Measures of central tendency — mean (grouped - ungrouped), median (grouped), mode (grouped): Worked examples

This page is the "throw everything at it" drill for Measures of Central Tendency. The parent note built the three tools (mean, median, mode) and their formulas. Here we make sure that no situation surprises you — odd counts, even counts, inclusive classes with hidden gaps, a class that ties for "most crowded", data pulled by an extreme outlier, and a real word problem.

Before touching numbers, we lay out every kind of case this topic can throw at you, then solve one example per case.

Symbol reminder (used all over this page)

Every symbol here was defined in the parent note; keep these four in view because Ex 5–7 lean on them constantly.

The scenario matrix

#	Case class	What makes it tricky	Example
A	Ungrouped, odd $n$	one clean middle value	Ex 1
B	Ungrouped, even $n$	no single middle — average two	Ex 2
C	Outlier present	mean lies; median/mode don't	Ex 3
D	Grouped mean, three methods agree	must give the SAME number	Ex 4
E	Inclusive classes (hidden gap)	must convert limits → boundaries	Ex 5
F	Grouped median + ogive picture	the "middle person" geometry	Ex 6
G	Grouped mode, peak leans left vs right	neighbour frequencies decide	Ex 7
H	Degenerate: all values equal / tie for mode	limiting behaviour, no unique peak	Ex 8
I	Word problem + empirical relation	translate words → table → estimate	Ex 9

Ex 1 — Case A: ungrouped, odd count

Worked example Odd

$n$ Runs scored in 7 matches: $12, 45, 30, 8, 22, 30, 19$ . Find the mean, median, mode.

Forecast: with $n = 7$ (odd) there is one middle value after sorting. Guess which number lands dead centre.

Mean. Add all: $12 + 45 + 30 + 8 + 22 + 30 + 19 = 166$ . Divide by $n = 7$ : $\overset{x}{ˉ} = 166/7 \approx 23.71$ . Why this step? Mean = fair share = total ÷ count. Nothing grouped, so use raw values.
Median — sort first. $8, 12, 19, 22, 30, 30, 45$ . Why this step? The median is defined on ordered data; the unsorted middle is meaningless.
Pick the middle. $n = 7$ odd → position $\frac{n + 1}{2} = 4$ . The 4th value is $22$ . Why this step? Odd count has exactly one central slot; 3 below, 3 above.
Mode. The value $30$ appears twice, all others once → mode $= 30$ . Why this step? Mode = most crowded value = highest count.

Verify: 3 values ( $8, 12, 19$ ) sit below $22$ and 3 ( $30, 30, 45$ ) sit above → median genuinely splits the set in half. ✓ Mean $23.71$ sits slightly above the median because the big $45$ tugs it up.

Ex 2 — Case B: ungrouped, even count

Worked example Even

$n$ Daily temperatures (°C): $18, 21, 17, 24, 20, 19$ . Find the median.

Forecast: $n = 6$ is even — there is no single middle. Which two values will you average?

Sort. $17, 18, 19, 20, 21, 24$ . Why this step? The median is defined as the middle of ordered data — "50% below, 50% above" only makes sense once the values are in increasing order.
Locate the two central slots. $\frac{n}{2} = 3$ and $\frac{n}{2} + 1 = 4$ → the 3rd and 4th values are $19$ and $20$ . Why this step? Even count straddles the centre between two items; both have equal claim.
Average them. Median $= \frac{19 + 20}{2} = 19.5$ °C. Why this step? The true "middle" lies halfway between the two central values.

Verify: Below $19.5$ are ${17, 18, 19}$ (3 values); above are ${20, 21, 24}$ (3 values). Perfect 50/50 split, and the answer carries °C just like the data. ✓

Ex 3 — Case C: the outlier that fools the mean

Worked example One extreme value

Monthly salaries (₹ thousand) of 5 staff: $20, 22, 25, 23, 210$ (the last is the owner). Which measure best describes a "typical" salary?

Forecast: one value ( $210$ ) is 8× the others. Guess: will the mean sit inside the cluster $20$ – $25$ , or get dragged far above it?

