MathsCBSE Class 10 · Statistics

Mean of Grouped Data by the Direct Method

Updated 13 October 2026 · Fact-checked

The direct method finds the mean of grouped data by taking each class mark, multiplying it by its frequency, adding these products and dividing by the total frequency. Board questions ask you to find the mean of a table, find one or two missing frequencies when the mean is given, and interpret the mean in real-life data.

Treat every value in a class as its class mark, multiply by frequency, add these products, then divide by total frequency.

Illustration drawn for QuestionBuddy.

Understand Mean of Grouped Data: Direct Method

In Class 9 you found the mean by adding every observation and dividing by how many there were. In grouped data you cannot do that, because you no longer see each observation. You only see classes such as 20–30 and a frequency that tells you how many observations fall in each class.

So you make one sensible assumption: every observation in a class sits at the centre of that class. This centre is the class mark, $x_{i} = \frac{lower limit + upper limit}{2}$ . If a class has $f_{i}$ observations, all treated as equal to $x_{i}$ , that class adds $f_{i} x_{i}$ to the grand total. Add these over all classes to get $\sum f_{i} x_{i}$ , then divide by the number of observations $\sum f_{i}$ . That gives $\overset{x}{ˉ} = \frac{\sum f _{i} x _{i}}{\sum f _{i}}$ .

Why is this a fair estimate? Inside a class, some values are below the class mark and some are above. The assumption is that these roughly cancel out. The result is an estimate of the true mean, and it is the value the board expects.

The method holds for any grouped table: classes of equal or unequal width, and inclusive classes like 10–14, 15–19. For inclusive classes you do not need to convert to continuous classes, because the class marks stay the same either way. If data comes as a cumulative frequency table, first subtract consecutive cumulative frequencies to get ordinary frequencies.

Use the direct method when class marks and frequencies are small, so the products are easy. It is also the neatest method when a frequency is missing and the mean is given: write $\sum f_{i}$ and $\sum f_{i} x_{i}$ in terms of the unknown, equate to the given mean and solve a linear equation. The assumed mean and step deviation methods are shortcuts built on this same formula, and all three give the same answer.

Mean of Grouped Data: Direct Method: formulas to remember

Class mark: $x_{i} = \frac{lower limit + upper limit}{2}$; The representative value of a class. Unchanged when inclusive classes are made continuous.
Mean by the direct method: $\overset{x}{ˉ} = \frac{\sum f _{i} x _{i}}{\sum f _{i}}$; Divide by the total frequency, never by the number of classes.
Total of all observations (estimate): $Total = \sum f_{i} x_{i} = \overset{x}{ˉ} \times \sum f_{i}$; Useful in case studies asking for a total amount, and for checking a missing-frequency answer.
Frequency from a 'less than' cumulative table: $f_{i} = c f_{i} - c f_{i - 1}, c f_{0} = 0$; Only for a less than table: subtract the previous cumulative frequency. The first class frequency equals its own cf.
Frequency from a 'more than' cumulative table: $f_{i} = c f_{i} - c f_{i + 1}$; Only for a more than table: subtract the next cumulative frequency. The last class frequency equals its own cf. Example: more than 0: 40, more than 10: 37, more than 20: 30 gives $f = 40 - 37 = 3$ for 0–10 and $f = 37 - 30 = 7$ for 10–20.

How to solve Mean of Grouped Data: Direct Method questions

This method works for any direct-method question, including missing frequencies.

1Read the classes. If a cumulative table is given, convert it to ordinary frequencies by subtraction.
2Draw a table with columns: class, $f_{i}$ , $x_{i}$ , $f_{i} x_{i}$ .
3Find each class mark $x_{i}$ as the average of the two limits.
4Multiply to get each $f_{i} x_{i}$ , writing any unknown frequency as it is (for example $35 k$ ).
5Add the columns to get $\sum f_{i}$ and $\sum f_{i} x_{i}$ .
6Write the formula $\overset{x}{ˉ} = \frac{\sum f _{i} x _{i}}{\sum f _{i}}$ , then substitute. If the mean is given, equate and solve for the unknown.
7Check: the mean must lie between the smallest and largest class marks, and an unknown frequency must be a whole number that satisfies the equation.
8Write the final answer with units.

