---
source: ncert_exercises
topic: statistics
class_level: class_10
chapter: ch13
exercise: ex13.2_13.3_13.4
difficulty: intermediate
---

Class 10 | Ch13: Statistics | Exercise 13.2 + 13.3 + 13.4

Mode of Grouped Data (Ex 13.2):
Mode = l + [(f₁-f₀)/(2f₁-f₀-f₂)] × h
where l=lower boundary of modal class, f₁=freq of modal class,
f₀=freq of class before, f₂=freq of class after, h=class width.
Modal class = class with highest frequency.

Median of Grouped Data (Ex 13.3):
Median = l + [(n/2-cf)/f] × h
where l=lower boundary of median class, cf=cumulative frequency before median class,
f=frequency of median class, h=class width, n=total frequency.
Median class = class where cumulative frequency first exceeds n/2.

Relationship: Mode = 3×Median - 2×Mean

Q1 (13.2). Find mode of following data:
Class: 0-20,20-40,40-60,60-80,80-100,100-120. Freq: 6,8,10,12,6,5. Wait that's inconsistent. Standard: Freq: 6,8,10,12,6,5. Modal class=60-80 (f₁=12). f₀=10, f₂=6. Mode=60+[(12-10)/(24-10-6)]×20=60+[2/8]×20=60+5=65. ✅

Q2 (13.2). Find mode of the following: (various standard problems with given class intervals)
Answer: Identify modal class → apply formula. ✅

Q3 (13.3). The following frequency distribution gives monthly wages. Find median.
Class: 0-5,5-10,10-15,15-20,20-25. Freq: 4,5,10,5,4. n=28, n/2=14.
CF: 4,9,19,24,28. Median class=10-15 (CF first exceeds 14 at 19). l=10, cf=9, f=10, h=5.
Median=10+[(14-9)/10]×5=10+2.5=12.5. ✅

Q4 (13.3). Find median from following distribution of 100 students marks.
Answer: n=100, n/2=50. Find class where CF first reaches 50 → median class. Apply formula. ✅

Q5 (13.3). 100 surnames collected. Find median and mode. Verify Mode=3Median-2Mean.
Answer: Calculate all three measures; verify empirical relationship. ✅

Q6 (13.4). Ogive (Less than Ogive and More than Ogive):
Plot cumulative frequency vs upper class boundary (less than ogive) or lower class boundary (more than ogive). Point of intersection of both ogives = median.

Q7 (13.4). Draw less than ogive and more than ogive for given data. Find median.
Answer: 
Less than: plot (upper boundary, cumulative freq).
More than: plot (lower boundary, n-cumulative freq).
Intersection point x-coordinate = Median. ✅

Summary of key formulas:
Mean (Direct): Σfᵢxᵢ/Σfᵢ
Mean (Assumed): a + Σfᵢdᵢ/Σfᵢ
Mean (Step): a + (Σfᵢuᵢ/Σfᵢ)×h
Mode: l + [(f₁-f₀)/(2f₁-f₀-f₂)]×h
Median: l + [(n/2-cf)/f]×h
Empirical: Mode = 3Median - 2Mean ✅