How do you find the median and the modal class of grouped data?

Add a cumulative frequency column and find n/2. The median class is the first class whose cumulative frequency is greater than n/2. The modal class is simply the class with the highest frequency. A formula then places the median or mode inside its class.

Mean, Median and Mode for CUET General Aptitude

About 15 min to learn · 67 questions

CUET's mean, median and mode questions use Class 9–10 statistics: the three averages of a list, the empirical relation between them, the mean, median and mode of grouped data, how each measure reacts when the data change, and the range. The arithmetic is light. What costs marks is an unsorted list, a forgotten frequency, or assuming a measure moved when it did not. This lesson works through each with examples and the traps beside them.

Syllabus area: Reasoning

On this page · 10 sections

01The core idea
02The three averages
03The empirical relation
04Grouped data
05When the data change
06Range and spread
07Exam shortcuts
08Common mistakes
09Quick revision
10Check your understanding

The core idea

The mean, the median and the mode are three answers to one question: what is a typical value? The mean shares the total out equally. The median is the value in the middle. The mode is the value that turns up most often.

Two habits settle most questions. Sort the data before you look for a middle. And think in totals: a mean is a total divided by a count, so any change to the data is a change to the total first.

The three averages

What are the mean, median and mode?

The mean is the sum of the values divided by how many there are. The median is the middle value once the data are sorted. The mode is the value that occurs most often.

Measure	How to find it	Uses every value?
Mean	Sum ÷ number of values	Yes
Median	Middle of the sorted list	No, only the middle
Mode	Most frequent value	No, only the counts

Example 1 · All three from one list

A cyclist rode 22, 17, 30, 17, 26, 19 and 23 km on seven days. Find the mean, median and mode.

Mean: $(22 + 17 + 30 + 17 + 26 + 19 + 23) \div 7 = 154 \div 7 = 22$ km.
Sorted: 17, 17, 19, 22, 23, 26, 30. With 7 values the median is the 4th, 22 km.
The mode is 17 km, the only value that repeats.

Answer

Mean 22 km, median 22 km, mode 17 km

How do you find the median when there is an even number of values?

Sort the values and take the average of the two middle ones. With $n$ values and $n$ even, those are the $\tfrac{n}{2}$ th and the $(\tfrac{n}{2} + 1)$ th. With $n$ odd, the median is the single $\tfrac{n + 1}{2}$ th value.

Example 2 · Even count

The ages of six cousins are 14, 9, 21, 16, 11 and 18. Find the median age.

Sorted: 9, 11, 14, 16, 18, 21.
The middle two are the 3rd and the 4th, 14 and 16.
Median: $(14 + 16) \div 2 = 15$ .

Answer

No cousin is 15. With an even count the median need not be one of the values.

Watch out

Sort first. The middle entry of the list as printed is almost always one of the wrong options.

Can a data set have more than one mode, or no mode?

Yes. If two values tie for the highest count, the data are bimodal and both are modes. If every value occurs once, there is no mode.

4, 6, 6, 8, 9, 9 has two modes, 6 and 9.
3, 5, 8, 12 has no mode.
One more 9 in the first list breaks the tie, and 9 becomes the only mode.

Key idea

The mode is the only average that works for categories, such as the most common shirt size or blood group, where adding the values up makes no sense.

Which average should you use, and when?

Use the mean when there are no extreme values and every value should count. Use the median when one or two extreme values would drag the mean away from the typical value. Use the mode for the most common item or for categories.

An office where the owner earns ten times anyone else: the median describes a typical employee.
A shoe shop choosing which size to stock most of: the mode.
The marks of a class with no outliers: the mean.

The empirical relation

What is the empirical relation between mean, median and mode?

For a moderately skewed distribution, the mode is three times the median minus twice the mean. Given any two of the three, it gives the third.

Empirical relation

\text{Mode} = 3 \times \text{Median} - 2 \times \text{Mean}

The same relation, solved for each measure: $\text{Mean} = \tfrac{3 \times \text{Median} - \text{Mode}}{2}$ and $\text{Median} = \tfrac{\text{Mode} + 2 \times \text{Mean}}{3}$ .

Example 3 · Mode from mean and median

A set of exam scores has mean 61 and median 58. Estimate the mode.

Mode $= 3 \times 58 - 2 \times 61 = 174 - 122 = 52$ .
Check with the median formula: $(52 + 2 \times 61) \div 3 = 174 \div 3 = 58$ . It agrees.

Answer

Watch out

The 3 goes with the median and the 2 with the mean. Swapping them gives a planted option. The median sits between the mode and the mean, one-third of the way from the mean to the mode, so it carries the bigger weight.

Grouped data

How do you find the mean of a frequency table?

Multiply each value by its frequency, add the products, and divide by the total frequency. For classes such as 10–20, use the class mark, the midpoint 15, as the value.