Mean of Grouped Data: Direct 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 of grouped data by the direct method

Suppose your housing society surveys how much mobile data each family uses per day, grouped into classes of 4 GB. One count got smudged on the survey sheet, but the secretary remembers that the mean came out as 8.4 GB per day. The society is considering an imagined shared plan that gives 10 GB a day per family.

The data is: 0–4 GB: 6 families, 4–8 GB: 10, 8–12 GB: $k$ , 12–16 GB: 8. The mean daily use is 8.4 GB. Find $k$ and the number of families surveyed. Then estimate the total daily use, and find how much data the society would have spare each day if every surveyed family got 10 GB.

Try it first, then show the solutionSolution

Class marks: $2, 6, 10, 14$ .
$f_{i} x_{i}$ : $12, 60, 10 k, 112$ , so $\sum f_{i} x_{i} = 184 + 10 k$ .
$\sum f_{i} = 24 + k$ .
$\frac{184 + 10 k}{24 + k} = 8.4 \Rightarrow 184 + 10 k = 201.6 + 8.4 k$ .
$1.6 k = 17.6 \Rightarrow k = 11$ .
Families surveyed $= 24 + 11 = 35$ .
Estimated total use $= 184 + 110 = 294$ GB per day. Check: $\frac{294}{35} = 8.4$ .
Allowance $= 35 \times 10 = 350$ GB, so spare $= 350 - 294 = 56$ GB per day.

Answer: $k = 11$ ; 35 families surveyed; total use about 294 GB per day; about 56 GB per day spare.

Memory hook: A smudged count hides in both the top and the bottom of the mean: put $k$ in $\sum f_{i}$ and in $\sum f_{i} x_{i}$ .

Tricks to remember Mean of Grouped Data: Direct Method

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

Middle, Multiply, Divide
M-M-D: find the Middle of each class, Multiply by its frequency, Divide the total by total frequency.
Use it when: Recall the three columns of the table in order.
Divide by people, not boxes
The denominator counts observations ( $\sum f_{i}$ ), never the number of classes.
Use it when: Stops the most common division error.
Unknown goes in twice
A missing frequency $k$ appears in both $\sum f_{i}$ and $\sum f_{i} x_{i}$ .
Use it when: Setting up missing-frequency equations correctly.

Mean of Grouped Data: Direct Method: solved examples

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

Example 1: Mean pocket money of a class · 3 marks · Short answerBoard level

The weekly pocket money (in Rs) of 40 students is given below. Find the mean weekly pocket money using the direct method.
Pocket money (Rs): 0–10, 10–20, 20–30, 30–40, 40–50
Number of students: 4, 8, 14, 10, 4

Try it first, then show the solutionSolution

Class marks $x_{i}$ : $5, 15, 25, 35, 45$ . (½ mark)
$f_{i} x_{i}$ : $20, 120, 350, 350, 180$ . (1 mark)
$\sum f_{i} = 40$ , $\sum f_{i} x_{i} = 1020$ . (½ mark)
$\overset{x}{ˉ} = \frac{\sum f _{i} x _{i}}{\sum f _{i}} = \frac{1020}{40} = 25.5$ . (1 mark)

Answer: Mean weekly pocket money = Rs 25.50

Example 2: One missing frequency · 3 marks · Short answerBoard level

The mean of the following distribution is 33. Find the missing frequency $k$ .
Classes: 10–20, 20–30, 30–40, 40–50, 50–60
Frequencies: 8, 12, $k$ , 10, 5

