MathsCBSE Class 10 · Statistics

Assumed Mean Method for the Mean of Grouped Data

Updated 14 October 2026 · Fact-checked

The assumed mean method finds the mean of grouped data by picking a convenient class mark as an assumed mean, working with small deviations from it, and then correcting the guess. Board questions ask you to find a mean by this method, often with a missing frequency, unequal class widths or a real-life context.

Deviations from a are negative on the left and positive on the right; the correction Σfd/Σf moves your guess a to the true mean.

Illustration drawn for QuestionBuddy.

Understand Assumed Mean Method

The mean of grouped data is $\overset{x}{ˉ} = \frac{\sum f _{i} x _{i}}{\sum f _{i}}$ , where $x_{i}$ is the class mark. When class marks are large, such as 450 or 1100, the products $f_{i} x_{i}$ become big and easy to get wrong. The assumed mean method gives the same answer with much smaller numbers.

You pick a value $a$ , called the assumed mean. It is usually a class mark near the middle of the table. For every class you find the deviation $d_{i} = x_{i} - a$ . Classes below $a$ give negative deviations, the class at $a$ gives $0$ , and classes above give positive ones. Then $\overset{x}{ˉ} = a + \frac{\sum f _{i} d _{i}}{\sum f _{i}}$ .

Why it works. Every class mark can be written as $x_{i} = a + d_{i}$ . So $\sum f_{i} x_{i} = a \sum f_{i} + \sum f_{i} d_{i}$ . Divide both sides by $\sum f_{i}$ and you get $\overset{x}{ˉ} = a + \frac{\sum f _{i} d _{i}}{\sum f _{i}}$ . The second term is simply the correction to your guess. If $\sum f_{i} d_{i}$ is positive, the true mean is above $a$ . If it is negative, the mean is below $a$ .

**Any $a$ works.** The algebra above never used a special value of $a$ , so every choice gives the same mean. A middle class mark just keeps the deviations smallest and splits them into positive and negative values that partly cancel. Choosing a class mark (not a random number) also keeps one deviation equal to $0$ .

Where it fits. This method sits between the direct method and the step deviation method. Step deviation needs every $d_{i}$ to share a common factor $h$ . With equal widths this is automatic. With unequal widths it may not hold, and even when it does the assumed mean method is the safe choice. It still rests on the same assumption as the direct method: every observation in a class is taken to be equal to its class mark. So the mean you get is an estimate of the true mean of the raw data, and all three methods give the same value for the same table.

It is also a good tool for missing frequency questions. With small deviations, the equation in the unknown frequency stays short and easy to solve.

Assumed Mean Method: formulas to remember

Class mark: $x_{i} = \frac{lower limit + upper limit}{2}$; Find this first for every class. Deviations are always taken from class marks, never from class limits.
Deviation from the assumed mean: $d_{i} = x_{i} - a$; Always class mark minus assumed mean. Classes below a give negative d_i.
Mean by the assumed mean method: $\overset{x}{ˉ} = a + \frac{\sum f _{i} d _{i}}{\sum f _{i}}$; Take a as a middle class mark. Works for equal and unequal class widths.
Why the formula works: $\sum f_{i} x_{i} = a \sum f_{i} + \sum f_{i} d_{i}$; Comes from x_i = a + d_i. Dividing by the sum of frequencies gives the method's formula.
Sign check: $\sum f_{i} d_{i} > 0 \Rightarrow \overset{x}{ˉ} > a, \sum f_{i} d_{i} < 0 \Rightarrow \overset{x}{ˉ} < a$; Use this to check whether your mean lies on the correct side of a.

How to solve Assumed Mean Method questions

Use this routine for any assumed mean question, including ones with a missing frequency. The table itself earns marks, so always draw it.

1Check the classes. If the question gives inclusive classes like 10–19, the class marks are still the midpoints, so you can work with them directly.
2Find the class mark $x_{i}$ of every class and write it in a column.
3Choose the assumed mean $a$ : a class mark near the middle of the table, ideally one with a large frequency. State it clearly, for example 'Let $a = 35$ '.
4Write the column $d_{i} = x_{i} - a$ , keeping signs: classes with a smaller class mark than $a$ (listed before it) give negative $d_{i}$ , the class at $a$ gives $0$ , and classes with a larger class mark (listed after it) give positive $d_{i}$ .
5Write the column $f_{i} d_{i}$ , then add to get $\sum f_{i}$ and $\sum f_{i} d_{i}$ . Add the negatives and the positives separately to avoid sign slips.
6Write the formula $\overset{x}{ˉ} = a + \frac{\sum f _{i} d _{i}}{\sum f _{i}}$ , substitute and simplify. If a frequency is unknown, set this equal to the given mean and solve the linear equation.
7Check: the mean must lie on the side of $a$ shown by the sign of $\sum f_{i} d_{i}$ , and within the range of the data. Write the units in the final answer.

