Mean, median, mode — calculation for raw and grouped data: Question bank

This page is a concept gym for Mean, median, mode — calculation for raw and grouped data. No heavy arithmetic here — every item hunts a misunderstanding or a boundary case. Read the prompt, answer out loud, then reveal.

Terms you should already own from the parent note: mean (sum ÷ count), median (the middle value by position), mode (the most frequent value), class mark (midpoint of an interval), cumulative frequency (running total of counts), and modal / median class (the interval a formula lands in).

Definition Symbol dictionary for the grouped-data items below

Before you meet these letters in the questions, pin them down — every grouped-data trap on this page uses exactly these:

$x_{i}$ = the class mark (midpoint) of the $i$ -th interval, $x_{i} = \frac{lower + upper}{2}$ . In the grouped-mean formula $\overset{x}{ˉ} = \frac{\sum f _{i} x _{i}}{N}$ , the symbol $x$ always means this midpoint, never a raw observation.
$f$ (or $f_{i}$ ) = the frequency of a class — how many observations fall in that interval.
$N = \sum f_{i}$ = the total frequency (total number of observations).
$C F$ = the cumulative frequency of the class just before the class you are working in (the running count of everyone below it).
$L$ = the lower boundary of the class a formula lands in (median class or modal class).
$h$ = the class width = upper limit − lower limit of that class.
For the mode: $f_{1}$ = frequency of the modal class, $f_{0}$ = frequency of the class immediately before it, $f_{2}$ = frequency of the class immediately after it.

The two grouped formulas these letters build: $Median = L + (\frac{\frac{N}{2} - C F}{f}) \times h, Mode = L + (\frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}}) \times h$

Look at the number-line below before starting — it shows why the same dataset can have three different "centres" and how they slide under transformations.

Figure — Mean, median, mode — calculation for raw and grouped data

True or false — justify

Every reveal is a full reason, not a bare verdict.

The mean must be one of the actual data values.

False — the mean of

{1, 2}

1.5

, which never occurs. Mean is a balance point of magnitude, not a value that has to appear.

The median must be one of the actual data values.

False for even

n

— with

{4, 10}

the median is

7

, the average of the two middle values, which is absent from the data.

The mode must be one of the actual data values.

True — the mode is defined as the most frequently occurring value, so it can only ever be something that actually occurs.

Adding a huge outlier changes the mean far more than the median.

True — the mean sums every value, so one extreme number drags the total; the median only cares about position, so a far-away value just sits at one end and barely moves the middle.

If you add the same constant

c

to every data value, all three of mean, median and mode increase by

c

True — shifting the whole dataset slides its centre, its middle value and its peak all rigidly by

c

; the shape is unchanged.

If you multiply every value by 2, the median doubles but its position in the sorted list changes.

False on the second half — every value doubles so the median value doubles, but the sorted order and hence the middle position is identical.

For grouped data, the computed mean is exact.

False — grouping replaces every value in a class by its class mark

x_{i}

, an assumption that all values sit at the midpoint, so the grouped mean is an approximation of the raw mean.

A dataset where every value is unique has no mode.

True — no value repeats, so no value has the highest frequency over the others; by convention this counts as "no mode."

A perfectly symmetric single-peaked distribution has mean = median = mode.

True — symmetry balances magnitude (mean) and count (median) at the same centre, and a single peak puts the mode there too.

The median is always exactly halfway between the smallest and largest value.

False — that describes the midrange. The median is halfway by count of observations, not halfway across the range of values.

The simple grouped formulas assume every class has the same width

h

True — the median and mode formulas each multiply by a single

h

, so if class widths differ the raw comparison of frequencies is distorted; you must adjust to frequency density (frequency ÷ width) first.

Spot the error

Each prompt contains one planted mistake; the reveal names and fixes it.

"Data

{5, 1, 3}

n = 3

odd, so median = the

\frac{n + 1}{2} = 2

nd term =

1

The error is not sorting first. Sorted it is

{1, 3, 5}

; the 2nd term is

3

, so the median is

3

, not

1

"Grouped mean: I used the class upper limits as the

x_{i}

\overset{x}{ˉ} = \frac{\sum f x}{N}

Wrong point — here

x

means the class mark

x_{i} = \frac{lower + upper}{2}

, not the raw values or the upper edge, since the assumption is that values cluster at the centre of each interval.

"In the median formula

L + (\frac{N /2 - C F}{f}) h

I used

C F

= the cumulative frequency of the median class itself."

Error —

C F

is the cumulative frequency of the class just before the median class, because you've already accounted for everyone below the median class before you interpolate into it.

"Modal class chosen as 60–80 because its class mark 70 is the biggest midpoint."

The modal class is the one with the highest frequency

f_{1}

, not the highest midpoint or class mark. Pick by the tallest bar, never by where it sits on the axis.

"For median I plugged

L = 40

but wrote

\frac{N}{2} = 40

for

N = 40

Slip — the median position is

\frac{N}{2} = 20

, not

40

N

is the total count, and half of it is where the middle observation lies.

"Mode formula denominator: I wrote

f_{1} - f_{0} - f_{2}

Missing a factor — the denominator is

2 f_{1} - f_{0} - f_{2}

, where

f_{1}

is the modal-class frequency and

f_{0}

f_{2}

its neighbours. It measures how the peak beats both neighbours, so

f_{1}

appears twice.