Try it first, then show the solutionSolution

Class marks: $15, 25, 35, 45, 55$ . (½ mark)
$f_{i} x_{i}$ : $120, 300, 35 k, 450, 275$ ; so $\sum f_{i} x_{i} = 1145 + 35 k$ . (½ mark)
$\sum f_{i} = 35 + k$ . (½ mark)
$\frac{1145 + 35 k}{35 + k} = 33 \Rightarrow 1145 + 35 k = 1155 + 33 k$ . (½ mark)
$2 k = 10 \Rightarrow k = 5$ . (½ mark)
Check: $\frac{1320}{40} = 33$ . (½ mark)

Answer: $k = 5$

Example 3: Mean from a cumulative frequency table · 3 marks · Short answerBoard level

The times (in minutes) taken by 40 students to finish a quiz are: less than 10: 3, less than 20: 10, less than 30: 22, less than 40: 34, less than 50: 40. Find the mean time by the direct method.

Try it first, then show the solutionSolution

Frequencies by subtraction: 0–10: 3, 10–20: 7, 20–30: 12, 30–40: 12, 40–50: 6. (1 mark)
Class marks: $5, 15, 25, 35, 45$ ; $f_{i} x_{i}$ : $15, 105, 300, 420, 270$ . (1 mark)
$\sum f_{i} x_{i} = 1110$ , $\sum f_{i} = 40$ , so $\overset{x}{ˉ} = \frac{1110}{40} = 27.75$ . (1 mark)

Answer: Mean time = 27.75 minutes

Example 4: Two missing frequencies · 5 marks · Long answerHarder board level

The mean of the distribution below is 52 and the total frequency is 40. Find $p$ and $q$ .
Classes: 0–20, 20–40, 40–60, 60–80, 80–100
Frequencies: 5, $p$ , 10, $q$ , 5

Try it first, then show the solutionSolution

Total: $5 + p + 10 + q + 5 = 40 \Rightarrow p + q = 20$ . (1 mark)
Class marks: $10, 30, 50, 70, 90$ . (½ mark)
$f_{i} x_{i}$ : $50, 30 p, 500, 70 q, 450$ , so $\sum f_{i} x_{i} = 1000 + 30 p + 70 q$ . (1 mark)
$\frac{1000 + 30 p + 70 q}{40} = 52 \Rightarrow 30 p + 70 q = 1080 \Rightarrow 3 p + 7 q = 108$ . (1 mark)
Substitute $q = 20 - p$ : $3 p + 140 - 7 p = 108 \Rightarrow p = 8$ . (½ mark)
$q = 20 - 8 = 12$ . (½ mark)
Check: $50 + 240 + 500 + 840 + 450 = 2080$ and $\frac{2080}{40} = 52$ . (½ mark)

Answer: $p = 8$ , $q = 12$

Example 5: Case study: comparing two sections · 4 marks · Case-basedHOTS / competency

Suppose two sections of Class 10 take the same test out of 50 marks.
Marks: 0–10, 10–20, 20–30, 30–40, 40–50
Section A (30 students): 2, 6, 10, 8, 4
Section B (20 students): 2, 3, 5, 6, 4
(i) Which class has class mark 35, and how many students of both sections together lie in it? (1 mark)
(ii) What percentage of Section A scored 30 or more? (1 mark)
(iii) Find the mean marks of each section, decide with a reason which section did better on average, and find the mean of all 50 students. (2 marks)

Try it first, then show the solutionSolution

(i) Class mark 35 belongs to 30–40; students $= 8 + 6 = 14$ . (1 mark)
(ii) Section A with 30 or more: $8 + 4 = 12$ ; $\frac{12}{30} \times 100 = 40%$ . (1 mark)
(iii) Class marks $5, 15, 25, 35, 45$ . Section A: $f_{i} x_{i} = 10, 90, 250, 280, 180$ , sum $810$ , mean $= \frac{810}{30} = 27$ . (½ mark)
Section B: $f_{i} x_{i} = 10, 45, 125, 210, 180$ , sum $570$ , mean $= \frac{570}{20} = 28.5$ . (½ mark)
$28.5 > 27$ , so Section B did better on average, even though it has fewer students. (½ mark)
Combined mean $= \frac{810 + 570}{30 + 20} = \frac{1380}{50} = 27.6$ . It is not $\frac{27 + 28.5}{2} = 27.75$ , because the sections have different sizes. (½ mark)

