Class 10Maths · StatisticsFull chapter

Statistics

The whole chapter in one place — read it, then test yourself. Clear notes, formula sheet, a practice quiz, and worked NCERT solutions & PYQs.

In this chapter
Mean of Grouped Data — Direct MethodMean by the Assumed-Mean MethodMean by the Step-Deviation MethodMode of Grouped DataMedian of Grouped Data

Mean of Grouped Data — Direct Method

Quick answer For grouped data we replace each class by its class mark xᵢ and compute the mean as x̄ = Σfᵢxᵢ ÷ Σfᵢ.

When data is arranged in class intervals, we do not know the exact value of each observation. We assume every observation in a class equals the class mark (midpoint) of that class, where class mark = (lower limit + upper limit) ÷ 2.

The direct method multiplies each class mark xᵢ by its frequency fᵢ, adds these products, and divides by the total frequency:

x̄ = Σfᵢxᵢ ÷ Σfᵢ

Worked example. The marks of 30 students are grouped as: 10–25 (2), 25–40 (3), 40–55 (7), 55–70 (6), 70–85 (6), 85–100 (6).

  1. Class marks xᵢ: 17.5, 32.5, 47.5, 62.5, 77.5, 92.5.
  2. Products fᵢxᵢ: 35, 97.5, 332.5, 375, 465, 555.
  3. Σfᵢ = 30 and Σfᵢxᵢ = 35 + 97.5 + 332.5 + 375 + 465 + 555 = 1860.
  4. x̄ = 1860 ÷ 30 = 62.

So the mean marks = 62. The direct method is best when the class marks and frequencies are small numbers.

Mean (Direct Method) x̄ = Σfᵢxᵢ ÷ Σfᵢ fᵢ is the frequency of the class whose class mark is xᵢ
Class mark xᵢ = (lower limit + upper limit) ÷ 2 the midpoint that represents the whole class
Remember
  • Each observation in a class is taken as the class mark xᵢ = (lower + upper) ÷ 2.
  • Direct method: x̄ = Σfᵢxᵢ ÷ Σfᵢ.
  • Σfᵢ = n, the total number of observations.
  • Best used when class marks and frequencies are small, easy-to-multiply numbers.
  • The mean uses every value, so it is affected by extreme classes.

Mean by the Assumed-Mean Method

Quick answer Choosing an assumed mean a and working with deviations dᵢ = xᵢ − a makes the arithmetic lighter: x̄ = a + Σfᵢdᵢ ÷ Σfᵢ.

When class marks are large, multiplying fᵢxᵢ is tedious. We pick an assumed mean a (usually a class mark near the middle) and measure each class mark from it as a deviation dᵢ = xᵢ − a. The true mean is then:

x̄ = a + (Σfᵢdᵢ ÷ Σfᵢ)

The extra term Σfᵢdᵢ ÷ Σfᵢ is the average deviation, which corrects the assumed value back to the real mean. The answer does not depend on which a you pick.

Worked example. Using the same data: 10–25 (2), 25–40 (3), 40–55 (7), 55–70 (6), 70–85 (6), 85–100 (6). Take a = 47.5.

  1. Deviations dᵢ = xᵢ − 47.5: −30, −15, 0, 15, 30, 45.
  2. Products fᵢdᵢ: −60, −45, 0, 90, 180, 270.
  3. Σfᵢ = 30 and Σfᵢdᵢ = −60 − 45 + 0 + 90 + 180 + 270 = 435.
  4. x̄ = 47.5 + (435 ÷ 30) = 47.5 + 14.5 = 62.

The mean is again 62, confirming the direct method. Choosing a as a middle class mark keeps the deviations small and often symmetric about zero.

Mean (Assumed-Mean Method) x̄ = a + (Σfᵢdᵢ ÷ Σfᵢ) a = assumed mean; dᵢ = xᵢ − a
Deviation dᵢ = xᵢ − a measured from the assumed mean a
Remember
  • Assumed mean a is any convenient value, usually a central class mark.
  • Deviation dᵢ = xᵢ − a; deviations can be negative, zero or positive.
  • x̄ = a + Σfᵢdᵢ ÷ Σfᵢ.
  • The final mean is independent of the choice of a.
  • Reduces the size of the numbers you multiply compared with the direct method.

