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CBSE · Class 10 · Mathematics

Statistics 13.2: Mean of Grouped Data

5 concepts · 40 practice questions with answers · Free to read

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Class mark, class width and the mean formula

From daily life

A teacher has marks of 20 students in groups like 0–10, 10–20. She cannot see each mark, so she pretends every student in a group scored the middle value of the group.

Simple definition

In grouped data each class has a class mark x = (lower limit + upper limit) ÷ 2 and a class width h = upper − lower. The mean is Σfx ÷ Σf, where f is the frequency of the class.

Formula: mean = Σfx ÷ Σf

23654515253545mean 28class markx = (lower + upper)/2mean =Σfx ÷ Σf

In the figure: Classes of width 10 with frequencies 2, 3, 6, 5, 4; class marks 5 to 45; mean 28.

Rule 1: Class mark and class width — Add the two limits and halve; subtract them for the width.

  • Easy: Class 20–30: x = (20 + 30) ÷ 2 = 25.
  • Medium: Class 20–30: h = 30 − 20 = 10.
  • Exam-type: Class 15–25: x = 20, h = 10.

Rule 2: Why the mean is only an estimate — Within a class we do not know the exact values, so x stands for all of them.

  • Easy: Σf is the total number of observations, N.
  • Medium: Each product fx is “frequency times class mark”.
  • Exam-type: The result is between the first and the last class mark.
Remember

Check that the mean lies between the smallest and the largest class mark.

Common mistake

Using the class limits instead of the class mark in Σfx.

Memory clue

"Frequency times middle, divided by total."

Practice questions

Easy
  1. 1Find the class width (class size) of the class interval 25–31.
  2. 2Find the class mark of the class interval 34–36.
  3. 3Find the class width (class size) of the class interval 105–111.
Medium
  1. 4The class intervals of a distribution are 300–320, 320–340, 340–360, 360–380. Find the class mark of the first class.
  2. 5The class intervals of a distribution are 50–60, 60–70, 70–80, 80–90. Find the class mark of the second class.
  3. 6The class intervals of a distribution are 20–25, 25–30, 30–35, 35–40, 40–45. Find the class mark of the fourth class.
Challenging
  1. 7The class marks of three consecutive classes of equal width are 74, 82 and 90. Find the first class interval.
  2. 8For a grouped data, Σfᵢxᵢ = 2944 and Σfᵢ = 32. Find the mean.
Show answers (try them yourself first)
  1. 16
  2. 235
  3. 36
  4. 4310
  5. 565
  6. 637.5
  7. 770–78
  8. 892

Direct method

From daily life

A shopkeeper groups the bills of the day, multiplies each group’s middle bill by the number of bills, adds all, and divides by the number of bills.

Simple definition

In the direct method make columns x, f and fx. Add the f column to get Σf and the fx column to get Σfx, then divide: mean = Σfx ÷ Σf. It is best when the numbers are small.

Formula: mean = Σfx ÷ Σf

xffx521015345256150355175454180Σ20560mean = Σfx ÷ Σf= 560 ÷ 20= 28

In the figure: Σf = 20, Σfx = 560, so the mean is 28.

Live figure

Direct method

In the app you can move this figure with your finger.

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Rule 1: Build the table — Find x for every class, then fx.

  • Easy: Classes 0–10 … 40–50 give x = 5, 15, 25, 35, 45.
  • Medium: With f = 2, 3, 6, 5, 4: fx = 10, 45, 150, 175, 180.
  • Exam-type: Σf = 20 and Σfx = 560.

Rule 2: Divide and check — The mean must look reasonable for the data.

  • Easy: mean = 560 ÷ 20 = 28.
  • Medium: 28 lies in the class 20–30, where the biggest frequencies are.
  • Exam-type: Wrong total of f is the most common slip.
Remember

Add each column twice; a wrong Σf or Σfx spoils the whole answer.

Common mistake

Dividing Σfx by the number of classes instead of by Σf.

Memory clue

"Multiply, add, divide by total frequency."

Practice questions

Easy
  1. 1Marks of two sections (classes 0–10, 10–20, 20–30, 30–40): Section A frequencies 7, 4, 1, 8; Section B frequencies 8, 8, 3, 1. Which section has the higher mean?
  2. 2The mean of a grouped data is 274 and Σfᵢ = 40. Find Σfᵢxᵢ.
  3. 3Marks of two sections (classes 0–10, 10–20, 20–30, 30–40): Section A frequencies 5, 5, 7, 3; Section B frequencies 2, 7, 2, 9. Which section has the higher mean?
Medium
  1. 4The mean of a grouped data is 332 and Σfᵢ = 50. Find Σfᵢxᵢ.
  2. 5The mean of a grouped data is 97 and Σfᵢ = 50. Find Σfᵢxᵢ.
  3. 6Marks of two sections (classes 0–10, 10–20, 20–30, 30–40): Section A frequencies 2, 1, 2, 15; Section B frequencies 6, 7, 1, 6. Which section has the higher mean?
Challenging
  1. 7Marks of two sections (classes 0–10, 10–20, 20–30, 30–40): Section A frequencies 1, 4, 10, 5; Section B frequencies 3, 5, 4, 8. Which section has the higher mean?
  2. 8The mean of a grouped data is 47.3 and Σfᵢ = 100. Find Σfᵢxᵢ.
Show answers (try them yourself first)
  1. 1Section A
  2. 210960
  3. 3Section B
  4. 416600
  5. 54850
  6. 6Section A
  7. 7Section A
  8. 84730