Assumed Mean Method in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Mean by the assumed mean method

Suppose your family is choosing a mobile plan and you note the daily data use (in MB) of 40 students in your class for a project. One plan being considered gives 1000 MB (1 GB) per day. The class marks run into hundreds and thousands, so you decide to use the assumed mean method.

The data is: 400–600 MB: 4 students, 600–800 MB: 8, 800–1000 MB: 14, 1000–1200 MB: 10, 1200–1400 MB: 4. Find the mean daily data use. Would a student using the mean amount stay within the 1000 MB daily plan, and how many MB would that student leave unused in total over a 30-day month?

Try it first, then show the solutionSolution

Class marks: 500, 700, 900, 1100, 1300.
Take $a = 900$ . Deviations $d_{i} = x_{i} - 900$ : $- 400, - 200, 0, 200, 400$ .
$f_{i} d_{i}$ : $4 \times (- 400) = - 1600$ , $8 \times (- 200) = - 1600$ , $0$ , $10 \times 200 = 2000$ , $4 \times 400 = 1600$ .
$\sum f_{i} = 40$ , $\sum f_{i} d_{i} = - 3200 + 3600 = 400$ .
$\overset{x}{ˉ} = 900 + \frac{400}{40} = 900 + 10 = 910$ MB.
$910 < 1000$ , so the plan covers the mean use, with $1000 - 910 = 90$ MB left each day.
Over 30 days the unused data adds up to $90 \times 30 = 2700$ MB in total (each day's limit resets, so this is the total unused, not a carried-over balance).

Answer: Mean daily use is 910 MB. Yes, it fits the 1000 MB plan, leaving 90 MB unused each day, 2700 MB unused in total over 30 days.

Memory hook: Guess the middle data class, then let the deviations correct your guess.

Tricks to remember Assumed Mean Method

Short memory aids. Say each one aloud once, then use it on a question the same day.

Guess, then correct
Mean = guess + average error. The assumed mean $a$ is your guess; $\frac{\sum f _{i} d _{i}}{\sum f _{i}}$ is how far off it was.
Use it when: Reminds you to add the correction to $a$ , not to report the correction itself as the mean.
Mark minus guess
$d = x - a$ : 'mark minus guess'. Below the guess is negative, above is positive.
Use it when: Stops you from writing $a - x_{i}$ and getting every sign flipped.
Sign tells the side
Positive $\sum f_{i} d_{i}$ pushes the mean up from $a$ ; negative pulls it down.
Use it when: A two-second check on your final answer before moving on.

Assumed Mean Method: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Mean of test marks by the assumed mean method · 3 marks · Short answerBoard level

The marks of 40 students in a class test are given below. Find the mean marks using the assumed mean method.
Marks: 10–20, 20–30, 30–40, 40–50, 50–60
Number of students: 5, 8, 12, 10, 5

Try it first, then show the solutionSolution

Class marks $x_{i}$ : 15, 25, 35, 45, 55. Let $a = 35$ . ( $\frac{1}{2}$ mark)
$d_{i} = x_{i} - 35$ : $- 20, - 10, 0, 10, 20$ . ( $\frac{1}{2}$ mark)
$f_{i} d_{i}$ : $- 100, - 80, 0, 100, 100$ . ( $\frac{1}{2}$ mark)
$\sum f_{i} = 40$ , $\sum f_{i} d_{i} = - 180 + 200 = 20$ . ( $\frac{1}{2}$ mark)
$\overset{x}{ˉ} = a + \frac{\sum f _{i} d _{i}}{\sum f _{i}} = 35 + \frac{20}{40} = 35.5$ . (1 mark)
Check: $\sum f_{i} d_{i} > 0$ , so the mean is above 35. Correct.

