CBSE • Class 10 • Mathematics • Chapter 13 (Statistics) • Exercise 13.3

Exercise 13.3: Statistics — NCERT Solutions

Median of grouped data and cumulative-frequency interpretation.

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What this exercise covers

Cumulative frequencyMedian classMedian formula

Step-by-step solutions — Exercise 13.3

4 solved questions • Each solution includes a Socratic hint, full working and a common-mistake callout • Last reviewed 2026-09-03

Exercise 13.3 Q1 • 3 marks

Find the median of the following distribution of weekly wages (in ₹): 100–120 (f=12), 120–140 (f=14), 140–160 (f=8), 160–180 (f=6), 180–200 (f=10).
Hint (Socratic — try this first)▾
What is n2\frac{n}{2}, and in which class does the cumulative frequency first reach or exceed it?
Step-by-step solution▾

Median formula: Median=l+(n2−cff)×h\text{Median} = l + \left(\dfrac{\frac{n}{2} - cf}{f}\right)\times h.

Build the cumulative frequency (cf) table:

| Class | ff | cf | |---|---|---| | 100–120 | 12 | 12 | | 120–140 | 14 | 26 | | 140–160 | 8 | 34 | | 160–180 | 6 | 40 | | 180–200 | 10 | 50 |

n=50n = 50, so n2=25\frac{n}{2} = 25.

The cf just reaching 25 is 26, so the median class is 120–140.

  • l=120l = 120, cf=12cf = 12 (cf of class before), f=14f = 14, h=20h = 20

Median=120+(25−1214)×20=120+1314×20\text{Median} = 120 + \left(\frac{25 - 12}{14}\right)\times 20 = 120 + \frac{13}{14}\times 20 =120+18.57=138.57= 120 + 18.57 = 138.57

The median wage is approximately ₹138.57.

Common mistake:
Using the cf of the median class instead of the cf of the class immediately preceding it for cfcf.
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Exercise 13.3 Q2 • 4 marks

The median of the following distribution is 28.5 and the total frequency is 60. Find the missing frequencies xx and yy: 0–10 (f=5), 10–20 (f=xx), 20–30 (f=20), 30–40 (f=15), 40–50 (f=yy), 50–60 (f=5).
Hint (Socratic — try this first)▾
You have two unknowns — can the total-frequency condition and the median formula give you two equations?
Step-by-step solution▾

Equation 1 (total frequency = 60): 5+x+20+15+y+5=60  ⟹  x+y=155 + x + 20 + 15 + y + 5 = 60 \implies x + y = 15

Since median =28.5= 28.5 lies in 20–30, that is the median class.

cf before median class =5+x= 5 + x, f=20f = 20, l=20l = 20, h=10h = 10, n2=30\frac{n}{2} = 30.

28.5=20+(30−(5+x)20)×1028.5 = 20 + \left(\frac{30 - (5+x)}{20}\right)\times 10 8.5=25−x28.5 = \frac{25 - x}{2} 17=25−x  ⟹  x=817 = 25 - x \implies x = 8

From x+y=15x + y = 15: y=15−8=7y = 15 - 8 = 7.

So x=8x = 8, y=7y = 7.

Common mistake:
Forgetting to use both equations, or plugging the wrong cumulative frequency into the median formula.
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Exercise 13.3 Q3 • 4 marks

The distribution of marks of 100 students is given below. Draw the 'less than' cumulative frequency table and use it to identify the median class: marks 0–10 (f=10), 10–20 (f=15), 20–30 (f=25), 30–40 (f=30), 40–50 (f=20).
Hint (Socratic — try this first)▾
For a 'less than' ogive, what value do you accumulate up to the upper boundary of each class?
Step-by-step solution▾

'Less than' cumulative frequency table (accumulate up to upper class boundaries):

| Marks less than | Cumulative frequency | |---|---| | 10 | 10 | | 20 | 25 | | 30 | 50 | | 40 | 80 | | 50 | 100 |

Here n=100n = 100, so n2=50\frac{n}{2} = 50.

The cumulative frequency first reaches 50 at 'less than 30', meaning entries beyond cf = 50 lie in 30–40. Since n2=50\frac{n}{2}=50 is reached exactly at cf of class 20–30, the median class is 30–40 (the class where cf first exceeds 50 for the next value).

Computing the median with l=30l = 30, cf=50cf = 50, f=30f = 30, h=10h = 10: Median=30+(50−5030)×10=30\text{Median} = 30 + \left(\frac{50 - 50}{30}\right)\times 10 = 30

So the median is 30 marks, and the 'less than' table shows how cf helps locate it.

Common mistake:
Plotting or accumulating against lower boundaries instead of upper boundaries in a 'less than' cumulative table.
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Exercise 13.3 Q4 • 3 marks

A survey of the lifetimes (in hours) of 40 electric bulbs gave: 200–300 (f=4), 300–400 (f=8), 400–500 (f=12), 500–600 (f=10), 600–700 (f=6). Find the median lifetime.
Hint (Socratic — try this first)▾
Once you have the cumulative frequencies, which interval contains the n2\frac{n}{2}-th observation?
Step-by-step solution▾

Cumulative frequency table:

| Class | ff | cf | |---|---|---| | 200–300 | 4 | 4 | | 300–400 | 8 | 12 | | 400–500 | 12 | 24 | | 500–600 | 10 | 34 | | 600–700 | 6 | 40 |

n=40n = 40, n2=20\frac{n}{2} = 20. The cf first reaching 20 is 24, so the median class is 400–500.

  • l=400l = 400, cf=12cf = 12, f=12f = 12, h=100h = 100

Median=400+(20−1212)×100=400+812×100\text{Median} = 400 + \left(\frac{20 - 12}{12}\right)\times 100 = 400 + \frac{8}{12}\times 100 =400+66.67=466.67= 400 + 66.67 = 466.67

The median lifetime is approximately 466.67 hours.

Common mistake:
Taking h=10h = 10 out of habit instead of the actual class width h=100h = 100 for this data.
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How to approach Exercise 13.3

  1. Re-read the chapter summary first. Open Statistics and refresh the key concepts: Grouped data, Mean, Median, Mode.
  2. Try each problem yourself before opening the solution. Ten focused minutes per problem usually beats reading two finished solutions.
  3. Use Guru AI for the questions you get stuck on. The Socratic AI tutor walks you through it with hints instead of dictating the answer.
  4. Mark the questions you got wrong and revisit them after 24 hours — the spacing is what locks the method into long-term memory.

All exercises in Statistics

  1. Exercise 13.1 — Mean of grouped data using the direct, assumed-mean and step-deviation methods.
  2. Exercise 13.2 — Mode of grouped data using the standard formula.
  3. Exercise 13.3 — Median of grouped data and cumulative-frequency interpretation.

← Exercise 13.2

Mode of grouped data using the standard formula.

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