Answer: (i) 30–40, 14 students (ii) 40% (iii) Section A: 27 marks, Section B: 28.5 marks; Section B did better on average; combined mean = 27.6 marks

Example 6: Unequal class widths and a range check · 3 marks · Short answerHarder board level

The time (in minutes) 40 students spend on homework daily is given below.
Time (minutes): 0–10, 10–30, 30–40, 40–60, 60–80
Number of students: 4, 10, 12, 10, 4
(i) Find the mean time by the direct method.
(ii) A classmate divided $\sum f_{i} x_{i}$ by the number of classes and got 284 minutes. Without redoing the sum, explain why this must be wrong.

Try it first, then show the solutionSolution

The classes have unequal widths, so find each class mark separately: $5, 20, 35, 50, 70$ . (½ mark)
$f_{i} x_{i}$ : $20, 200, 420, 500, 280$ . (½ mark)
$\sum f_{i} = 40$ , $\sum f_{i} x_{i} = 1420$ . (½ mark)
$\overset{x}{ˉ} = \frac{1420}{40} = 35.5$ . (½ mark)
(ii) The mean must lie between the smallest class mark $5$ and the largest class mark $70$ . Since $284 > 70$ , it is impossible; the classmate divided by $5$ classes instead of $\sum f_{i} = 40$ . (1 mark)

Answer: (i) Mean time = 35.5 minutes (ii) 284 lies outside the range 5 to 70, so it cannot be the mean.

Example 7: Inclusive classes with two missing frequencies · 5 marks · Long answerHarder board level

The ages (in years) of 40 people at a yoga camp are given below. The mean age is 22 years. Find $p$ and $q$ .
Age (years): 10–14, 15–19, 20–24, 25–29, 30–34
Number of people: 4, $p$ , 10, $q$ , 6

Try it first, then show the solutionSolution

Total: $4 + p + 10 + q + 6 = 40 \Rightarrow p + q = 20$ . (1 mark)
Class marks: $12, 17, 22, 27, 32$ . No need to make the classes continuous, since class marks do not change. (½ mark)
$f_{i} x_{i}$ : $48, 17 p, 220, 27 q, 192$ , so $\sum f_{i} x_{i} = 460 + 17 p + 27 q$ . (1 mark)
$\frac{460 + 17 p + 27 q}{40} = 22 \Rightarrow 17 p + 27 q = 420$ . (1 mark)
Substitute $q = 20 - p$ : $17 p + 540 - 27 p = 420 \Rightarrow 10 p = 120 \Rightarrow p = 12$ . (½ mark)
$q = 20 - 12 = 8$ . (½ mark)
Check: $48 + 204 + 220 + 216 + 192 = 880$ and $\frac{880}{40} = 22$ . (½ mark)

Answer: $p = 12$ , $q = 8$

Where students lose marks in Mean of Grouped Data: Direct Method

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

Mistake: Using a class limit instead of the class mark
Students take the lower or upper limit as the class value.
Wrong working
For 20–30 with $f = 5$ : $5 \times 30 = 150$
Right working
$x = \frac{20 + 30}{2} = 25$ , so $5 \times 25 = 125$
Do this: Always write an $x_{i}$ column first.
Usually costs: Most of the calculation marks in the question.
Mistake: Dividing by the number of classes
Confusing classes with observations.
Wrong working
$\overset{x}{ˉ} = \frac{1020}{5} = 204$
Right working
$\overset{x}{ˉ} = \frac{1020}{40} = 25.5$
Do this: Check the mean lies between the smallest and largest class marks.
Usually costs: The final answer mark.
Mistake: Leaving the unknown out of the total frequency
Students add only the known frequencies.
Wrong working
$\frac{1145 + 35 k}{35} = 33$
Right working
$\frac{1145 + 35 k}{35 + k} = 33 \Rightarrow k = 5$
Do this: The unknown appears in both sums.
Usually costs: Equation and answer marks, often 2 marks.
Mistake: Treating cumulative frequencies as frequencies
Not noticing 'less than' in the table.
Wrong working
Using 3, 10, 22, 34, 40 as $f_{i}$
Right working
$f_{i}$ : 3, 7, 12, 12, 6
Do this: Subtract consecutive cumulative frequencies first.
Usually costs: Almost the whole question.

