Find the Median of the marks obtained by 5 students in an economics test: 15, 22, 9, 30, 18.

Numerical Problems · Q3

Q. Find the Median of the marks obtained by 5 students in an economics test: 15, 22, 9, 30, 18.

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Concept understanding — Measures of Central Tendency

Measures of central tendency are single values that summarise the centre of a data set. The three mathematical averages are the arithmetic mean (AM), geometric mean (GM) and harmonic mean (HM); the two positional averages are the median and mode.

Each average captures "the typical value" differently — pick the one that suits the data, and use the fixed inequality between AM, GM and HM to sanity-check answers.

How it works

The arithmetic mean adds and divides; the geometric mean multiplies and takes a root (ideal for ratios and growth rates); the harmonic mean handles rates like speed. The median is the middle value in order, and the mode is the most frequent value — both unaffected by extreme outliers.

For two positive numbers a neat identity ties the three mathematical means together, so any two of them determine the third.

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Key properties

$\text{AM}\ge\text{GM}\ge\text{HM}$ for any set of positive values, with equality only when all values are equal.
The identity $\text{GM}^{2}=\text{AM}\times\text{HM}$ holds exactly for two observations.
Median and mode ignore extreme values; the arithmetic mean does not.

Quick example

Given $\text{AM}=64$ and $\text{HM}=16$ , find the GM.

Use $\text{GM}=\sqrt{\text{AM}\times\text{HM}}$ .
$\text{GM}=\sqrt{64\times16}=\sqrt{1024}$ .
$\sqrt{1024}=32$ .
Check the ordering: $64\ge32\ge16$ holds. Result: GM $=32$ .

Watch out

Applying $\text{GM}^{2}=\text{AM}\times\text{HM}$ to more than two observations — it is exact only for two.
Using the arithmetic mean for average speed or average rates, where the harmonic mean is correct.
Forgetting the mean is dragged toward outliers, while median/mode are not.

Tip

Memorise $\text{AM}\ge\text{GM}\ge\text{HM}$ — it instantly flags an impossible answer (a GM larger than the AM). For "average speed over equal distances" questions, reach for the harmonic mean.

Step 1 — arrange in ascending order: 9, 15, 18, 22, 30.

Step 2 — locate the middle item: with $n=5$ (odd), the median is the value of the $\left(\frac{n+1}{2}\right)^{th} = \left(\frac{6}{2}\right)^{th} = 3^{rd}$ item in the ordered series.

The 3rd value in 9, 15, 18, 22, 30 is 18, with exactly two values below it (9, 15) and two above it (22, 30).

Independent re-check: counting from the top of the sorted list also gives the 3rd position as 18 — confirmed from both ends.

✓Final answer

Median = 18 marks.

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2025Set MARCH1 markMCQ
$Q. The correct relationship among A.M., G.M., and H.M., is :$
(a) $H.M \geq G.M \geq A.M$
(b) $A.M < G.M < H.M$
(c) $A.M \geq G.M \geq H.M$
(d) $G.M \geq A.M \geq H.M$
›Reveal solutionSolution

The standard inequality among the three means is $A.M \ge G.M \ge H.M$ .

For any set of positive real numbers, the arithmetic mean (A.M.), geometric mean (G.M.) and harmonic mean (H.M.) always satisfy

$A.M \ge G.M \ge H.M,$

with equality if and only if all the observations are equal.

For two positive numbers $a,b$ this is easy to see: $A.M=\dfrac{a+b}{2}$ , $G.M=\sqrt{ab}$ , $H.M=\dfrac{2ab}{a+b}$ , and one can verify $\dfrac{a+b}{2}\ge\sqrt{ab}\ge\dfrac{2ab}{a+b}$ . A useful related identity is $G.M=\sqrt{A.M\times H.M}$ .

This relationship among the averages is a key result in the measures-of-central-tendency section of the TN HSC Class-11 Business Statistics syllabus.

✓Final answer
$A.M \ge G.M \ge H.M$
CBSE 2023Set MARCH1 markMCQ
$Q. Which of the following represents Median ?$
(a) $Q_3$
(b) $Q_1$
(c) $D_2$
(d) $Q_2$
›Reveal solutionSolution
The median is the value below which half the data lie, which is exactly the second quartile $Q_2$ , so the answer is option (d).

Quartiles divide an ordered data set into four equal parts:

$Q_1$ (first quartile) cuts off the lowest 25%,

$Q_2$ (second quartile) cuts off the lowest 50%,

$Q_3$ (third quartile) cuts off the lowest 75%.

The median is the middle value that splits the data into two equal halves — that is precisely the point with 50% below it, so

$\text{Median} = Q_2.$

(Equivalently, Median $= D_5 = P_{50}$ .)

✓Final answer
$Q_2$

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Q.Find the Median of the marks obtained by 5 students in an economics test: 15, 22, 9, 30, 18.

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