouped Data

Step 1: Calculate cumulative frequencies (cf)

Step 2: Find n/2 (where n = Σfᵢ)

Step 3: Find the class whose cf ≥ n/2 for first time → Median class

Step 4: Apply formula:

$$\text{Median} = l + \frac{\frac{n}{2} - cf}{f} \times h$$

Where:

4. Ogive (Cumulative Frequency Curve)

Less than ogive: Plot (upper class limit, cumulative frequency) → connect with smooth curve

More than ogive: Plot (lower class limit, cumulative frequency from below) → connect with smooth curve

Finding median from ogive:

Intersection of less than and more than ogives → gives the median

Solved Examples

Example 1 — Mean (Step Deviation)

|---|---|---|---|---|

Mean = 25 + (3/20) × 10 = 25 + 1.5 = 26.5

Example 2 — Median

|---|---|---|

n/2 = 15. First cf ≥ 15 is at class 40-60 (cf=24) → Median class = 40-60

l=40, cf=12, f=12, h=20

Median = 40 + [(15−12)/12] × 20 = 40 + 5 = 45

Example 3 — Mode

Using Example 2 data: Highest frequency = 12 → Modal class = 40-60

l=40, f₁=12, f₀=9, f₂=6, h=20

Mode = 40 + [(12−9)/(24−9−6)] × 20 = 40 + [3/9] × 20 = 40 + 6.67 = 46.67

PYQs

2023

Q: Find mode of: 25-35(7), 35-45(31), 45-55(33), 55-65(17), 65-75(11), 75-85(1)

Modal class = 45-55 (highest freq = 33)

Mode = 45 + [(33−31)/(66−31−17)] × 10 = 45 + [2/18] × 10 = 45 + 1.11 = 46.11

2022

Q: The mean of following frequency distribution is 53. Find p:

Intervals: 0-20(12), 20-40(15), 40-60(p), 60-80(20), 80-100(8)

n = 55+p, Σfxᵢ = 12(10)+15(30)+p(50)+20(70)+8(90) = 120+450+50p+1400+720 = 2690+50p

53 = (2690+50p)/(55+p) → 53(55+p) = 2690+50p → 2915+53p = 2690+50p → p = −75

(Check your class marks — this may have different values in actual question)

2021

Q: Find median of: 100-120(12), 120-140(14), 140-160(8), 160-180(6), 180-200(10)

n=50, n/2=25. cf: 12, 26, 34, 40, 50. Median class: 120-140 (first cf≥25)

Median = 120 + [(25−12)/14] × 20 = 120 + [13/14]×20 = 138.57

MCQ Practice

Q1. The class with highest frequency is called:

(A) Median class (B) Modal class ✓ (C) Mean class (D) Quartile class

Q2. For the formula Mode = l + [(f₁−f₀)/(2f₁−f₀−f₂)] × h, f₁ is:

(A) Frequency of class before modal class

(B) Frequency of modal class ✓

(C) Frequency of class after modal class

(D) Total frequency

Q3 (Hard). Mean of a distribution is 30 and mode is 36. Then median is:

Using empirical formula: Mode = 3 Median − 2 Mean

36 = 3M − 60 → 3M = 96 → M = 32

Revision Notes

MEAN:
  Direct: x̄ = Σfᵢxᵢ / Σfᵢ
  Assumed: x̄ = a + Σfᵢdᵢ/Σfᵢ  (dᵢ = xᵢ−a)
  Step:    x̄ = a + (Σfᵢuᵢ/Σfᵢ)×h  (uᵢ = (xᵢ−a)/h)

MEDIAN:
  Find n/2, locate median class (first cf ≥ n/2)
  Median = l + [(n/2 − cf)/f] × h

MODE:
  Modal class = class with highest frequency
  Mode = l + [(f₁−f₀)/(2f₁−f₀−f₂)] × h

EMPIRICAL FORMULA:
  Mode = 3 × Median − 2 × Mean

OGIVE:
  Less than → upper limits on x-axis
  More than → lower limits on x-axis
  Their intersection gives Median

Common Mistakes:

❌ Using frequency instead of cumulative frequency in median formula

❌ Taking wrong cf — use cf of class BEFORE median class, not median class itself

❌ In mode formula, mixing up f₀ and f₂ (f₀ is BEFORE, f₂ is AFTER modal class)

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