Exercise 14.3 — Median of Grouped Data
Median of the grouped data.
Exercise 14.3 — Statistics: Median of Grouped Data
Class 10 Mathematics · CBSE, Telangana & Andhra Pradesh Syllabus · Chapter 14
Exercise 14.3 is the final exercise of Chapter 14, Statistics, in Class 10 Mathematics (CBSE, Telangana & Andhra Pradesh syllabus). It focuses entirely on the median — the value of the middle-most observation in a data set — and shows how to calculate it for both small lists of raw numbers and large grouped frequency tables. Several questions also bring back mean and mode from the earlier exercises so that all three measures of central tendency can be compared side by side.
Below is a complete walkthrough of the median formula for ungrouped and grouped data, followed by fully worked, step-by-step solutions to all 7 problems in this exercise — covering electricity bills, policy holders' ages, leaf lengths, lamp lifetimes, surnames, and students' weights — plus a quick-reference table of every final answer for fast revision.
Median of Ungrouped Data
When the raw list of observations is small enough to write out, finding the median is straightforward: arrange the values in ascending order, count how many values there are (n), and then pick out (or average) the middle value depending on whether n is odd or even.
n odd → Median = [(n+1)/2]th observation | n even → Median = mean of (n/2)th and [(n/2)+1]th observations
Odd Number of Values
Data: 3, 7, 5, 11, 6, 18, 9 (n = 7)
Sorted: 3, 5, 6, 7, 9, 11, 18
Middle (4th) value → Median = 7
Even Number of Values
Data: 3, 7, 5, 11, 6, 18, 9, 16 (n = 8)
Sorted: 3, 5, 6, 7, 9, 11, 16, 18
Average of 4th & 5th values: (7+9)/2 → Median = 8
Median of Grouped Data — Step by Step
With grouped (class-interval) data, individual values are not visible, so the median has to be estimated using a cumulative frequency table and a formula. Follow these steps for every problem in this exercise:
- Step 1: Build a table with the class intervals, their frequency (f), and the running total — the cumulative frequency (cf).
- Step 2: Add up all frequencies to get the total number of observations, n, and calculate n/2.
- Step 3: Find the median class — the first class interval whose cumulative frequency is greater than or equal to n/2.
- Step 4: Apply the median formula using the lower boundary of that class.
Median = l + [(n/2 − cf) / f] × h
- l — lower boundary of the median class
- n — total number of observations
- cf — cumulative frequency of the class just before the median class
- f — frequency of the median class itself
- h — size (width) of the median class
A cumulative-frequency curve (ogive) rises step by step — the median is the class where the curve crosses the n/2 mark on the vertical axis.
Quick Recap: Mean and Mode Formulas (Grouped Data)
Several problems in this exercise ask you to also compute the mean and mode so all three measures can be compared. Here are the formulas you'll need, carried over from the earlier exercises of this chapter:
Mean
a + (Σfᵢuᵢ/Σfᵢ) × h
Step-deviation / assumed-mean method, where a is the assumed mean and uᵢ = (xᵢ − a)/h
Median
l + [(n/2 − cf)/f] × h
Based on the cumulative frequency table and the median class
Mode
l + [(f₁−f₀)/(2f₁−f₀−f₂)] × h
f₁ = modal class frequency, f₀ = preceding, f₂ = succeeding
Problem 1 — Median, Mean and Mode of Electricity Consumption
Question: The table below gives the monthly electricity consumption of 68 consumers. Find the median, mean, and mode of the data, and compare them.
| Monthly Consumption | 65–85 | 85–105 | 105–125 | 125–145 | 145–165 | 165–185 | 185–205 |
|---|---|---|---|---|---|---|---|
| No. of consumers (f) | 4 | 5 | 13 | 20 | 14 | 8 | 4 |
| Class Interval | f | Cumulative Frequency (cf) |
|---|---|---|
| 65 – 85 | 4 | 4 |
| 85 – 105 | 5 | 9 |
| 105 – 125 | 13 | 22 |
| 125 – 145 (median class) | 20 | 42 |
| 145 – 165 | 14 | 56 |
| 165 – 185 | 8 | 64 |
| 185 – 205 | 4 | 68 |
| Class Mark (xᵢ) | 75 | 95 | 115 | 135 (a) | 155 | 175 | 195 |
|---|---|---|---|---|---|---|---|
| uᵢ = (xᵢ−a)/h | −3 | −2 | −1 | 0 | 1 | 2 | 3 |
| fᵢuᵢ | −12 | −10 | −13 | 0 | 14 | 16 | 12 |
Problem 2 — Finding Unknown Frequencies x and y
Question: The median of the following 60 observations is 28.5. Find the missing frequencies x and y.
| Class Interval | f | Cumulative Frequency (cf) |
|---|---|---|
| 0 – 10 | 5 | 5 |
| 10 – 20 | x | 5+x |
| 20 – 30 (median class) | 20 | 25+x |
| 30 – 40 | 15 | 40+x |
| 40 – 50 | y | 40+x+y |
| 50 – 60 | 5 | 45+x+y |
Problem 3 — Median Age of Policy Holders
Question: A life insurance agent recorded the ages of 100 policy holders as a "less than" cumulative table. Find the median age (policies are issued from age 18 up to, but not including, 60).