Answer: Mean marks = 35.5

Example 2: Missing frequency with the mean given · 3 marks · Short answerBoard level

The weights (in kg) of some parcels at a courier office are given below. The mean weight is 75.8 kg. Using the assumed mean method, find the missing frequency $k$ .
Weight (kg): 50–60, 60–70, 70–80, 80–90, 90–100
Number of parcels: 6, $k$ , 15, 12, 7

Try it first, then show the solutionSolution

Class marks: 55, 65, 75, 85, 95. Let $a = 75$ , so $d_{i}$ : $- 20, - 10, 0, 10, 20$ . ( $\frac{1}{2}$ mark)
$f_{i} d_{i}$ : $- 120, - 10 k, 0, 120, 140$ . ( $\frac{1}{2}$ mark)
$\sum f_{i} = 40 + k$ , $\sum f_{i} d_{i} = 140 - 10 k$ . ( $\frac{1}{2}$ mark)
$75 + \frac{140 - 10 k}{40 + k} = 75.8$ , so $\frac{140 - 10 k}{40 + k} = 0.8$ . ( $\frac{1}{2}$ mark)
$140 - 10 k = 32 + 0.8 k \Rightarrow 108 = 10.8 k \Rightarrow k = 10$ . ( $\frac{1}{2}$ mark)
Check: $\sum f_{i} = 50$ , $\sum f_{i} d_{i} = 40$ , mean $= 75 + 0.8 = 75.8$ kg. ( $\frac{1}{2}$ mark)

Answer: $k = 10$

Example 3: Unequal class widths · 3 marks · Short answerBoard level

A library records how long (in minutes) 50 visitors stay. Find the mean time using the assumed mean method.
Time (min): 0–20, 20–30, 30–50, 50–60, 60–80
Number of visitors: 8, 12, 15, 10, 5

Try it first, then show the solutionSolution

Widths are unequal (20, 10, 20, 10, 20). Step deviation needs every $d_{i}$ to share a common factor $h$ ; with unequal widths this is not guaranteed, so the assumed mean method is the safe choice.
Class marks: 10, 25, 40, 55, 70. Let $a = 40$ . ( $\frac{1}{2}$ mark)
$d_{i}$ : $- 30, - 15, 0, 15, 30$ . ( $\frac{1}{2}$ mark)
$f_{i} d_{i}$ : $- 240, - 180, 0, 150, 150$ . ( $\frac{1}{2}$ mark)
$\sum f_{i} = 50$ , $\sum f_{i} d_{i} = - 420 + 300 = - 120$ . ( $\frac{1}{2}$ mark)
$\overset{x}{ˉ} = 40 + \frac{- 120}{50} = 40 - 2.4 = 37.6$ minutes. (1 mark)

Answer: Mean time = 37.6 minutes

Example 4: Two missing frequencies from total and mean · 5 marks · Long answerHarder board level

The scores of 50 players in a quiz league are given below. The mean score is 52. Using the assumed mean method, find $p$ and $q$ .
Score: 0–20, 20–40, 40–60, 60–80, 80–100
Number of players: 7, $p$ , 12, $q$ , 8

Try it first, then show the solutionSolution

Total: $7 + p + 12 + q + 8 = 50 \Rightarrow p + q = 23$ ... (1). (1 mark)
Class marks: 10, 30, 50, 70, 90. Let $a = 50$ , $d_{i}$ : $- 40, - 20, 0, 20, 40$ . (1 mark)
$f_{i} d_{i}$ : $- 280, - 20 p, 0, 20 q, 320$ , so $\sum f_{i} d_{i} = 40 - 20 p + 20 q$ . (1 mark)
$52 = 50 + \frac{40 - 20 p + 20 q}{50} \Rightarrow 40 - 20 p + 20 q = 100 \Rightarrow q - p = 3$ ... (2). (1 mark)
Adding (1) and (2): $2 q = 26 \Rightarrow q = 13$ , so $p = 10$ . ( $\frac{1}{2}$ mark)
Check by direct method: $\sum f_{i} x_{i} = 70 + 300 + 600 + 910 + 720 = 2600$ , and $\frac{2600}{50} = 52$ . ( $\frac{1}{2}$ mark)