Exam tips

Write the formula $\overset{x}{ˉ} = \frac{\sum f _{i} x _{i}}{\sum f _{i}}$ before substituting; it carries a step mark.
Show the full table with $x_{i}$ and $f_{i} x_{i}$ columns, including totals.
If a question names another method, use that one; use the direct method when none is named or when a frequency is missing.
In missing-frequency questions, verify by substitution in a final line.
Give units (Rs, minutes, years) in the final answer, especially in case studies.

Practice questions on Mean of Grouped Data: Direct Method

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

Q1 · 2 marks · Very short answerBoard level
Find the mean of the following data by the direct method.
Classes: 0–6, 6–12, 12–18, 18–24
Frequencies: 2, 5, 8, 5
Hint
Class marks are 3, 9, 15, 21.
Answer
$\sum f_{i} x_{i} = 6 + 45 + 120 + 105 = 276$ , $\sum f_{i} = 20$ , mean $= 13.8$
Q2 · 3 marks · Short answerBoard level
The mean of the distribution is 27. Find $f$ .
Classes: 0–10, 10–20, 20–30, 30–40, 40–50
Frequencies: 7, 10, $f$ , 13, 10
Hint
Write $\sum f_{i} x_{i} = 1090 + 25 f$ and $\sum f_{i} = 40 + f$ .
Answer
$1090 + 25 f = 27 (40 + f) = 1080 + 27 f \Rightarrow f = 5$ . Check: $\frac{1215}{45} = 27$ .
Q3 · 3 marks · Short answerBoard level
The ages (in years) of 50 participants in a sports camp are: 10–14: 5, 15–19: 10, 20–24: 15, 25–29: 12, 30–34: 8. Find the mean age. Do you need to make the classes continuous? Give a reason.
Hint
Class marks are 12, 17, 22, 27, 32 whether or not you adjust by 0.5.
Answer
$\sum f_{i} x_{i} = 60 + 170 + 330 + 324 + 256 = 1140$ ; mean $= \frac{1140}{50} = 22.8$ years. No, because class marks do not change when 0.5 is subtracted from lower limits and added to upper limits.
Q4 · 5 marks · Long answerHarder board level
The total frequency is 50 and the mean is 25.4. Find $a$ and $b$ .
Classes: 0–10, 10–20, 20–30, 30–40, 40–50
Frequencies: 6, $a$ , 15, $b$ , 7
Hint
One equation from the total ( $a + b = 22$ ), one from the mean.
Answer
$720 + 15 a + 35 b = 1270 \Rightarrow 3 a + 7 b = 110$ . With $b = 22 - a$ : $154 - 4 a = 110$ , so $a = 11$ , $b = 11$ . Check: $\frac{1270}{50} = 25.4$ .
Q5 · 3 marks · Short answerHOTS / competency
For classes 0–10, 10–20, 20–30 with frequencies 2, 3, 5, a student used the upper limits 10, 20, 30 in place of class marks and got a mean of 23. Find the correct mean and explain why the error is exactly the amount it is.
Hint
Each upper limit is half a class width above the class mark.
Answer
Correct: $\frac{10 + 45 + 125}{10} = 18$ . Every value used was $5$ (half of $h = 10$ ) too large, so the mean is $5$ too large: $23 - 5 = 18$ .
Q6 · 1 mark · MCQBoard level
For a grouped distribution, $\sum f_{i} = 20 + k$ and $\sum f_{i} x_{i} = 500 + 35 k$ . If the mean is 27, the value of $k$ is
(a) 4
(b) 5
(c) 6
(d) 8
Hint
Equate $\frac{500 + 35 k}{20 + k}$ to 27.
Answer
(b) 5, since $500 + 35 k = 540 + 27 k \Rightarrow 8 k = 40$ . Check: $\frac{675}{25} = 27$ .
Q7 · 1 mark · MCQHarder board level