Assumed mean method

From daily life

To add 28, 31, 29 quickly you start from 30 and add −2, +1, −1; small numbers are easier.

Simple definition

Pick one class mark in the middle as the assumed mean A. Find the deviation d = x − A for each class. Then mean = A + Σfd ÷ Σf. Choosing the class with the largest frequency makes the d column small.

Formula: mean = A + Σfd ÷ Σf

xd = x−25ffd5-202-4015-103-302506035105504520480Σ2060A = 25mean = A + Σfd/Σf= 25 + 60/20= 28

In the figure: A = 25: Σfd = 60, so the mean is 25 + 60/20 = 28.

Rule 1: Deviations and fd — The deviation of the assumed class is always 0.

  • Easy: A = 25: d = −20, −10, 0, 10, 20.
  • Medium: fd = −40, −30, 0, 50, 80; Σfd = 60.
  • Exam-type: Keep the minus signs; they cancel part of the sum.

Rule 2: Any A gives the same mean — The method is only a shortcut.

  • Easy: A = 15: d = −10, 0, 10, 20, 30; Σfd = −20 + 0 + 60 + 100 + 120 = 260.
  • Medium: mean = 15 + 260/20 = 28 again.
  • Exam-type: Pick A from the middle to keep numbers small.
Remember

Check: the deviation of the class you chose as A must be zero.

Common mistake

Forgetting to add A back at the end.

Memory clue

"Start from a guess, add the average slip."

Practice questions

Easy
  1. 1For a distribution with class marks 110, 130, 150, 170, 190, 210, the assumed mean is a = 170. Find the deviation d = x − a for the class mark 150.
  2. 2For a distribution with class marks 30, 50, 70, 90, the assumed mean is a = 70. Find the deviation d = x − a for the class mark 50.
  3. 3For a distribution with class marks 2.5, 7.5, 12.5, 17.5, the assumed mean is a = 12.5. Find the deviation d = x − a for the class mark 2.5.
Medium
  1. 4For a distribution with class marks 12.5, 17.5, 22.5, 27.5, 32.5, 37.5, the assumed mean is a = 32.5. Find the deviation d = x − a for the class mark 22.5.
  2. 5For a distribution with class marks 55, 65, 75, 85, 95, 105, the assumed mean is a = 85. Find the deviation d = x − a for the class mark 55.
  3. 6With assumed mean 55, Σfᵢdᵢ = −15 and Σfᵢ = 25. If the assumed mean is changed to 75, what is the new Σfᵢdᵢ?
Challenging
  1. 7By the assumed mean method, mean = 20.24, Σfᵢdᵢ = 62 and Σfᵢ = 50. Find the assumed mean a.
  2. 8By the assumed mean method, mean = 38.92, Σfᵢdᵢ = 73 and Σfᵢ = 25. Find the assumed mean a.
Show answers (try them yourself first)
  1. 1−20
  2. 2−20
  3. 3−10
  4. 4−10
  5. 5−30
  6. 6−515
  7. 719
  8. 836

Step-deviation method

From daily life

When all deviations are multiples of 10, dividing them by 10 gives tiny numbers like −2, −1, 0, 1, 2.

Simple definition

When all class widths are equal to h, use the step deviation u = (x − A) ÷ h. Then mean = A + h × Σfu ÷ Σf. It is the quickest method for large class marks.

Formula: mean = A + h × Σfu ÷ Σf

xu = (x−25)/10ffu5-22-415-13-3250603515545248Σ206A = 25, h = 10A + h × Σfu/Σf= 25 + 10 × 6/20= 28

In the figure: A = 25, h = 10: Σfu = 6, mean = 25 + 10 × 6/20 = 28.

Rule 1: Make the u column — Divide each deviation by h.

  • Easy: d = −20, −10, 0, 10, 20 ÷ 10 gives u = −2, −1, 0, 1, 2.
  • Medium: fu = −4, −3, 0, 5, 8; Σfu = 6.
  • Exam-type: u of the assumed class is 0.