Answer: $p = 10$ , $q = 13$

Example 5: Case study: monthly savings · 4 marks · Case-basedBoard level

Suppose a school savings club records the monthly savings (in Rs) of 60 students.
Savings (Rs): 200–300, 300–400, 400–500, 500–600, 600–700
Number of students: 6, 12, 20, 14, 8
The treasurer uses the assumed mean method with $a = 450$ .
(i) Write the deviation $d_{i}$ for the class 600–700. (1 mark)
(ii) Find $\sum f_{i} d_{i}$ . (1 mark)
(iii) Find the mean savings. A friend uses $a = 550$ instead. Show that she gets the same mean. (2 marks)

Try it first, then show the solutionSolution

(i) Class mark of 600–700 is 650, so $d = 650 - 450 = 200$ . (1 mark)
(ii) $d_{i}$ : $- 200, - 100, 0, 100, 200$ ; $f_{i} d_{i}$ : $- 1200, - 1200, 0, 1400, 1600$ ; $\sum f_{i} d_{i} = 600$ . (1 mark)
(iii) $\overset{x}{ˉ} = 450 + \frac{600}{60} = 460$ . (1 mark)
With $a = 550$ : $d_{i}$ : $- 300, - 200, - 100, 0, 100$ ; $f_{i} d_{i}$ : $- 1800, - 2400, - 2000, 0, 800$ ; sum $= - 5400$ . ( $\frac{1}{2}$ mark)
$\overset{x}{ˉ} = 550 - \frac{5400}{60} = 550 - 90 = 460$ , the same, since the formula holds for any $a$ . ( $\frac{1}{2}$ mark)

Answer: (i) 200 (ii) 600 (iii) Mean savings = Rs 460 with either assumed mean.

Example 6: Inclusive classes with a frequency in terms of p · 5 marks · Long answerHarder board level

The marks scored by students in a quiz are given below. The mean mark is 34.25. Using the assumed mean method, find $p$ and hence the frequencies of the classes 20–29 and 40–49.
Marks: 10–19, 20–29, 30–39, 40–49, 50–59
Number of students: 5, $p + 3$ , 11, $2 p$ , 3

Try it first, then show the solutionSolution

The classes are inclusive, but the class mark is still the midpoint: $\frac{10 + 19}{2} = 14.5$ . Class marks: 14.5, 24.5, 34.5, 44.5, 54.5 (converting to 9.5–19.5 and so on gives the same midpoints). Let $a = 34.5$ . (1 mark)
$d_{i} = x_{i} - 34.5$ : $- 20, - 10, 0, 10, 20$ . $f_{i} d_{i}$ : $- 100$ , $- 10 p - 30$ , $0$ , $20 p$ , $60$ . (1 mark)
$\sum f_{i} = 5 + p + 3 + 11 + 2 p + 3 = 3 p + 22$ , $\sum f_{i} d_{i} = 10 p - 70$ . (1 mark)
$34.5 + \frac{10 p - 70}{3 p + 22} = 34.25 \Rightarrow \frac{10 p - 70}{3 p + 22} = - \frac{1}{4}$ . (1 mark)
$40 p - 280 = - 3 p - 22 \Rightarrow 43 p = 258 \Rightarrow p = 6$ . Frequencies: $p + 3 = 9$ and $2 p = 12$ . ( $\frac{1}{2}$ mark)
Check: $\sum f_{i} = 40$ , $\sum f_{i} d_{i} = 60 - 70 = - 10$ , mean $= 34.5 - 0.25 = 34.25$ . ( $\frac{1}{2}$ mark)

Answer: $p = 6$ ; the classes 20–29 and 40–49 have frequencies 9 and 12.

Example 7: Correcting wrongly recorded data · 5 marks · Long answerHOTS / competency

A teacher records the marks of 50 students.
Marks: 0–20, 20–40, 40–60, 60–80, 80–100
Number of students: 6, 10, 18, 12, 4
(i) Find the mean by the assumed mean method.
(ii) Later she finds that one student's mark was written as 45 instead of 54. Does the mean of this grouped data change? Give a reason.
(iii) She also finds that 5 students who scored in 60–80 had been entered in 20–40 by mistake. Find the corrected mean.