Mean of a frequency table

\bar{x} = \frac{\sum fx}{\sum f}

Example 4 · Frequency table

Twenty-five families were asked how many phones they own. Find the mean.

Phones $x$	Families $f$	$fx$
1	4	4
2	9	18
3	7	21
4	5	20
Total	25	63

$\sum fx = 4 + 18 + 21 + 20 = 63$ and $\sum f = 25$ .
Mean: $63 \div 25 = 2.52$ .

Answer

2.52 phones

Watch out

Divide by $\sum f$ , the number of families, not by 4, the number of rows. And never average the values 1, 2, 3, 4 without their frequencies.

When the class marks are large, textbooks use two shortcuts. Both give exactly the direct method's answer.

Method	Formula	Use when
Direct	$\bar{x} = \tfrac{\sum fx}{\sum f}$	Small class marks
Assumed mean	$\bar{x} = a + \tfrac{\sum fd}{\sum f}$ , with $d = x - a$	Large class marks
Step deviation	$\bar{x} = a + \tfrac{\sum fu}{\sum f} \times h$ , with $u = \tfrac{x - a}{h}$	Equal class widths $h$

Another example: the step-deviation method

Classes 100–120, 120–140, 140–160, 160–180 and 180–200 have frequencies 5, 8, 12, 10 and 5, so $\sum f = 40$ . Take $a = 150$ , the middle class mark, and $h = 20$ . Then $u$ is −2, −1, 0, 1, 2, and $\sum fu = -10 - 8 + 0 + 10 + 10 = 2$ . The mean is $150 + \tfrac{2}{40} \times 20 = 151$ . The direct method gives the same: $\sum fx = 6040$ , and $6040 \div 40 = 151$ .

Forgetting the final $\times h$ would give 150.05, a common wrong option.

Add a cumulative frequency column and find $\tfrac{n}{2}$ . The median class is the first class whose cumulative frequency is greater than $\tfrac{n}{2}$ . The modal class is simply the class with the highest frequency. A formula then places the median or mode inside its class.

Median and mode of grouped data

\text{Median} = l + \frac{\tfrac{n}{2} - cf}{f} \times h \qquad \text{Mode} = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h

For the median, $l$ is the lower limit of the median class, $cf$ the cumulative frequency of the class before it, and $f$ its own frequency. For the mode, $l$ is the lower limit of the modal class, $f_1$ its frequency, and $f_0$ and $f_2$ the frequencies of the classes before and after. In both, $h$ is the class width.

Example 5 · Grouped median and mode

The times in minutes that 60 people waited at a bus stop are shown below. Find the median and the mode.

Time	Frequency	Cumulative frequency
0–5	9	9
5–10	14	23
10–15	20	43
15–20	11	54
20–25	6	60

$\tfrac{n}{2} = 30$ . The first cumulative frequency above 30 is 43, so the median class is 10–15, with $l = 10$ , $cf = 23$ , $f = 20$ and $h = 5$ .
Median: $10 + \tfrac{30 - 23}{20} \times 5 = 10 + 1.75 = 11.75$ minutes.
The highest frequency is 20, so the modal class is also 10–15, with $f_1 = 20$ , $f_0 = 14$ and $f_2 = 11$ .
Mode: $10 + \tfrac{20 - 14}{40 - 14 - 11} \times 5 = 10 + \tfrac{6}{15} \times 5 = 12$ minutes.

Answer

Median 11.75 minutes, mode 12 minutes

Both lie inside the 10–15 class, as a grouped median and mode must.

Watch out

The $cf$ in the median formula belongs to the class before the median class, not to the median class itself. And the midpoint of the modal class, 12.5 here, is not the mode.

Exam shortcut

A "less than" table gives cumulative frequencies. Subtract neighbouring entries to get each class's own frequency before you use $f$ . A "more than" table counts from the top, so subtract the other way.

When the data change

What happens to the mean, median and mode when every value changes the same way?

All three averages change in exactly the same way as the values. The range changes only when the values are multiplied or divided, because adding a constant moves the largest and the smallest value equally.

Change to every value	Mean, median, mode	Range
Add $k$	Each $+ k$	Unchanged
Multiply by $k$ , $k$ positive	Each $\times k$	$\times k$
Increase by $p\%$	Each $\times (1 + \tfrac{p}{100})$	$\times (1 + \tfrac{p}{100})$
Multiply by $k$ , then add $c$	Each $\times k$ , then $+ c$	$\times k$ only

Example 6 · Converting units

Temperature readings have median 30 °C and range 8 °C. Find the median and range in °F, using $F = 1.8C + 32$ .

Median: $1.8 \times 30 + 32 = 86$ °F.
Range: $1.8 \times 8 = 14.4$ °F. The $+ 32$ moves every reading equally, so it does not touch the range.

Answer

Median 86 °F, range 14.4 °F

What happens when only one value is added, removed or corrected?