The "Below" values are already cumulative frequencies, so the class-wise frequencies are found by subtracting consecutive totals:
| Age (years) | f | Cumulative Frequency (cf) |
|---|---|---|
| 15 – 20 | 2 | 2 |
| 20 – 25 | 4 | 6 |
| 25 – 30 | 18 | 24 |
| 30 – 35 | 21 | 45 |
| 35 – 40 (median class) | 33 | 78 |
| 40 – 45 | 11 | 89 |
| 45 – 50 | 3 | 92 |
| 50 – 55 | 6 | 98 |
| 55 – 60 | 2 | 100 |
Problem 4 — Median Length of Leaves
Question: The lengths of 40 leaves are grouped into classes such as 118–126, 127–135, etc. Find the median length.
Since the data is rounded to the nearest mm, the classes are first converted into continuous boundaries (118–126 becomes 117.5–126.5, and so on) before applying the median formula:
| Length (mm), continuous | f | Cumulative Frequency (cf) |
|---|---|---|
| 117.5 – 126.5 | 3 | 3 |
| 126.5 – 135.5 | 5 | 8 |
| 135.5 – 144.5 | 9 | 17 |
| 144.5 – 153.5 (median class) | 12 | 29 |
| 153.5 – 162.5 | 5 | 34 |
| 162.5 – 171.5 | 4 | 38 |
| 171.5 – 180.5 | 2 | 40 |
Problem 5 — Median Lifetime of Neon Lamps
Question: The lifetimes (in hours) of 400 neon lamps are grouped into class intervals of width 500. Find the median lifetime.
| Lifetime (hours) | f | Cumulative Frequency (cf) |
|---|---|---|
| 1500 – 2000 | 14 | 14 |
| 2000 – 2500 | 56 | 70 |
| 2500 – 3000 | 60 | 130 |
| 3000 – 3500 (median class) | 86 | 216 |
| 3500 – 4000 | 74 | 290 |
| 4000 – 4500 | 62 | 352 |
| 4500 – 5000 | 48 | 400 |
Problem 6 — Median, Mean and Mode of Surname Lengths
Question: 100 surnames were picked from a telephone directory and grouped by the number of letters. Find the median, mean, and modal number of letters.
| No. of Letters | f | Cumulative Frequency (cf) |
|---|---|---|
| 1 – 4 | 6 | 6 |
| 4 – 7 | 30 | 36 |
| 7 – 10 (median class) | 40 | 76 |
| 10 – 13 | 16 | 92 |
| 13 – 16 | 4 | 96 |
| 16 – 19 | 4 | 100 |
| Class Mark (xᵢ) | 2.5 | 5.5 | 8.5 (a) | 11.5 | 14.5 | 17.5 |
|---|---|---|---|---|---|---|
| dᵢ = xᵢ−a | −6 | −3 | 0 | 3 | 6 | 9 |
| fᵢdᵢ | −36 | −90 | 0 | 48 | 24 | 36 |
Problem 7 — Median Weight of Students
Question: The weights of 30 students in a class are grouped as shown below. Find the median weight.
| Weight (kg) | f | Cumulative Frequency (cf) |
|---|---|---|
| 40 – 45 | 2 | 2 |
| 45 – 50 | 3 | 5 |
| 50 – 55 | 8 | 13 |
| 55 – 60 (median class) | 6 | 19 |
| 60 – 65 | 6 | 25 |
| 65 – 70 | 3 | 28 |
| 70 – 75 | 2 | 30 |
Common Mistakes to Avoid
- Choosing the wrong median class: The median class is the first class whose cumulative frequency is greater than or equal to n/2 — not the class containing n/2 itself as a frequency value.
- Using f instead of cf in the formula, or vice-versa: Remember cf in the formula is the cumulative frequency of the class before the median class, while f is the plain (non-cumulative) frequency of the median class.
- Forgetting to convert inclusive classes: As seen in Problem 4, classes like 118–126, 127–135 must first be converted to continuous boundaries (117.5–126.5, etc.) before the formula can be applied.
- Mixing up "below" cumulative tables with frequency tables: In Problem 3, the given numbers were already cumulative ("Below 20", "Below 25"...) — they must be subtracted from each other first to recover the class-wise frequencies.
- Sign errors when solving for unknown frequencies: In problems like Problem 2, carefully substitute the median class data into the formula and simplify step by step rather than rearranging too many terms at once.
- Rounding too early: Carry at least 2 decimal places through intermediate steps (especially in the mean's step-deviation method) to avoid small errors compounding into the final answer.
Quick Reference — All 7 Answers at a Glance
| Q.No | Scenario | Key Result |
|---|---|---|
| 1 | Electricity consumption (68 consumers) | Median = 137, Mean ≈ 137, Mode = 135.7 |
| 2 | Unknown frequencies x, y (median 28.5) | x = 8, y = 7 |
| 3 | Median age of 100 policy holders | 35.76 years |
| 4 | Median length of 40 leaves | 146.75 mm |
| 5 | Median lifetime of 400 neon lamps | 3406.98 hours |
| 6 | Surname letters (100 surnames) | Median = 8.05, Mean = 8.32, Mode = 7.88 |
| 7 | Median weight of 30 students | 56.67 kg |
What This Exercise Prepares You For
Exercise 14.3 completes the picture started by Exercise 14.1 (Mean of Grouped Data) and Exercise 14.2 (Mode of Grouped Data) — together, these three exercises let you compute and compare all three measures of central tendency for any frequency distribution, a skill that is tested almost every year in board exams.
The cumulative frequency tables built in this exercise are also the foundation for plotting an ogive (cumulative frequency curve) and for graphically estimating the median, which is often covered as a follow-up topic. Once this exercise feels comfortable, it connects naturally to the Probability chapter, where organising data into frequency tables is just as essential.