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

Before we start, here is the symbol dictionary every grouped-data problem on this page uses — meet each one on the picture first, then in words:

Figure — Measures of central tendency — mean (grouped - ungrouped), median (grouped), mode (grouped)

Figure 1 — Table $T$ drawn as a histogram. Each coloured bar is one class; its height is the frequency $f$ (printed on top: 4, 9, 15, 11, 6). The coral bar (20–30) is used to label the four working symbols: the coral arrow marks $l$ , the lower boundary (left edge, $= 20$ ); the double-headed arrow along the base marks $h$ , the class width ( $= 10$ ); the top arrow marks $f$ , the bar's height ( $= 15$ ); and $n$ is the sum of all heights ( $= 45$ ). The horizontal axis shows the class boundaries; the vertical axis shows frequency. Read this picture whenever a symbol below feels abstract.

Definition The six grouped-data symbols (read the figure alongside)

A class is one bar (a range of values, e.g. $20$ – $30$ ). Its two vertical edges are the class boundaries.

$x_{i}$ = the class midpoint — the value at the centre of a bar, found as $\frac{lower boundary + upper boundary}{2}$ (e.g. for 20–30 it is $\frac{20 + 30}{2} = 25$ ). We treat every item in a class as if it sat exactly at this midpoint, so $x_{i}$ is the single number that stands in for the whole class when we compute the mean.
$l$ = the lower class boundary of the class we are working in — the left edge of that bar.
$h$ = the class width — how wide one bar is (right edge minus left edge, e.g. $30 - 20 = 10$ ).
$f$ = the frequency of that one class — the height of that single bar (how many items it holds).
$n = \sum f_{i}$ = the total number of items — add up every bar's height.
$c f$ = the cumulative frequency — a running total swept left to right: how many items you have passed by the time you reach a given boundary (see Cumulative Frequency & Ogives). In the median formula, $c f$ means the running total just before entering the median class.

The bars themselves are a histogram (see Histograms & Frequency Distributions); their heights are the frequencies. Keep this figure in mind — every formula below is just "how far into one bar does the answer sit?"

$f_{1}, f_{0}, f_{2}$ (used only for the mode) are three neighbouring bar heights: $f_{1}$ the tallest (modal) bar, $f_{0}$ the bar just left of it, $f_{2}$ the bar just right. We build those on the figure again when we reach the mode.

Level 1 — Recognition

Recall Solution

WHAT is asked: the value splitting the data into two equal halves → that is the median (not mean, not mode). Formula: $Median = l + (\frac{\frac{n}{2} - c f}{f}) \times h$ where (from the symbol dictionary above) $l$ = lower boundary of the median class, $n$ = total count, $c f$ = cumulative frequency of the class before the median class, $f$ = frequency of the median class, $h$ = class width. WHY these words map to it: "50% below" is literally the definition of the middle position $\frac{n}{2}$ , and the formula finds where inside a class that middle person lands.

Recall Solution

WHAT: the modal class is the class with the largest individual frequency — never cumulative. Highest $f$ is $15$ → modal class = 20–30.

$f_{1} = 15$ (modal class itself)
$f_{0} = 9$ (class before, 10–20)
$f_{2} = 11$ (class after, 30–40) WHY: the mode formula measures how sharply the peak rises above its two neighbours, so we need the frequencies on both sides.

Level 2 — Application

Running table $T$ (already continuous, so limits = boundaries) — this is the very table drawn in the figure above:

Class	$f_{i}$	Midpoint $x_{i}$
0–10	4	5
10–20	9	15
20–30	15	25
30–40	11	35
40–50	6	45

$\sum f_{i} = n = 45$ .

Recall Solution

Rule: sort first, then the median is the middle position.

If $n$ is odd, there is one middle: the $(\frac{n + 1}{2})$ th value.
If $n$ is even, there is no single middle, so we average the two central values: the $\frac{n}{2}$ th and the $(\frac{n}{2} + 1)$ th.

(a) Sort: $3, 4, 7, 8, 9$ . Here $n = 5$ (odd), middle position $= \frac{5 + 1}{2} = 3$ rd value $= 7$ . Median $= 7$ . (b) Sort: $2, 3, 4, 7, 8, 9$ . Here $n = 6$ (even), the two central positions are the $3$ rd and $4$ th values: $4$ and $7$ . Average $= \frac{4 + 7}{2} = 5.5$ . Median $= 5.5$ . WHY average for even $n$ : with an even count the data splits cleanly between two neighbours, and any value between them has 50% below and 50% above; we pick their midpoint by convention.

Recall Solution

WHAT is $u_{i}$ ? It is the class's midpoint measured in steps of width $h$ away from our guess $a$ : $u_{i} = \frac{x _{i} - a}{h}$ . This turns the midpoints into tiny integers. $u_{i} = - 2, - 1, 0, 1, 2$ WHY do this? Adding $- 2, - 1, 0, 1, 2$ weighted by frequency is far easier than adding $5, 15, 25, 35, 45$ — and the answer is identical because $x_{i} = a + h u_{i}$ . $\sum f_{i} u_{i} = 4 (- 2) + 9 (- 1) + 15 (0) + 11 (1) + 6 (2) = - 8 - 9 + 0 + 11 + 12 = 6$ $\overset{x}{ˉ} = a + h (\frac{\sum f _{i} u _{i}}{\sum f _{i}}) = 25 + 10 \cdot \frac{6}{45} = 25 + 1.333 \dots = 26.33$ Answer: $\overset{x}{ˉ} \approx 26.33$ .