Mean by the Step-Deviation Method

Quick answer When all classes have equal width h, scaling deviations to uᵢ = (xᵢ − a) ÷ h gives the quickest formula: x̄ = a + h(Σfᵢuᵢ ÷ Σfᵢ).

If every class has the same width h, the deviations dᵢ are all multiples of h. Dividing them by h gives small whole numbers uᵢ, the step deviations:

uᵢ = (xᵢ − a) ÷ h

The mean is then recovered by multiplying back by h:

x̄ = a + h × (Σfᵢuᵢ ÷ Σfᵢ)

Worked example. Same data: 10–25 (2), 25–40 (3), 40–55 (7), 55–70 (6), 70–85 (6), 85–100 (6). Here h = 15 and take a = 47.5.

  1. Step deviations uᵢ = (xᵢ − 47.5) ÷ 15: −2, −1, 0, 1, 2, 3.
  2. Products fᵢuᵢ: −4, −3, 0, 6, 12, 18.
  3. Σfᵢ = 30 and Σfᵢuᵢ = −4 − 3 + 0 + 6 + 12 + 18 = 29.
  4. x̄ = 47.5 + 15 × (29 ÷ 30) = 47.5 + 14.5 = 62.

All three methods give x̄ = 62. The step-deviation method involves the smallest arithmetic and is preferred whenever the class widths are equal.

Mean (Step-Deviation Method) x̄ = a + h × (Σfᵢuᵢ ÷ Σfᵢ) h = common class width; a = assumed mean
Step deviation uᵢ = (xᵢ − a) ÷ h scales deviations to small whole numbers
Remember
  • Applicable only when all class widths h are equal.
  • Step deviation uᵢ = (xᵢ − a) ÷ h are small integers.
  • x̄ = a + h(Σfᵢuᵢ ÷ Σfᵢ).
  • Direct, assumed-mean and step-deviation methods always give the same mean.
  • Fastest method for large class marks with equal class size.

Mode of Grouped Data

Quick answer The mode lies in the class of highest frequency (the modal class) and is found by Mode = l + [(f₁ − f₀) ÷ (2f₁ − f₀ − f₂)] × h.

The mode is the value that occurs most often. For grouped data we first find the modal class — the class with the highest frequency. The mode is then estimated inside that class using:

Mode = l + [(f₁ − f₀) ÷ (2f₁ − f₀ − f₂)] × h

where l = lower limit of the modal class, f₁ = frequency of the modal class, f₀ = frequency of the class before it, f₂ = frequency of the class after it, and h = class width.

Worked example. Family sizes of 20 families: 1–3 (7), 3–5 (8), 5–7 (2), 7–9 (2), 9–11 (1).

  1. Highest frequency is 8, so the modal class is 3–5. Thus l = 3, f₁ = 8, f₀ = 7, f₂ = 2, h = 2.
  2. Mode = 3 + [(8 − 7) ÷ (2×8 − 7 − 2)] × 2.
  3. = 3 + [1 ÷ (16 − 9)] × 2 = 3 + (1 ÷ 7) × 2 = 3 + 0.286.
  4. Mode ≈ 3.29.

So the most typical family size is about 3.29 members. The formula weighs how much the modal frequency stands out above its neighbours.

Mode of grouped data Mode = l + [(f₁ − f₀) ÷ (2f₁ − f₀ − f₂)] × h l, f₁, f₀, f₂, h refer to the modal class and its neighbours
Remember
  • Modal class = the class with the greatest frequency.
  • Mode = l + [(f₁ − f₀) ÷ (2f₁ − f₀ − f₂)] × h.
  • f₀ and f₂ are the frequencies just before and just after the modal class.
  • The mode represents the most frequently occurring value.
  • If two classes tie for highest frequency, the data is bimodal.