Mean. $\frac{20 + 22 + 25 + 23 + 210}{5} = \frac{300}{5} = 60$ . Why this step? The mean must balance total deviation, so a huge value pulls it far off. $60$ is above every ordinary worker.
Median. Sort: $20, 22, 23, 25, 210$ ; $n = 5$ odd → 3rd value $= 23$ . Why this step? Median only cares about position, not size — the giant $210$ just sits at one end and never moves the middle.
Mode. All values distinct → no mode (no value repeats). Why this step? Mode needs a genuine pile-up; here every salary is unique.

Verify: $23$ (median) describes the typical worker; $60$ (mean) describes nobody. This is exactly why Skewness matters — a long right tail drags the mean rightward while the median stays put. ✓

Ex 4 — Case D: grouped mean by all three methods (they must agree)

Worked example Direct = assumed-mean = step-deviation

Class	$f_{i}$	Midpoint $x_{i}$
0–10	4	5
10–20	6	15
20–30	10	25
30–40	6	35
40–50	4	45

Forecast: the table is symmetric about $25$ . Guess the mean before computing — symmetry is a strong hint.

Direct method. $\sum f_{i} = 30$ . $\sum f_{i} x_{i} = 4 (5) + 6 (15) + 10 (25) + 6 (35) + 4 (45) = 20 + 90 + 250 + 210 + 180 = 750$ . So $\overset{x}{ˉ} = 750/30 = 25$ . Why this step? Each class contributes $f_{i} x_{i}$ (count × its representative midpoint); total ÷ count = fair share.
Assumed-mean. Take $a = 25$ , $d_{i} = x_{i} - 25 = - 20, - 10, 0, 10, 20$ . $\sum f_{i} d_{i} = 4 (- 20) + 6 (- 10) + 0 + 6 (10) + 4 (20) = - 80 - 60 + 0 + 60 + 80 = 0$ . $\overset{x}{ˉ} = 25 + \frac{0}{30} = 25$ . Why this step? Shifting by a guess $a$ shrinks the arithmetic; the total correction here is $0$ because the table balances around $25$ .
Step-deviation. $h = 10$ , $u_{i} = \frac{x _{i} - 25}{10} = - 2, - 1, 0, 1, 2$ . $\sum f_{i} u_{i} = 4 (- 2) + 6 (- 1) + 0 + 6 (1) + 4 (2) = - 8 - 6 + 0 + 6 + 8 = 0$ . $\overset{x}{ˉ} = 25 + 10 \cdot \frac{0}{30} = 25$ . Why this step? Dividing deviations by $h$ turns them into tiny integers — least error-prone.

Verify: All three give $25$ , which equals the centre of symmetry we forecast. Same mean, different-sized arithmetic. ✓

Ex 5 — Case E: inclusive classes with a hidden gap

Worked example Convert limits to boundaries first

Marks recorded in inclusive classes:

Class	$f_{i}$
10–19	3
20–29	7
30–39	12
40–49	8

Find the median class boundary $l$ and the median.

Forecast: there's a gap between 19 and 20. If you naively use $l = 30$ you'll be wrong. Guess the true lower boundary of the 30–39 class.

Spot the gap. $19$ to $20$ has a width-1 gap. Split it: subtract $0.5$ from each lower limit, add $0.5$ to each upper limit → continuous boundaries $9.5 - 19.5, 19.5 - 29.5, 29.5 - 39.5, 39.5 - 49.5$ . Why this step? Interpolation assumes classes touch. A gap would make the width wrong. Now every class width is $h = 10$ .
Cumulative frequencies. $3, 10, 22, 30$ . $n = 30$ , so $\frac{n}{2} = 15$ . Why this step? Median class = first class whose $c f \geq 15$ . Here $c f = 22$ (the 29.5–39.5 class) is first to cross 15.
Read off the pieces. $l = 29.5$ (the boundary, not 30), $c f = 10$ (class before), $f = 12$ , $h = 10$ . Why this step? The grouped-median formula needs exactly these four inputs — the starting edge $l$ , how many items are already used up ( $c f$ ), how many sit in the median class ( $f$ ), and how wide that class is ( $h$ ). Gathering them now means the interpolation in Step 4 is pure substitution with no re-reading of the table.
Apply the formula. $Median = 29.5 + \frac{15 - 10}{12} \times 10 = 29.5 + \frac{5}{12} \cdot 10 = 29.5 + 4.17 = 33.67.$ Why this step? We need $15 - 10 = 5$ more items into a class of 12 spread over width 10; each item occupies $10/12$ of width.