Try it first, then show the solutionSolution

(i) Class marks: 10, 30, 50, 70, 90. Let $a = 50$ ; $d_{i}$ : $- 40, - 20, 0, 20, 40$ . (1 mark)
$f_{i} d_{i}$ : $- 240, - 200, 0, 240, 160$ ; $\sum f_{i} d_{i} = - 40$ , so $\overset{x}{ˉ} = 50 - \frac{40}{50} = 49.2$ . (1 mark)
(ii) Both 45 and 54 lie in the class 40–60, so no frequency changes. The grouped mean stays 49.2, because every observation in a class is taken as its class mark. (1 mark)
(iii) Corrected frequencies: 20–40 becomes 5 and 60–80 becomes 17; the total is still 50. Each moved student's $d$ changes from $- 20$ to $20$ , a rise of 40, so $\sum f_{i} d_{i}$ rises by $5 \times 40 = 200$ to $160$ . (1 mark)
Check: $- 240 - 100 + 0 + 340 + 160 = 160$ . Corrected mean $= 50 + \frac{160}{50} = 53.2$ , which is $\frac{200}{50} = 4$ marks higher, as expected. (1 mark)

Answer: (i) 49.2 (ii) No change; the mean stays 49.2 because both values are in the same class. (iii) Corrected mean = 53.2 marks

Example 8: Reading a half-filled assumed mean table · 3 marks · Short answerHarder board level

Part of an assumed mean table is shown below. Some entries are missing and the assumed mean is not stated. Find the assumed mean, complete the table and find the mean.
Class: 0–10, 10–20, 20–30, 30–40, 40–50
$f_{i}$ : 4, ?, 17, 10, ?
$d_{i}$ : ?, $- 10$ , 0, ?, ?
$f_{i} d_{i}$ : $- 80$ , $- 60$ , ?, ?, 60

Try it first, then show the solutionSolution

The class mark of 10–20 is 15 and its $d_{i} = - 10$ , so $a = 15 + 10 = 25$ , which is why 20–30 has $d_{i} = 0$ . ( $\frac{1}{2}$ mark)
$d_{i}$ : $- 20, - 10, 0, 10, 20$ . Check: $4 \times (- 20) = - 80$ , matching the table. ( $\frac{1}{2}$ mark)
Missing frequencies: $f_{2} = \frac{- 60}{- 10} = 6$ and $f_{5} = \frac{60}{20} = 3$ . ( $\frac{1}{2}$ mark)
Missing $f_{i} d_{i}$ : $17 \times 0 = 0$ and $10 \times 10 = 100$ . ( $\frac{1}{2}$ mark)
$\sum f_{i} = 4 + 6 + 17 + 10 + 3 = 40$ , $\sum f_{i} d_{i} = - 140 + 160 = 20$ . ( $\frac{1}{2}$ mark)
$\overset{x}{ˉ} = 25 + \frac{20}{40} = 25.5$ . Check by direct method: $\frac{20 + 90 + 425 + 350 + 135}{40} = \frac{1020}{40} = 25.5$ . ( $\frac{1}{2}$ mark)

Answer: $a = 25$ , missing frequencies 6 and 3, missing deviations $- 20, 10, 20$ , and mean = 25.5

Where students lose marks in Assumed Mean Method

The slips examiners see most often, and what to write instead.

Mistake: Taking deviations the wrong way round
Writing $a - x_{i}$ flips every sign, so the correction moves the mean the wrong way.
Wrong working
$a = 35$ , $x = 15$ : $d = 35 - 15 = 20$
Right working
$d = x_{i} - a = 15 - 35 = - 20$
Do this: Always 'class mark minus assumed mean'. Classes before $a$ must have negative $d_{i}$ .
Usually costs: The final answer and its mark, often 1 to 2 marks.
Mistake: Reporting the correction as the mean
The student stops at $\frac{\sum f _{i} d _{i}}{\sum f _{i}}$ and forgets to add $a$ .
Wrong working
$\frac{20}{40} = 0.5$ , so mean $= 0.5$
Right working
Mean $= 35 + \frac{20}{40} = 35.5$
Do this: Write the full formula with $a$ before substituting, and check the mean lies inside the data range.
Usually costs: The final answer mark.
Mistake: Dividing by the number of classes instead of the total frequency
The student confuses 'how many classes' with 'how many observations'.
Wrong working
$\overset{x}{ˉ} = 35 + \frac{20}{5} = 39$
Right working
$\overset{x}{ˉ} = 35 + \frac{20}{40} = 35.5$
Do this: The denominator is always $\sum f_{i}$ , which you write at the foot of the frequency column.
Usually costs: 1 to 2 marks.
Mistake: Taking deviations from class limits
The student uses lower limits as $x_{i}$ instead of class marks.
Wrong working
Class 10–20 with $a = 35$ : $d = 10 - 35 = - 25$ (the lower limit used)
Right working
Class mark of 10–20 is 15, so $d = 15 - 35 = - 20$ , using class marks throughout
Do this: Make the $x_{i}$ column of midpoints first, and choose $a$ from that column.
Usually costs: The whole calculation, up to 3 marks.