Recall Solution

Step 1 — cumulative frequency (cf): sweep left to right, keep a running total. $4, 13, 28, 39, 45$ Step 2 — find the median class. $\frac{n}{2} = \frac{45}{2} = 22.5$ . The first cf that reaches or passes $22.5$ is $28$ → median class = 20–30. Step 3 — read the pieces: $l = 20$ , $c f = 13$ (the cf before the median class), $f = 15$ , $h = 10$ . Step 4 — interpolate. We already used $13$ items before this class; we need $22.5 - 13 = 9.5$ more, out of $15$ items spread over width $10$ : $Median = 20 + (\frac{22.5 - 13}{15}) \times 10 = 20 + \frac{9.5}{15} \cdot 10 = 20 + 6.333 = 26.33$ Answer: Median $\approx 26.33$ .

Recall Solution

Modal class = 20–30 (largest $f = 15$ ). Read: $l = 20$ , $f_{1} = 15$ , $f_{0} = 9$ , $f_{2} = 11$ , $h = 10$ . WHAT the fraction means: rise over the left neighbour, $f_{1} - f_{0} = 6$ , divided by the total excess over both neighbours, $2 f_{1} - f_{0} - f_{2} = 30 - 9 - 11 = 10$ . $Mode = 20 + (\frac{15 - 9}{2 ( 15 ) - 9 - 11}) \times 10 = 20 + \frac{6}{10} \cdot 10 = 20 + 6 = 26$ Answer: Mode $= 26$ . WHY it lands where it does: since $f_{0} = 9 < f_{2} = 11$ , the class after is heavier, so the peak leans right of the class centre ( $25$ ). Indeed $26 > 25$ . ✓

Level 3 — Analysis

Recall Solution

Step 1 — recover $f$ from the total. $4 + f + 15 + 11 + 6 = 45 \Rightarrow f = 9$ . Good — but the problem wants us to verify this is consistent with median $= 25$ , so let's treat the second frequency as unknown $f$ and solve from the median condition instead, then confirm.

Let the second frequency be $f$ . Then $n = 36 + f$ and $\frac{n}{2} = \frac{36 + f}{2}$ . With median class 20–30: $l = 20$ , $c f = 4 + f$ , freq of median class $= 15$ , $h = 10$ . $25 = 20 + (\frac{\frac{36 + f}{2} - ( 4 + f )}{15}) \times 10$ Step 2 — simplify the numerator. $\frac{36 + f}{2} - (4 + f) = 18 + \frac{f}{2} - 4 - f = 14 - \frac{f}{2}$ . $25 - 20 = \frac{14 - \frac{f}{2}}{15} \times 10 \Rightarrow 5 = \frac{10 ( 14 - \frac{f}{2} )}{15} = \frac{2 ( 14 - \frac{f}{2} )}{3}$ Step 3 — solve. $5 \cdot 3 = 2 (14 - \frac{f}{2}) \Rightarrow 15 = 28 - f \Rightarrow f = 13.$

But check total: with $f = 13$ , $n = 49$ , not $45$ . The two conditions ( $n = 45$ and median $= 25$ ) are inconsistent — you cannot have both. This is the analysis lesson. Answer: If $n = 45$ is fixed, $f = 9$ and the resulting median is $26.33$ (from L2·2), not $25$ . If median $= 25$ is fixed with these other frequencies, $f = 13$ and $n = 49$ . The data as posed is over-constrained. Report the contradiction — do not force a wrong number.

Recall Solution

WHAT's the trap here: inclusive classes have gaps (10 to 11 is missing), so the printed limit is not the true lower boundary. Step 1 — convert to continuous boundaries. Gap $= 11 - 10 = 1$ ; subtract $\frac{1}{2} = 0.5$ from each lower limit and add $0.5$ to each upper limit: $0.5 - 10.5, 10.5 - 20.5, 20.5 - 30.5, 30.5 - 40.5$ Step 2 — cf: $6, 16, 30, 40$ . Here $n = 40$ , $\frac{n}{2} = 20$ . Step 3 — median class: first cf $\geq 20$ is $30$ → class 20.5–30.5. Read: $l = 20.5$ (the boundary, not $21$ ), $c f = 16$ , $f = 14$ , $h = 10$ . $Median = 20.5 + (\frac{20 - 16}{14}) \times 10 = 20.5 + \frac{4}{14} \cdot 10 = 20.5 + 2.857 = 23.357$ Answer: Median $\approx 23.36$ . Using $l = 21$ instead would have given $23.86$ — off by exactly the $0.5$ boundary shift.