For classes 5–15, 15–25, 25–35, 35–45 with some frequencies, the mean is 24. If the classes are changed to 10–20, 20–30, 30–40, 40–50 with the same frequencies, the new mean is
(a) 19
(b) 24
(c) 29
(d) 34
Hint
Compare the class mark of 5–15 with that of 10–20.
Answer
(c) 29. Every class mark rises by 5 (for example 10 becomes 15), so the mean rises by 5: $24 + 5 = 29$ .
Q8 · 1 mark · MCQBoard level
A distribution has classes 0–10, 10–20, 20–30, 30–40, 40–50, each with a non-zero frequency. Which of these cannot be its mean?
(a) 4.5
(b) 18
(c) 27.5
(d) 40
Hint
Find the smallest and largest class marks.
Answer
(a) 4.5. The mean must lie between the smallest class mark 5 and the largest class mark 45, and 4.5 is below 5.
Q9 · 1 mark · Assertion-reasonHarder board level
Classes: 0–10, 10–20, 20–30 with frequencies 4, $k$ , 6.
Assertion (A): If the mean of this distribution is 16, then $k = 10$ .
Reason (R): The mean of grouped data lies between the smallest and largest class marks.
(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
Solve for $k$ first, then ask whether R is what gives that value.
Answer
(b). $\sum f_{i} x_{i} = 20 + 15 k + 150 = 170 + 15 k$ and $\sum f_{i} = 10 + k$ ; $170 + 15 k = 160 + 16 k$ gives $k = 10$ , so A is true (check: $\frac{320}{20} = 16$ ). R is true, but $k$ comes from solving the mean equation, not from the range fact.
Q10 · 2 marks · Very short answerBoard level
The mean of the distribution below is 6.5. Find $p$ .
Classes: 0–4, 4–8, 8–12
Frequencies: 3, $p$ , 5
Hint
Class marks are 2, 6, 10.
Answer
$\sum f_{i} x_{i} = 56 + 6 p$ , $\sum f_{i} = 8 + p$ ; $56 + 6 p = 6.5 (8 + p) = 52 + 6.5 p \Rightarrow 0.5 p = 4 \Rightarrow p = 8$ . Check: $\frac{104}{16} = 6.5$ .
Q11 · 3 marks · Short answerHarder board level
The mean of 50 observations was found to be 32. Later it was found that the value 45 had been wrongly read as 54 and the value 20 had been wrongly read as 2. Find the correct mean. Explain why the mean goes up even though one value was read too high.
Hint
Work with the total, not the mean.
Answer
Wrong total $= 50 \times 32 = 1600$ . Correct total $= 1600 - 54 - 2 + 45 + 20 = 1609$ . Correct mean $= \frac{1609}{50} = 32.18$ . The total falls by 9 for the first error but rises by 18 for the second, a net rise of 9, so the mean rises by $\frac{9}{50} = 0.18$ .
Q12 · 3 marks · Short answerHarder board level
Suppose Section A of 30 students has a mean score of 40 marks. The marks of the 20 students of Section B are:
Marks: 0–20, 20–40, 40–60, 60–80
Number of students: 2, 5, 8, 5
Find the mean of Section B and the mean of all 50 students.
Hint
Combine totals, not means.
Answer
Section B: $\sum f_{i} x_{i} = 20 + 150 + 400 + 350 = 920$ , mean $= \frac{920}{20} = 46$ marks. Combined mean $= \frac{30 \times 40 + 920}{50} = \frac{2120}{50} = 42.4$ marks.
Q13 · 3 marks · Short answerHOTS / competency
The class marks of a distribution with equal class widths are $x - 20$ , $x - 10$ , $x$ , $x + 10$ , $x + 20$ with frequencies 2, 4, 6, 5, 3. The mean is 41.5. Find $x$ and write down all the classes.
Hint
$\sum f_{i} x_{i}$ simplifies to $20 x$ plus a number.