Rule 2: Multiply back by h — The shortcut shrinks the numbers, so h must be multiplied back.

  • Easy: mean = 25 + 10 × 6/20 = 25 + 3 = 28.
  • Medium: All three methods give 28, so any may be used.
  • Exam-type: It needs equal class widths; otherwise use the assumed mean method.
Remember

If your answer differs from the direct method, check the sign of u and the factor h.

Common mistake

Forgetting to multiply Σfu ÷ Σf by h.

Memory clue

"Shrink by h, then stretch by h."

Practice questions

Easy
  1. 1For a distribution with class width 10 and assumed mean a = 45, find uᵢ = (xᵢ − a)/h for the class mark 55.
  2. 2For a distribution with class width 15 and assumed mean a = 82.5, find uᵢ = (xᵢ − a)/h for the class mark 67.5.
  3. 3For a distribution with class width 15 and assumed mean a = 72.5, find uᵢ = (xᵢ − a)/h for the class mark 87.5.
Medium
  1. 4For a distribution with class width 10 and assumed mean a = 65, find uᵢ = (xᵢ − a)/h for the class mark 55.
  2. 5By the step-deviation method, mean = 44, a = 40, Σfᵢuᵢ = 4 and Σfᵢ = 20. Find the class width h.
  3. 6In a step-deviation calculation, a = 20, h = 15, Σfᵢuᵢ = 40 and Σfᵢ = 20. Find the mean.
Challenging
  1. 7In a step-deviation calculation, a = 50, h = 25, Σfᵢuᵢ = 18 and Σfᵢ = 25. Find the mean.
  2. 8In a step-deviation calculation, a = 90, h = 25, Σfᵢuᵢ = 12 and Σfᵢ = 20. Find the mean.
Show answers (try them yourself first)
  1. 11
  2. 2−1
  3. 31
  4. 4−1
  5. 520
  6. 650
  7. 768
  8. 8105

Missing frequencies and applications

From daily life

If a register of 20 students is torn and one frequency is lost, the mean written on the cover still lets us find it.

Simple definition

When the mean is given and one frequency is missing, call it x, write Σfx and Σf in terms of x, and form the equation Σfx ÷ Σf = mean. Solve for x. The frequency must come out a whole number.

Formula: Σfx ÷ Σf = mean

Σfx = 515 + 15xN = 17 + x(515 + 15x) ÷ (17 + x) = 28x = 3

In the figure: Frequencies 2, x, 6, 5, 4 and mean 28 give x = 3.

Rule 1: Find the missing frequency — Keep the x terms and the numbers separate.

  • Easy: Classes 0–10 … 40–50, f = 2, x, 6, 5, 4: Σf = 17 + x.
  • Medium: Σfx = 10 + 15x + 150 + 175 + 180 = 515 + 15x.
  • Exam-type: 515 + 15x = 28(17 + x) = 476 + 28x, so 13x = 39 and x = 3.

Rule 2: Applications — Averages of wages, heights and runs are all grouped-mean problems.

  • Easy: Check the answer: put x = 3 back, Σf = 20, Σfx = 560, mean 28.
  • Medium: A frequency like 3.5 means a mistake: students are whole numbers.
  • Exam-type: Use the same steps for wages, heights and ages.
Remember

Put the answer back into the table and recompute the mean.

Common mistake

Putting x in the Σf but forgetting it in the Σfx.

Memory clue

"Same unknown in both the sums."

Practice questions

Easy
  1. 1For class intervals 30–35, 35–40, 40–45, 45–50, the frequencies are 11, 36, 2, p respectively, and the total frequency is 50. Find p.
  2. 2For class intervals 20–30, 30–40, 40–50, 50–60, 60–70, 70–80, the frequencies are p, 13, 2, 12, 4, 8 respectively, and the total frequency is 50. Find p.
  3. 3For class intervals 10–20, 20–30, 30–40, 40–50, 50–60, the frequencies are 1, 14, p, 1, 2 respectively, and the total frequency is 20. Find p.
Medium
  1. 4For class intervals 120–160, 160–200, 200–240, 240–280, 280–320, 320–360, the frequencies are 3, 14, 3, 3, 15, p respectively, and the total frequency is 40. Find p.
  2. 5A group of 30 students has mean marks 89 and another group of 10 students has mean marks 65. Find the mean of all 40 students.
  3. 6A group of 30 students has mean marks 41 and another group of 30 students has mean marks 75. Find the mean of all 60 students.
Challenging
  1. 7A group of 40 students has mean marks 85 and another group of 60 students has mean marks 65. Find the mean of all 100 students.
  2. 8The mean of some observations is 103. If every observation is increased by 7, what is the new mean?
Show answers (try them yourself first)
  1. 11
  2. 211
  3. 32
  4. 42
  5. 583
  6. 658
  7. 773
  8. 8110