Recall Solution

Median: already sorted, $n = 7$ (odd) → the $\frac{n + 1}{2} = 4$ th value $= 5$ . Mean: $\frac{2 + 4 + 4 + 5 + 5 + 6 + 40}{7} = \frac{66}{7} = 9.43$ . WHY they disagree: the single value $40$ is an outlier. The mean is a balance point — one heavy value drags it far right (to $9.43$ ). The median only cares about position, so the outlier merely sits at the far end and doesn't move the middle. Answer: Median $= 5$ is the better "typical" value here; the mean $9.43$ is inflated by the outlier. This is exactly why we keep three measures (see Skewness).

Level 4 — Synthesis

Recall Solution

Empirical estimate: $3 Median - 2 Mean = 3 (26.33) - 2 (26.33) = 79.00 - 52.67 = 26.33$ Compare to true Mode $= 26$ : agreement within $0.33$ . ✓ Interpretation — WHY they're so close: mean $\approx$ median $\approx$ mode means the distribution is nearly symmetric (little Skewness). When mean $=$ median exactly, the relation predicts Mode $=$ Median too, and the tiny gap ( $26$ vs $26.33$ ) just reflects the mode formula's peak-leaning correction.

Recall Solution

Step 1 — class midpoints. Using $x_{i} = \frac{lower + upper}{2}$ : the midpoints are $10, 30, 50$ . Step 2 — set up the direct grouped-mean equation. The mean is the total of $f_{i} x_{i}$ divided by the total count: $\overset{x}{ˉ} = \frac{\sum f _{i} x _{i}}{\sum f _{i}} = \frac{8 ( 10 ) + 12 ( 30 ) + x ( 50 )}{8 + 12 + x} = 34$ Step 3 — compute the known parts. Numerator $= 80 + 360 + 50 x = 440 + 50 x$ ; denominator $= 20 + x$ . $\frac{440 + 50 x}{20 + x} = 34 \Rightarrow 440 + 50 x = 34 (20 + x) = 680 + 34 x$ Step 4 — solve for $x$ . $50 x - 34 x = 680 - 440 \Rightarrow 16 x = 240 \Rightarrow x = 15$ Answer: $x = 15$ students. Check: mean $= \frac{440 + 750}{35} = \frac{1190}{35} = 34$ . ✓

Level 5 — Mastery

Recall Solution

Is 20–30 correct for $T$ ? Yes — but for the right reason, not the student's reason. The correct test uses cumulative frequency: cf $= 4, 13, 28, 39, 45$ ; $\frac{n}{2} = 22.5$ ; the first cf $\geq 22.5$ is $28$ , sitting in class 20–30. It happens to coincide with the modal class, but that is a coincidence, not a rule.

A clean counterexample. Take five classes 0–10, 10–20, 20–30, 30–40, 40–50 with frequencies $1, 40, 2, 3, 40.$

Modal classes: the largest frequency is $40$ , and it appears twice — in 10–20 and in 40–50. So the data is bimodal; neither single class is "the" mode.
Median class (by cf): cf $= 1, 41, 43, 46, 86$ . Total $n = 86$ , so $\frac{n}{2} = 43$ . The first cf $\geq 43$ is $43$ , which belongs to class 20–30.
The mismatch: the median class is 20–30, which is neither modal class (10–20 or 40–50).

Lesson: the median class must always come from the cumulative frequency reaching $\frac{n}{2}$ , never from "which bar is tallest". For symmetric single-peaked data the two often coincide, but this example proves they need not.

Recall Solution

Target ordering: mean furthest right, mode furthest left → a right (positive) tail. A working dataset: $2, 2, 3, 5, 18$ .

Mode $= 2$ (appears twice; every other value once — the most crowded value).
Median $=$ middle of the $5$ sorted values $2, 2, 3, 5, 18$ → the $\frac{5 + 1}{2} = 3$ rd value $= 3$ .
Mean $= \frac{2 + 2 + 3 + 5 + 18}{5} = \frac{30}{5} = 6$ . Check the ordering: $6 > 3 > 2$ , so $Mean > Median > Mode$ . ✓ WHY this signals positive skew: the lone large value $18$ forms a right tail. The mode sits at the crowded left, the median at the positional middle, and the mean is dragged rightward by the tail — so they line up mean → median → mode from right to left. That is the fingerprint of positive skew (see Skewness). The same far value also inflates the spread, connecting to Measures of Dispersion — Variance & Standard Deviation.

Recap

Recall Which column finds the median class?

Cumulative frequency — the first class whose cf reaches $\frac{n}{2}$ . ::: Cumulative frequency; first cf $\geq n /2$ .

Recall Which frequency finds the modal class?

The largest individual frequency $f_{i}$ , never cumulative. ::: Largest individual $f_{i}$ .

Recall For even

$n$ , how is the ungrouped median found? Sort, then average the two central values (the $\frac{n}{2}$ th and $(\frac{n}{2} + 1)$ th). ::: Average the two middle values.

Recall Before finding a grouped median with inclusive classes, what must you fix first?

Convert printed limits to continuous boundaries (shift by half the gap). ::: Convert to continuous class boundaries.