Exam tips

Write 'Let assumed mean $a = \dots$ ' as a separate line. Examiners look for it before giving method marks.
Draw the full table with columns class, $f_{i}$ , $x_{i}$ , $d_{i}$ , $f_{i} d_{i}$ , and show the totals. The table carries step marks even if the final arithmetic slips.
If the question says 'by the assumed mean method', using the direct method can lose method marks even with the correct answer.
In missing frequency questions, choose $a$ as the class mark of the unknown class when possible: its $f_{i} d_{i}$ becomes 0 and the equation gets simpler.
Finish with a quick check: compare the sign of $\sum f_{i} d_{i}$ with the side of $a$ your mean is on, and write units.

Practice questions on Assumed Mean Method

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 3 marks · Short answerBoard level
Find the mean of the following data using the assumed mean method.
Classes: 0–10, 10–20, 20–30, 30–40, 40–50
Frequencies: 3, 7, 15, 10, 5
Hint
Take $a = 25$ , the middle class mark.
Answer
$d_{i}$ : $- 20, - 10, 0, 10, 20$ ; $f_{i} d_{i}$ : $- 60, - 70, 0, 100, 100$ ; $\sum f_{i} d_{i} = 70$ , $\sum f_{i} = 40$ . Mean $= 25 + 1.75 = 26.75$ .
Q2 · 3 marks · Short answerBoard level
The mean of the distribution below is 34.5. Using the assumed mean method, find $k$ .
Classes: 10–20, 20–30, 30–40, 40–50, 50–60
Frequencies: 4, 6, $k$ , 8, 2
Hint
Choose $a$ carefully.
Answer
Take $a = 35$ . $f_{i} d_{i}$ : $- 80, - 60, 0, 80, 40$ ; $\sum f_{i} d_{i} = - 20$ . $35 - \frac{20}{20 + k} = 34.5 \Rightarrow 20 + k = 40 \Rightarrow k = 20$ .
Q3 · 3 marks · Short answerHarder board level
The rainfall (in mm) on 50 days is: 0–10: 5 days, 10–30: 15 days, 30–60: 20 days, 60–80: 6 days, 80–100: 4 days. Find the mean rainfall using the assumed mean method. Can the step deviation method, with h equal to the class width, be used here? Why or why not?
Hint
Class marks are 5, 20, 45, 70, 90. Take $a = 45$ .
Answer
$d_{i}$ : $- 40, - 25, 0, 25, 45$ ; $f_{i} d_{i}$ : $- 200, - 375, 0, 150, 180$ ; sum $= - 245$ . Mean $= 45 - \frac{245}{50} = 45 - 4.9 = 40.1$ mm. No: the class widths are 10, 20, 30, 20 and 20, which are not equal, so there is no single class width to use as h. The assumed mean method works whatever the widths.
Q4 · 5 marks · Long answerHarder board level
The total frequency of the distribution below is 40 and its mean is 26. Using the assumed mean method, find $x$ and $y$ .
Classes: 0–10, 10–20, 20–30, 30–40, 40–50
Frequencies: 5, $x$ , 10, $y$ , 7
Hint
One equation comes from the total, the other from the mean with $a = 25$ .
Answer
$x + y = 18$ . $\sum f_{i} d_{i} = - 100 - 10 x + 10 y + 140 = 40 - 10 x + 10 y$ . $25 + \frac{40 - 10 x + 10 y}{40} = 26 \Rightarrow y - x = 0$ . So $x = y = 9$ . Check: $\sum f_{i} x_{i} = 1040$ , $\frac{1040}{40} = 26$ .
Q5 · 3 marks · Short answerHOTS / competency
For the data 40–50: 5, 50–60: 10, 60–70: 15, 70–80: 12, 80–90: 8, a student took $a = 65$ and wrote deviations $20, 10, 0, - 10, - 20$ , getting $\sum f_{i} d_{i} = - 80$ and mean $63.4$ . Find the error, give the correct mean, and explain how the frequencies alone suggest the mean should exceed 65.
Hint
Check the order of subtraction in $d_{i}$ .
Answer
