Chapter 13 · Kerala SSLC Class 10 Maths
Statistics
Two ways to learn — a crisp formula reference, or a story that builds from a skewed household income to the median's deep link with probability.
Five scenes from a skewed mean to the median's link with probability.
How this works
Five scenes build from one skewed mean all the way to the median's hidden link with probability. Read the story, then tap "The Maths" for the formal result.
The Household That Skewed Everything
Ten households have these monthly incomes (in rupees): 16500, 21700, 18600, 21050, 19500, 17000, 21000, 18000, 22000, 17500. Add them up and divide by 10 — the mean is 19285.
That single number tells a useful story: incomes cluster around 19285, roughly half the households earn more and half less, and none is wildly different from the rest.
Someone earning 175000 rupees a month moves into the neighbourhood. What's the new mean, now with 11 households?
New mean = (19285×10 + 175000) / 11
= (192850+175000)/11
≈ 33441 rupees
The mean jumped to almost 33441 — nearly double the original! But does that number actually describe the neighbourhood well? Ten of the eleven households still earn far less than that.
The mean's job is to compress a whole collection of numbers into one representative number — but a single very large (or very small) value can drag it far from where most of the data actually sits.
Finding the True Middle
Instead, arrange all 11 incomes in increasing order: 16500, 17000, 17500, 18000, 18600, 19500, 21000, 21050, 21700, 22000, 175000.
Take the number exactly in the middle — the 6th of these 11 values. That's 19500. It's called the median: 5 households earn less than this, and 5 earn more.
Median (11 values, odd count) = the ((11+1)/2)th = 6th value = 19500
Unlike the mean, the huge income of 175000 barely disturbs this — it's just the last value in the line, however far out it sits.
With only 10 values (an even count), there's no single middle position — instead there are TWO middle values.
Ordered (10 values): 16500,17000,17500,18000,18600,19500,21000,21050,21700,22000
Middle two (5th, 6th): 18600, 19500
The median is the average of these two:
Median = (18600+19500)/2 = 19050
Kids in a Line
25 children are tested for haemoglobin level and sorted into groups. To find the median, imagine lining up all 25 children in order of their level — we want the value of the 13th child (12 would be below, 12 above).
n=25 → median is the ((25+1)/2)=13th value in order
Add up the frequencies group by group to see where the 13th child falls.
| Haemoglobin level | Frequency | Cumulative |
|---|---|---|
| 12.0 | 2 | 2 |
| 12.4 | 3 | 5 |
| 12.7 | 5 | 10 |
| 13.1 | 6 | 16 |
| 13.3 | 4 | 20 |
| 13.6 | 3 | 23 |
| 14.0 | 2 | 25 |
The 13th child falls within positions 11–16 — all of whom are at level 13.1. So that's the median.
Splitting a Class in Half
Now the data is grouped into class INTERVALS, not exact values. 41 factory workers are sorted by daily wage into classes 400–500, 500–600, ..., 900–1000, with frequencies 6, 7, 10, 9, 5, 4.
With 41 workers, the median is the wage of the 21st worker in order. Cumulative frequencies: 6, 13, 23, 32, 37, 41 — the 21st worker falls in the 600–700 class (which spans positions 14 to 23).
We don't know each worker's EXACT wage within that class — so we assume they're evenly spaced, forming an arithmetic sequence across the class width.
Class width 100, split into 10 equal steps (one per worker): each step = 10 rupees
14th worker's assumed wage = midpoint of the first step = 605
Each subsequent worker's wage increases by 10 (an arithmetic sequence)
The 21st worker is 7 positions after the 14th:
21st worker's wage = 605 + (7×10) = 675 rupees
46 office employees are sorted by age into 5-year classes. With an even count, we need BOTH the 23rd and 24th workers — and this time, both happen to fall in the SAME class, 40–45 (10 workers, positions 20–29).
step = 5/10 = 0.5 years
20th worker's assumed age = 40.25 (midpoint of first step)
23rd worker: 40.25 + 3(0.5) = 41.75
24th worker: 40.25 + 4(0.5) = 42.25
Median age = (41.75+42.25)/2 = 42
One more subtlety: sometimes the cumulative frequency lands EXACTLY on a class boundary. In a class of 40 children sorted by marks (classes of width 10), the cumulative frequency reaches EXACTLY 20 — half of 40 — right at "below 30".
The Median's Hidden Symmetry
Draw a histogram of the age data from Scene 4, and draw a vertical line through the median, 42.
That line splits the histogram into two regions — and remarkably, both regions have EXACTLY the same area, even though the bars around 42 aren't a clean shape.
This gives the median a lovely connection to probability: if you pick a random point from that histogram, it is exactly as likely to land on the left of the median as on the right.
The Big Picture
Every idea in this chapter is really about ONE measure — the median — and how to pin it down exactly, no matter how the data is presented.
Mean misled by outliers?
check the median too — it's not affected
Median, odd count?
the ((n+1)/2)th ordered value
Median, even count?
average of the two middle ordered values
Median from a frequency table?
use cumulative frequency to find the position
Median inside a class interval?
step = class width / frequency
Median's deeper meaning?
P(below)=P(above)=1/2
Want to go deeper?
Our tutors cover the full Kerala SSLC syllabus live, 1-on-1. If Chapter 13 still feels shaky after this, one session usually fixes it.
Book a Free Trial Class