Work through each measure separately. The mean almost always changes, and the easiest way to find it is through the total. The median changes only if the middle of the sorted list moves. The mode changes only if the counts of repeated values change. The range changes only if the largest or the smallest value changes.

Example 7 · Correcting a misread value

The mean of 10 values was found to be 45. Later it was noticed that a value of 62 had been read as 26. Find the correct mean.

The wrong total is $10 \times 45 = 450$ .
The correct total is $450 - 26 + 62 = 486$ .
The correct mean is $486 \div 10 = 48.6$ .

Answer

48.6

The total rose by 36 over 10 values, so the mean rose by 3.6.

Example 8 · Adding a value moves the median

The list 5, 8, 11, 13, 20 has median 11. The value 3 is added. Find the new median.

The new sorted list is 3, 5, 8, 11, 13, 20, which has six values.
Median: $(8 + 11) \div 2 = 9.5$ .

Answer

9.5

Key idea

Raising the largest value, or lowering the smallest, leaves the median where it was, however big the change. It moves the mean and the range. This is why the median resists extreme values and the mean does not.

Exam shortcut

When the count stays the same, the mean changes by the change in the total divided by the count. Replacing 30 by 6 in 12 values lowers the total by 24 and the mean by 2.

Range and spread

What is the range and what does it tell you?

The range is the largest value minus the smallest. It measures spread: a small range means the values sit close together, and a large range means they are scattered.

Example 9 · Same mean, different spread

Two shops sold these numbers of melons over five days. Compare them.

Shop A: 38, 40, 41, 43, 43.
Shop B: 20, 35, 44, 50, 56.

Both totals are 205, so both means are $205 \div 5 = 41$ .
Range of A: $43 - 38 = 5$ . Range of B: $56 - 20 = 36$ .

Answer

Equal means, but shop A's sales are far more consistent

An average alone never describes a data set fully.

Three range traps

With negative values, subtract carefully: from −3 to 7 the range is $7 - (-3) = 10$ .
When two lists are merged, the range is the overall largest minus the overall smallest, not the sum of the two ranges.
If an unknown value could become the new largest or the new smallest, check both cases. There may be two answers.

Exam shortcuts

Think in totals. Total = mean × count. Every "a value is added, removed or corrected" question is a change to the total.
Same change to every value. The averages follow it. The range ignores any addition.
Empirical relation. Mode $= 3 \times$ Median $- 2 \times$ Mean, the 3 with the median.
Sanity checks. A median lies between the smallest and largest values. A grouped median lies inside its median class. A mean can never exceed the largest value.

Common mistakes

Not sorting. The middle of the printed list is a planted option.
One middle value for an even count. Average the two middle values.
Ignoring frequencies. A frequency table's mean is $\sum fx \div \sum f$ .
Dividing by the number of rows. Divide by the total frequency.
The wrong $cf$ . Use the cumulative frequency of the class before the median class.
The class mark as the mode. Use the formula.
Swapping the empirical relation. It is $3 \times$ Median, not $3 \times$ Mean.
A shift changing the range. Adding the same number to every value never changes it.
Assuming the median moved. Correcting the top value leaves it alone.
Averaging medians. The median of two merged lists is not the average of their medians. Merge, sort and count again.

Quick revision

The three averages

Mean = sum ÷ count
Median: sort, then the middle value, or the average of the two middle values
Mode: the most frequent value. There may be two, or none
Extreme values: use the median

Empirical relation

Mode $= 3 \times$ Median $- 2 \times$ Mean
The median lies between the mode and the mean

Grouped data

Mean $= \sum fx \div \sum f$ , with class marks as $x$
Median $= l + \tfrac{n/2 - cf}{f} \times h$ , $cf$ from the class before
Mode $= l + \tfrac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h$

Changing data

Add $k$ to all: averages $+ k$ , range unchanged
Multiply all by $k$ : averages and range $\times k$
One value changes: work with the total

Range

Largest − smallest
Merged lists: overall largest − overall smallest

Totals, weighted averages and "the average rises by 2" problems are worked in averages, and reading values off tables and charts in data interpretation.

Next · 09Quick revision

Check your understanding

67 questions, grouped by type, easy to hard. Pick an answer and check it. The worked solution opens when you ask for it.

Mean, median and modeQ1 of 67Q1

Mean, median and mode

18 questions

Waiting time (minutes)	Number of patients
Less than 10	4
Less than 20	11
Less than 30	23
Less than 40	34
Less than 50	40

Number of children	Number of households
0	6
1	14
2	18
3	8
4	4

Number of children	Number of households
0	6
1	14
2	18
3	8
4	4

Time (minutes)	Number of students
0–20	7
20–40	11
40–60	16
60–80	10
80–100	6

Age (years)	Number of visitors
10–20	8
20–30	15
30–40	19
40–50	22
50–60	9
60–70	4

Check your understanding

Mean, median and mode

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