The student used $a - x_{i}$ instead of $x_{i} - a$ . Correct $d_{i}$ : $- 20, - 10, 0, 10, 20$ ; $f_{i} d_{i}$ : $- 100, - 100, 0, 120, 160$ ; sum $= 80$ . Mean $= 65 + 1.6 = 66.6$ . The classes sit at equal distances (10 and 20) on both sides of 65, and each class above has at least as many students as its partner below (12 against 10, 8 against 5), so $\sum f_{i} d_{i} > 0$ and the mean is above 65.
Q6 · 1 mark · MCQBoard level
In the assumed mean method, $a = 47.5$ , $\sum f_{i} = 40$ and $\sum f_{i} d_{i} = - 150$ . The mean is
(a) 51.25
(b) 43.75
(c) 47.5
(d) 44.25
Hint
The correction is negative, so the mean is below $a$ .
Answer
(b) 43.75, since $47.5 - \frac{150}{40} = 47.5 - 3.75 = 43.75$ .
Q7 · 1 mark · MCQHarder board level
For a grouped distribution, $\sum f_{i} = 50$ , and a student using $a = 60$ finds the mean to be 62. If $a = 70$ is used instead for the same table, $\sum f_{i} d_{i}$ will be
(a) 100
(b) $- 400$
(c) 400
(d) $- 100$
Hint
The mean does not change when $a$ changes.
Answer
(b) $- 400$ . The mean stays 62, so $70 + \frac{\sum f _{i} d _{i}}{50} = 62$ gives $\sum f_{i} d_{i} = 50 \times (62 - 70) = - 400$ .
Q8 · 1 mark · Assertion-reasonBoard level
Assertion (A): For a grouped table, a student finds the mean as 36.2 using $a = 35$ . Using $a = 45$ for the same table, she will again get 36.2.
Reason (R): $x_{i} = a + d_{i}$ for every class, so $\sum f_{i} x_{i} = a \sum f_{i} + \sum f_{i} d_{i}$ for any choice of $a$ .
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Did the derivation of the formula use any special value of $a$ ?
Answer
(a) Both A and R are true and R is the correct explanation of A. Dividing R by $\sum f_{i}$ gives the same mean for every $a$ .
Q9 · 2 marks · Short answerBoard level
The mean of the distribution below is 19. Using the assumed mean method with $a = 15$ , find $k$ .
Classes: 0–10, 10–20, 20–30, 30–40
Frequencies: 4, 6, $k$ , 2
Hint
Write $\sum f_{i}$ and $\sum f_{i} d_{i}$ in terms of $k$ .
Answer
$d_{i}$ : $- 10, 0, 10, 20$ ; $\sum f_{i} d_{i} = - 40 + 10 k + 40 = 10 k$ , $\sum f_{i} = 12 + k$ (1 mark). $15 + \frac{10 k}{12 + k} = 19 \Rightarrow 10 k = 48 + 4 k \Rightarrow k = 8$ (1 mark). Check: $15 + \frac{80}{20} = 19$ .
Q10 · 3 marks · Short answerHOTS / competency
Section A of Class 10 has 30 students with marks: 20–30: 4, 30–40: 8, 40–50: 10, 50–60: 6, 60–70: 2. Section B has 20 students whose mean mark is 48. Find the mean of Section A by the assumed mean method, then the mean mark of all 50 students. Explain why the overall mean is not simply the average of the two section means.
Hint
Overall mean = total marks of both sections divided by 50.
Answer
Section A: $a = 45$ , $d_{i}$ : $- 20, - 10, 0, 10, 20$ ; $f_{i} d_{i}$ : $- 80, - 80, 0, 60, 40$ ; sum $= - 60$ ; mean $= 45 - 2 = 43$ . Totals: $30 \times 43 = 1290$ and $20 \times 48 = 960$ . Overall mean $= \frac{2250}{50} = 45$ . The simple average $\frac{43 + 48}{2} = 45.5$ is wrong because Section A has more students, so its mean must carry more weight.
Q11 · 4 marks · Case-basedHOTS / competency
Suppose your school surveys the one-way auto fare (in Rs) paid by 50 students to reach school.
Fare (Rs): 20–40, 40–60, 60–80, 80–100, 100–120
Number of students: 5, 12, 18, 10, 5
The school plans a travel allowance of Rs 70 per trip for each student.
(i) Which class mark would you choose as the assumed mean, and why? (1 mark)
(ii) Find $\sum f_{i} d_{i}$ with your choice. (1 mark)