Median of Grouped Data

Quick answer The median is the middle value; using cumulative frequencies, Median = l + [(n/2 − cf) ÷ f] × h, and it links to mean and mode by Mode = 3 Median − 2 Mean.

The median divides the data into two equal halves. For grouped data we build the cumulative frequency (running total) and locate the median class — the first class whose cumulative frequency is greater than or equal to n/2. Then:

Median = l + [(n/2 − cf) ÷ f] × h

where l = lower limit of the median class, n = Σfᵢ, cf = cumulative frequency of the class before the median class, f = frequency of the median class, and h = class width.

Worked example. Marks: 0–10 (5), 10–20 (8), 20–30 (20), 30–40 (15), 40–50 (7).

  1. Cumulative frequencies: 5, 13, 33, 48, 55. So n = 55 and n/2 = 27.5.
  2. First cumulative frequency ≥ 27.5 is 33, so the median class is 20–30. Thus l = 20, cf = 13, f = 20, h = 10.
  3. Median = 20 + [(27.5 − 13) ÷ 20] × 10.
  4. = 20 + (14.5 ÷ 20) × 10 = 20 + 7.25 = 27.25.

The median marks = 27.25. When two of the three measures are known, the third follows from the empirical relationship Mode = 3 Median − 2 Mean.

Median of grouped data Median = l + [(n/2 − cf) ÷ f] × h cf = cumulative frequency before the median class; n = Σfᵢ
Empirical relationship Mode = 3 Median − 2 Mean connects the three measures of central tendency
Remember
  • Build a cumulative frequency (less-than) column first.
  • Median class = first class with cumulative frequency ≥ n/2.
  • Median = l + [(n/2 − cf) ÷ f] × h.
  • cf is the cumulative frequency of the class just before the median class.
  • Empirical relation: Mode = 3 Median − 2 Mean, useful to find any missing measure.

The formula sheet

Every formula in this chapter, in one place — screenshot it before your exam.

x̄ = Σfᵢxᵢ ÷ Σfᵢ
Mean (Direct Method)
xᵢ = (lower limit + upper limit) ÷ 2
Class mark
x̄ = a + (Σfᵢdᵢ ÷ Σfᵢ)
Mean (Assumed-Mean Method)
dᵢ = xᵢ − a
Deviation
x̄ = a + h × (Σfᵢuᵢ ÷ Σfᵢ)
Mean (Step-Deviation Method)
uᵢ = (xᵢ − a) ÷ h
Step deviation
Mode = l + [(f₁ − f₀) ÷ (2f₁ − f₀ − f₂)] × h
Mode of grouped data
Median = l + [(n/2 − cf) ÷ f] × h
Median of grouped data
Mode = 3 Median − 2 Mean
Empirical relationship

Test yourself

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0 correct · 0/12 answered
Q1 Mean easy

The mean of a grouped frequency distribution by the direct method is given by:

Q2 Mean easy

The class mark (midpoint) of the class interval 25–35 is:

Q3 Mean easy

In the step-deviation method, the step deviation uᵢ is defined as:

Q4 Mean easy

In the assumed-mean formula x̄ = a + Σfᵢdᵢ ÷ Σfᵢ, the deviation dᵢ equals:

Q5 Mode easy

The modal class of a grouped frequency distribution is the class having the:

Q6 Mode medium

For a modal class 10–20 with l = 10, f₁ = 12, f₀ = 8, f₂ = 6, h = 10, the mode is:

Q7 Median easy

The median of a grouped frequency distribution is calculated using:

Q8 Relationship medium

The empirical relationship among mean, median and mode is:

Q9 Median medium

To locate the median class of a grouped distribution, we find the class whose cumulative frequency is:

Q10 Relationship medium

If the mean and median of a distribution are 22 and 24 respectively, the mode is:

Q11 Mean easy

In the step-deviation formula x̄ = a + h(Σfᵢuᵢ ÷ Σfᵢ), the symbol h denotes the:

Q12 Median medium

For classes 0–10, 10–20, 20–30, 30–40, 40–50 with cumulative frequencies 8, 20, 34, 50, 60, the median class is:

NCERT solutions & previous-year questions

Step-by-step model answers — tap a question to reveal the full solution.