Answer
$\sum f_{i} = 20$ ; $\sum f_{i} x_{i} = 20 x + (- 40 - 40 + 0 + 50 + 60) = 20 x + 30$ . $20 x + 30 = 20 \times 41.5 = 830 \Rightarrow x = 40$ . Class marks 20, 30, 40, 50, 60 with width 10, so the classes are 15–25, 25–35, 35–45, 45–55, 55–65. Check: $40 + 120 + 240 + 250 + 180 = 830$ .
Q14 · 5 marks · Long answerHarder board level
The number of books read in a year by 30 students is given below. The mean is 13.5. Find $a$ and $b$ .
Books: 1–5, 6–10, 11–15, 16–20, 21–25
Students: 3, $a$ , 8, $b$ , 4
Hint
Class marks are 3, 8, 13, 18, 23; $a + b = 15$ .
Answer
$\sum f_{i} x_{i} = 9 + 8 a + 104 + 18 b + 92 = 205 + 8 a + 18 b = 30 \times 13.5 = 405$ , so $4 a + 9 b = 100$ . With $b = 15 - a$ : $135 - 5 a = 100$ , so $a = 7$ , $b = 8$ . Check: $9 + 56 + 104 + 144 + 92 = 405$ and $\frac{405}{30} = 13.5$ .
Q15 · 3 marks · Short answerHarder board level
The mean of the distribution below is 33.5. Find $f$ .
Classes: 0–20, 20–30, 30–50, 50–60
Frequencies: 5, 8, $f$ , 5
Hint
Widths are unequal, so find each class mark separately: 10, 25, 40, 55.
Answer
$\sum f_{i} x_{i} = 50 + 200 + 40 f + 275 = 525 + 40 f$ , $\sum f_{i} = 18 + f$ . $525 + 40 f = 33.5 (18 + f) = 603 + 33.5 f \Rightarrow 6.5 f = 78 \Rightarrow f = 12$ . Check: $\frac{1005}{30} = 33.5$ .
Q16 · 3 marks · Short answerHOTS / competency
Riya says that for classes 0–10, 10–20, 20–30 with frequencies 5, $k$ , 5, the mean is 18 for a suitable $k$ . Show that she is wrong and explain why without heavy calculation.
Hint
Look at how the frequencies sit around the middle class.
Answer
$\sum f_{i} x_{i} = 25 + 15 k + 125 = 150 + 15 k = 15 (10 + k)$ and $\sum f_{i} = 10 + k$ , so the mean is 15 for every $k$ . Setting it to 18 gives $150 + 15 k = 180 + 18 k \Rightarrow k = - 10$ , which is impossible. Reason: the frequencies are symmetric about the middle class mark 15, so the mean is always 15.

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.Class mark of 35–45?Show answer
40
2.Direct method formula?Show answer
$\overset{x}{ˉ} = \frac{\sum f _{i} x _{i}}{\sum f _{i}}$
3.What is assumed about values inside a class?Show answer
They are centred at the class mark.
4.Class mark of 15–19 (inclusive)?Show answer
17, same as for 14.5–19.5
5.If $\sum f_{i} x_{i} = 600$ and $\sum f_{i} = 25$ , mean?Show answer
24

Mean of Grouped Data: Direct Method: frequently asked questions

When should I use the direct method?

Use it when class marks and frequencies are small or when a frequency is missing and the mean is given. If the question names the assumed mean or step deviation method, use that instead. All three methods give the same mean.

Do I need continuous classes for the mean?

No. Class marks do not change when you convert inclusive classes to continuous ones, so you can find them directly. You do need continuous classes for the median and mode.

Is the grouped mean exact?

It is an estimate, because each observation is replaced by its class mark. The board accepts this value as the mean.

Are ogives in the syllabus?

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