(iii) Find the mean fare. Is Rs 70 per trip enough for the average student? What does the sign of $\sum f_{i} d_{i}$ tell the school about the total fare for one trip of all 50 students? (2 marks)
Hint
Think about which class mark keeps the deviations small.
Answer
(i) 70: it is the middle class mark and has the largest frequency, so the deviations stay small and one of them is zero. (ii) $d_{i}$ : $- 40, - 20, 0, 20, 40$ ; $f_{i} d_{i}$ : $- 200, - 240, 0, 200, 200$ ; $\sum f_{i} d_{i} = - 40$ . (iii) Mean $= 70 - \frac{40}{50} = 69.2$ , that is Rs 69.20, so Rs 70 is enough on average, with Rs 0.80 to spare. Since $\sum f_{i} d_{i} = - 40$ is negative, the estimated total fare is Rs 40 less than the Rs 3500 allowance ( $50 \times 70 = 3500$ ), so the allowance covers the total with Rs 40 left.
Q12 · 5 marks · Long answerHarder board level
The mean of the distribution below is 28.75. Using the assumed mean method, find $y$ and the frequencies of the classes 15–24 and 35–44.
Classes: 5–14, 15–24, 25–34, 35–44, 45–54
Frequencies: 6, $2 y$ , 10, $y + 4$ , 5
Hint
Class marks of inclusive classes are still midpoints, for example 9.5.
Answer
Class marks: 9.5, 19.5, 29.5, 39.5, 49.5; take $a = 29.5$ , $d_{i}$ : $- 20, - 10, 0, 10, 20$ . $f_{i} d_{i}$ : $- 120, - 20 y, 0, 10 y + 40, 100$ , so $\sum f_{i} d_{i} = 20 - 10 y$ and $\sum f_{i} = 25 + 3 y$ . $29.5 + \frac{20 - 10 y}{25 + 3 y} = 28.75 \Rightarrow 4 (20 - 10 y) = - 3 (25 + 3 y) \Rightarrow 80 - 40 y = - 75 - 9 y \Rightarrow y = 5$ . Frequencies: 10 and 9. Check: $\sum f_{i} = 40$ , $\sum f_{i} d_{i} = - 30$ , mean $= 29.5 - 0.75 = 28.75$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.State the assumed mean formula.Show answer
$\overset{x}{ˉ} = a + \frac{\sum f _{i} d _{i}}{\sum f _{i}}$ , with $d_{i} = x_{i} - a$ .
2.What is $d_{i}$ for class mark 25 when $a = 45$ ?Show answer
$25 - 45 = - 20$
3.If $a = 60$ , $\sum f_{i} = 50$ , $\sum f_{i} d_{i} = - 100$ , what is the mean?Show answer
$60 - 2 = 58$
4.Does the mean depend on the choice of $a$ ?Show answer
No. Any $a$ gives the same mean; a middle class mark just keeps the numbers small.
5.Can this method be used with unequal class widths?Show answer
Yes, it works for any widths. Step deviation needs every $d_{i}$ to share a common factor $h$ , which is automatic only when widths are equal.

Assumed Mean Method: frequently asked questions

How do I choose the assumed mean?

Pick a class mark near the middle of the table, preferably one with a high frequency. Any value gives the correct mean, but a middle class mark keeps the deviations small and makes one of them zero.

Will I get the same answer as the direct method?

Yes. The assumed mean method is just the direct method rearranged, so both give exactly the same mean for the same table. You can use the direct method as a check if time allows.

When should I use the assumed mean method instead of step deviation?

Use it when the question names it, or when class widths are unequal. Step deviation needs every deviation to share a common factor h; with equal widths this is automatic, but with unequal widths it may not hold, and even when it does the assumed mean method is the safe choice. If widths are equal and class marks are large, step deviation is usually quicker unless another method is asked.

Are ogives needed for this topic?

No. Graphical representation of cumulative frequency (ogives) is not in the 2026-27 syllabus.