NCERT questions 6

1 Find the mean of the following distribution by the direct method: Marks 10–25, 25–40, 40–55, 55–70, 70–85, 85–100 with frequencies 2, 3, 7, 6, 6, 6.Mean (Direct Method)

Step 1. Find the class marks xᵢ = (lower + upper) ÷ 2:

  • 17.5, 32.5, 47.5, 62.5, 77.5, 92.5.

Step 2. Compute fᵢxᵢ:

  • 2×17.5 = 35, 3×32.5 = 97.5, 7×47.5 = 332.5, 6×62.5 = 375, 6×77.5 = 465, 6×92.5 = 555.

Step 3. Σfᵢ = 2+3+7+6+6+6 = 30 and Σfᵢxᵢ = 35+97.5+332.5+375+465+555 = 1860.

Step 4. x̄ = Σfᵢxᵢ ÷ Σfᵢ = 1860 ÷ 30 = 62.

2 The daily pocket allowance of a group of children is given below and the mean allowance is Rs 18. Find the missing frequency f. Classes 11–13, 13–15, 15–17, 17–19, 19–21, 21–23, 23–25 with frequencies 7, 6, 9, 13, f, 5, 4.Mean (Missing Frequency)

Step 1. Class marks xᵢ: 12, 14, 16, 18, 20, 22, 24.

Step 2. Σfᵢ = 7+6+9+13+f+5+4 = 44 + f.

Step 3. Σfᵢxᵢ = 84 + 84 + 144 + 234 + 20f + 110 + 96 = 752 + 20f.

Step 4. Mean x̄ = Σfᵢxᵢ ÷ Σfᵢ, so 18 = (752 + 20f) ÷ (44 + f).

Step 5. 18(44 + f) = 752 + 20f → 792 + 18f = 752 + 20f → 40 = 2f → f = 20.

3 Find the mean of the distribution 10–25, 25–40, 40–55, 55–70, 70–85, 85–100 with frequencies 2, 3, 7, 6, 6, 6 using the step-deviation method (take a = 47.5, h = 15).Mean (Step-Deviation)

Step 1. Class marks xᵢ: 17.5, 32.5, 47.5, 62.5, 77.5, 92.5.

Step 2. Step deviations uᵢ = (xᵢ − 47.5) ÷ 15: −2, −1, 0, 1, 2, 3.

Step 3. Products fᵢuᵢ: −4, −3, 0, 6, 12, 18.

Step 4. Σfᵢ = 30 and Σfᵢuᵢ = −4 − 3 + 0 + 6 + 12 + 18 = 29.

Step 5. x̄ = a + h(Σfᵢuᵢ ÷ Σfᵢ) = 47.5 + 15 × (29 ÷ 30) = 47.5 + 14.5 = 62.

4 A survey of 20 families gave the family sizes 1–3, 3–5, 5–7, 7–9, 9–11 with frequencies 7, 8, 2, 2, 1. Find the mode.Mode

Step 1. The highest frequency is 8, so the modal class is 3–5.

Step 2. Here l = 3, f₁ = 8, f₀ = 7, f₂ = 2, h = 2.

Step 3. Mode = l + [(f₁ − f₀) ÷ (2f₁ − f₀ − f₂)] × h = 3 + [(8 − 7) ÷ (16 − 7 − 2)] × 2.

Step 4. = 3 + (1 ÷ 7) × 2 = 3 + 0.286 = 3.29 (approximately 3.286).

The modal family size is about 3.29 members.

5 Find the median of the distribution: Marks 0–10, 10–20, 20–30, 30–40, 40–50 with frequencies 5, 8, 20, 15, 7.Median

Step 1. Build cumulative frequencies: 5, 13, 33, 48, 55. So n = 55 and n/2 = 27.5.

Step 2. The first cumulative frequency ≥ 27.5 is 33, so the median class is 20–30.

Step 3. Here l = 20, cf = 13, f = 20, h = 10.