Verify: The median $33.67$ lies inside the median class $(29.5, 39.5)$ ✓. Had we used $l = 30$ we'd get $34.17$ — off by exactly the $0.5$ boundary correction, showing why the fix matters. ✓

Ex 6 — Case F: grouped median as a geometric "middle person"

Worked example Median via the ogive

Reuse the parent's table: classes $0$ – $10, \dots, 40$ – $50$ with $f = 5, 8, 12, 7, 3$ ; $n = 35$ . Show the median $= 23.75$ geometrically.

Forecast: $\frac{n}{2} = 17.5$ . Guess: on a graph of cumulative frequency vs. class, where does the horizontal line " $c f = 17.5$ " cross the curve?

How to read it: the horizontal axis is marks (upper class boundary), the vertical axis is cumulative frequency

c f

. The black staircase-curve is the ogive; the red horizontal line sits at height

17.5 = \frac{n}{2}

, and the red dashed drop shows where that height meets the curve and falls to the marks axis at

23.75

Build cumulative frequencies. $5, 13, 25, 32, 35$ . Plot each against the upper boundary of its class: points $(10, 5), (20, 13), (30, 25), (40, 32), (50, 35)$ . Join them — this rising line is the ogive (see Cumulative Frequency & Ogives). Why we use upper boundaries: $c f$ counts everyone up to and including a class, so that running total is only fully "achieved" at the class's right edge. Plotting at the midpoint or lower edge would claim we'd finished the class before we actually had, shifting the whole curve left.
Draw the half-way line. The red horizontal line at height $\frac{n}{2} = 17.5$ (see figure). It cuts the ogive somewhere between $x = 20$ and $x = 30$ . Why this step? The person at cumulative rank $17.5$ is the median; the ogive tells us their data value.
Drop to the axis. From the crossing point, drop straight down. It lands at $x = 23.75$ . Why this step? Reading the $x$ -value converts "rank 17.5" into an actual marks value.
Confirm algebraically. $l = 20, c f = 13, f = 12, h = 10$ : $20 + \frac{17.5 - 13}{12} \cdot 10 = 20 + 3.75 = 23.75$ . Why this step? The formula is the algebra of that red-line crossing: between the points $(20, 13)$ and $(30, 25)$ the ogive is a straight segment, and the formula computes exactly where the horizontal line $c f = 17.5$ meets that segment. Getting the same number both ways confirms the picture and the algebra are one idea.

Verify: Graph reading ( $23.75$ ) and formula ( $23.75$ ) agree exactly — the formula is just the algebra of that red-line crossing. ✓

Ex 7 — Case G: mode leaning left vs right

Worked example Which side does the peak lean?

Same table $f = 5, 8, 12, 7, 3$ . Compare the mode against the modal-class centre.

Forecast: modal class is $20$ – $30$ (highest $f = 12$ ), centre $25$ . The class before has $f_{0} = 8$ , the class after has $f_{2} = 7$ . Since $f_{0} > f_{2}$ , will the true peak sit left of $25$ or right?

How to read it: the horizontal axis is marks (class), the vertical axis is frequency

f

. Each bar is a class; the red bar is the modal class

20

–

30

(tallest,

f = 12

). The dotted vertical line marks the class centre

25

; the black dashed line marks the computed mode

24.44

, sitting just to its left.

Identify pieces. $l = 20, f_{1} = 12, f_{0} = 8, f_{2} = 7, h = 10$ . Why this step? Mode uses the individual frequency, never cumulative — and needs both neighbours.
Compute the rise fractions. Rise over left neighbour: $f_{1} - f_{0} = 4$ . Total excess: $2 f_{1} - f_{0} - f_{2} = 24 - 8 - 7 = 9$ . Why this step? The fraction $\frac{4}{9}$ says the peak is $\frac{4}{9}$ of the way into the class — closer to the left because the left neighbour is heavy and "props up" that side.
Apply. $Mode = 20 + \frac{4}{9} \cdot 10 = 20 + 4.44 = 24.44$ .
Interpret the lean. $24.44 < 25$ (class centre) → peak leans left, exactly because $f_{0} = 8 > f_{2} = 7$ . The mode marker in the figure sits left of the dotted centre line. Why this step? Confirms the geometry: heavier left neighbour → mode nearer the left boundary.