"The two middle values were 18 and 20, so median

= 18 \times 20 = 360

For even

n

you average the two middles, not multiply:

\frac{18 + 20}{2} = 19

"Class widths were 0–10, 10–30, 30–35, but I compared raw frequencies to pick the modal and median classes anyway."

Error — with unequal widths

h

is not constant, so a wide class collects more counts just by being wide; convert to frequency density (frequency ÷ width) before comparing or the formulas mislead.

Why questions

Explain the reasoning, don't just state a rule.

Why do we divide by

\sum f = N

(total frequency) and not by the number of classes in the grouped mean

\overset{x}{ˉ} = \frac{\sum f _{i} x _{i}}{N}

Because each class stands for

f_{i}

real observations, not one; dividing by the number of classes would ignore that some intervals hold far more data than others.

Why is the median called "resistant to outliers" but the mean is not?

The median depends only on which value sits in the middle position, so pushing an extreme value further out doesn't move it; the mean adds every value's magnitude, so an outlier's size enters the total directly.

Why do we interpolate inside the median class using

L + (\frac{N /2 - C F}{f}) h

instead of just reporting the class mark?

Because the exact

\frac{N}{2}

-th observation usually isn't at the class centre — the term

\frac{N /2 - C F}{f}

is the fraction of the way into the class we must travel, and multiplying by width

h

turns that fraction into a real distance added to

L

Why does the mode formula use differences

f_{1} - f_{0}

and

f_{1} - f_{2}

rather than the frequencies alone?

The peak leans toward whichever neighbour it towers over less; comparing how much the modal class (

f_{1}

) beats the class before (

f_{0}

) and after (

f_{2}

) tells us which way inside the interval to nudge the mode.

Why can grouped data give a median class but never a single exact median value without the formula?

Grouping hides the individual values, so we only know the middle observation lies somewhere in that class; the interpolation formula is how we recover a specific estimate.

Why might a report quote the mode for shoe sizes but the median for house prices?

Shoe sizes need the most common size to stock (mode); house prices have extreme mansions that would inflate the mean, so the outlier-resistant median better describes a "typical" home.

Why does an open-ended interval like "80+" break the grouped-mean formula?

There is no upper limit, so its class mark

x_{i} = \frac{lower + upper}{2}

is undefined; you must either assume a sensible closing value or fall back on the median, which only needs positions and doesn't touch that unbounded top class.

Edge cases

Boundaries, degenerate inputs and traps the definitions invite.

A dataset

{4, 4, 7, 7}

— what is its mode?

It is bimodal: both

4

and

7

appear twice, tying for most frequent, so two modes are reported rather than none.

Every value in a dataset is identical, e.g.

{6, 6, 6, 6}

— do mean, median, mode agree?

Yes — sum ÷ count is

6

, the middle value is

6

, and

6

is the only (hence most frequent) value, so all three equal

6

In the grouped median formula

L + (\frac{N /2 - C F}{f}) h

, what happens if

\frac{N}{2}

lands exactly on a cumulative frequency boundary?

The median class is the next class where

C F

first reaches or exceeds

\frac{N}{2}

; if

C F

(of the class before) equals

\frac{N}{2}

exactly, then

\frac{N /2 - C F}{f} = 0

and the median sits right at that class's lower boundary

L

The modal class is the very first interval, so there is no class before it — is the mode formula still usable?

Yes, but treat the missing

f_{0}

0

; there are simply no observations "below," so the peak can only lean toward the class after it (via

f_{2}

Two intervals share the identical highest frequency

f_{1}

in grouped data — which is the modal class?

The distribution is bimodal at the interval level; the single-mode formula no longer cleanly applies, and you'd report two modal classes rather than force one number.

A frequency table has an open top class "80+" — can you still compute the median?

Usually yes — the median formula uses only positions, boundaries

L

C F

f

and width

h

, none of which needs the top class's upper limit, provided the median class isn't the open one; the mean, which needs every

x_{i}

, is the one that fails.

The classes are 0–10, 10–30, 30–60 (unequal widths) — can you compare frequencies directly to find the modal class?

No — a wider class collects more counts just by spanning more values; divide each frequency by its width

h

to get frequency density and compare those instead, otherwise the widest class falsely looks "most popular."

A class interval is written "20–40" and the next "40–60" — which class does an exact value of

40

belong to?

By the standard upper-limit-exclusive convention,

40

falls in

40

–

60

, so consistency of this rule matters when building the frequency table and cumulative frequencies.

For raw data with

n

even and the two middle values far apart, is the median unstable?

Its value is just their average and is well defined, but it can sit in a sparse gap where no data lives, which is why median alone can mislead about a bimodal shape — pair it with the mode.

Recall One-line self-test

Which measure is unchanged if you replace the largest value with an even larger one? ::: The median (and often the mode); only the mean moves, because only the mean sums magnitudes.

Where to go next

Distinguishing spread from centre: Measures of Dispersion
Reading the median off a graph: Cumulative Frequency Graphs
Building the tables these questions use: Frequency Distributions
When mean ≠ median ≠ mode and what it means: Skewness
Visual five-number summaries: Box Plots
Averages with unequal importance: Weighted Mean
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