Step 4. Median = l + [(n/2 − cf) ÷ f] × h = 20 + [(27.5 − 13) ÷ 20] × 10.

Step 5. = 20 + (14.5 ÷ 20) × 10 = 20 + 7.25 = 27.25.

6 If the median and mean of a distribution are 25.7 and 22.1 respectively, find the mode using the empirical relationship.Relationship

Step 1. The empirical relationship is Mode = 3 Median − 2 Mean.

Step 2. Substitute Median = 25.7 and Mean = 22.1:

Mode = 3(25.7) − 2(22.1) = 77.1 − 44.2 = 32.9.

So the mode of the distribution is 32.9.

Previous-year board questions 4

Q1 Find the mean of the following frequency distribution by the step-deviation method: Class 0–20, 20–40, 40–60, 60–80, 80–100 with frequencies 6, 8, 10, 9, 7. CBSE 2023 3 marks

Step 1. Class marks xᵢ: 10, 30, 50, 70, 90. Take a = 50 and h = 20.

Step 2. Step deviations uᵢ = (xᵢ − 50) ÷ 20: −2, −1, 0, 1, 2.

Step 3. Products fᵢuᵢ: −12, −8, 0, 9, 14.

Step 4. Σfᵢ = 6+8+10+9+7 = 40 and Σfᵢuᵢ = −12 − 8 + 0 + 9 + 14 = 3.

Step 5. x̄ = a + h(Σfᵢuᵢ ÷ Σfᵢ) = 50 + 20 × (3 ÷ 40) = 50 + 1.5 = 51.5.

Q2 Find the mode of the following distribution: Class 0–20, 20–40, 40–60, 60–80, 80–100 with frequencies 15, 20, 35, 25, 5. CBSE 2020 3 marks

Step 1. The highest frequency is 35, so the modal class is 40–60.

Step 2. Here l = 40, f₁ = 35, f₀ = 20, f₂ = 25, h = 20.

Step 3. Mode = l + [(f₁ − f₀) ÷ (2f₁ − f₀ − f₂)] × h = 40 + [(35 − 20) ÷ (70 − 20 − 25)] × 20.

Step 4. = 40 + (15 ÷ 25) × 20 = 40 + 12 = 52.

Q3 Find the median of the following data: Class 0–10, 10–20, 20–30, 30–40, 40–50 with frequencies 8, 12, 10, 11, 9. CBSE 2019 3 marks

Step 1. Cumulative frequencies: 8, 20, 30, 41, 50. So n = 50 and n/2 = 25.

Step 2. The first cumulative frequency ≥ 25 is 30, so the median class is 20–30.

Step 3. Here l = 20, cf = 20, f = 10, h = 10.

Step 4. Median = l + [(n/2 − cf) ÷ f] × h = 20 + [(25 − 20) ÷ 10] × 10.

Step 5. = 20 + (5 ÷ 10) × 10 = 20 + 5 = 25.

Q4 The mean of the following frequency distribution is 50 and the total frequency is 120. Find the missing frequencies f₁ and f₂: Class 0–20, 20–40, 40–60, 60–80, 80–100 with frequencies 17, f₁, 32, f₂, 19. CBSE 2022 5 marks

Step 1. Class marks xᵢ: 10, 30, 50, 70, 90.

Step 2. Total frequency: 17 + f₁ + 32 + f₂ + 19 = 120 → f₁ + f₂ = 52. …(i)

Step 3. Σfᵢxᵢ = 170 + 30f₁ + 1600 + 70f₂ + 1710 = 3480 + 30f₁ + 70f₂.

Step 4. Mean = 50, so (3480 + 30f₁ + 70f₂) ÷ 120 = 50 → 30f₁ + 70f₂ = 2520 → 3f₁ + 7f₂ = 252. …(ii)

Step 5. From (i), f₁ = 52 − f₂. Substitute in (ii): 3(52 − f₂) + 7f₂ = 252 → 156 + 4f₂ = 252 → 4f₂ = 96 → f₂ = 24.

Step 6. Then f₁ = 52 − 24 = 28. Hence f₁ = 28 and f₂ = 24.

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