Verify: $\frac{4}{9} \in (0, \frac{1}{2})$ confirms "less than halfway" → left of centre ✓. If we swapped to $f_{0} = 7, f_{2} = 8$ , the fraction would be $\frac{5}{9} > \frac{1}{2}$ → mode would lean right instead. ✓

Ex 8 — Case H: degenerate inputs

Worked example All equal, and a tie for the peak

(a) Data $7, 7, 7, 7, 7$ . (b) Data $2, 2, 5, 5, 9$ .

Forecast: in (a) every summary should collapse to $7$ . In (b) two values tie for "most frequent" — what happens to the mode?

(a) Mean. $\frac{7 \cdot 5}{5} = 7$ . Median: sorted middle $= 7$ . Mode: $7$ (occurs 5 times). Why this step? With zero spread, "typical" is unambiguous — all three tools must agree. This is the limiting case of a distribution with no variability (see Measures of Dispersion — Variance & Standard Deviation, where variance $= 0$ here).
(b) Mode. Both $2$ and $5$ occur twice → the data is bimodal; there is no single mode. Why this step? Mode is defined by the highest count; when two counts tie, both are modes. Reporting just one would hide structure.
(b) Mean & median for contrast. Mean $= \frac{2 + 2 + 5 + 5 + 9}{5} = \frac{23}{5} = 4.6$ . Median (3rd of $2, 2, 5, 5, 9$ ) $= 5$ . Why this step? Mean and median remain single-valued even when the mode splits.

Verify (a): all three equal $7$ and total deviation $\sum (x_{i} - 7) = 0$ ✓ (perfect balance point). Verify (b): counts ${2 : 2, 5 : 2, 9 : 1}$ — two maxima confirm bimodality; mean $4.6$ ≠ median $5$ shows mild left skew. ✓

Ex 9 — Case I: word problem + empirical relation

Worked example Estimate the mode from mean and median

A survey of daily screen-time (hours) gives a mean of $3.2$ and a median of $3.5$ for a moderately skewed distribution. The raw data is lost. Estimate the mode.

Forecast: median > mean here. Guess: does the mode come out above the median or below the mean?

Choose the tool. Use the empirical relation $Mode \approx 3 Median - 2 Mean$ . Why this step? It's the only formula that recovers one measure from the other two — perfect when raw data is unavailable, valid for moderate skew.
Substitute. $Mode \approx 3 (3.5) - 2 (3.2) = 10.5 - 6.4 = 4.1$ hours. Why this step? Plug the two knowns; the "3 and 2" weighting is empirical, tuned to typical skewed shapes.
Interpret. Ordering: Mean $3.2 <$ Median $3.5 <$ Mode $4.1$ . Mean below mode signals a left-skewed long tail of very-low-usage people pulling the mean down (compare Skewness).

Verify: Rearranging, $3 (3.5) - 2 (3.2) = 4.1$ ✓. Sanity: for moderate skew the three measures should be roughly evenly spaced; gaps here are $0.3$ (mean→median) and $0.6$ (median→mode) — same direction, consistent with one-sided skew. ✓

(The idea of a data value weighting a total also underlies Weighted Mean and the expected value, where each outcome is weighted by its probability instead of its frequency.)

Recall check

Recall Why does the median beat the mean in Ex 3?

Because the median depends only on position, so a single extreme value (the ₹210k owner) cannot drag it; the mean balances totals and gets pulled far above every ordinary worker.

Recall In Ex 5, why isn't

$l = 30$ ? Inclusive classes (20–29, 30–39) have a hidden gap. Converting to continuous boundaries gives $l = 29.5$ ; using $30$ would shift every answer by $0.5$ .

Recall In Ex 7, what makes the mode lean left?

The class before ( $f_{0} = 8$ ) is heavier than the class after ( $f_{2} = 7$ ), so the fraction $\frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} = \frac{4}{9} < \frac{1}{2}$ — less than halfway in, i.e. left of the class centre.

Which measure is unaffected by an extreme outlier?

The median (and often the mode); the mean is dragged toward the outlier.

For inclusive classes 20–29, 30–39, what is the lower boundary of the second class?

29.5

(subtract 0.5 to close the gap).

If two values tie for highest frequency, the data is?

Bimodal — there is no single mode.

Empirical relation to estimate the mode?

Mode \approx 3 Median - 2 Mean