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State Board · Class 10 · Mathematics Part 1

Lesson 6: Statistics

3 concepts · 24 practice questions with answers · Free to read

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Mean of grouped data

From daily life

Writing the marks of 100 students one by one is hard, so we write how many fall into groups like 0–10 and 10–20. From such a table we can still find the average of the class.

Simple definition

A group is a class, the number in it is its frequency (f), and the average of the two ends of a class is its class mark (x). The sum of all frequencies is N = Σf. Mean = Σfx ÷ Σf.

35845152535xfΣfx ÷ Σf

In the figure: Classes 0–10, 10–20, 20–30, 30–40; frequencies 3, 5, 8, 4; class marks 5, 15, 25, 35. Column height = frequency.

Rule 1: Direct method — Find x for each class, multiply f × x, add all fx and divide by Σf: mean = Σfx ÷ Σf.

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Mean of grouped data

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  • Easy: Classes 0–10, 10–20, 20–30; f = 2, 5, 3; x = 5, 15, 25 ⇒ Σf = 10, Σfx = 10 + 75 + 75 = 160; mean = 160 ÷ 10 = 16.
  • Medium: Classes 0–10 to 30–40; f = 3, 5, 8, 4 ⇒ Σfx = 15 + 75 + 200 + 140 = 430; Σf = 20; mean = 430 ÷ 20 = 21.5.
  • Exam-type: Classes 20–30 to 60–70; f = 4, 6, 10, 6, 4; x = 25, 35, 45, 55, 65 ⇒ Σfx = 100 + 210 + 450 + 330 + 260 = 1350; Σf = 30; mean = 45.

Rule 2: Assumed mean method — If x values are big, take one x as the assumed mean A and find d = x − A. Mean = A + Σfd ÷ Σf.

  • Easy: x = 10, 20, 30; f = 2, 6, 2; A = 20 ⇒ d = −10, 0, 10; Σfd = −20 + 0 + 20 = 0; mean = 20 + 0 ÷ 10 = 20.
  • Medium: x = 10, 30, 50, 70, 90; f = 3, 5, 8, 6, 3; A = 50 ⇒ d = −40, −20, 0, 20, 40; Σfd = −120 − 100 + 0 + 120 + 120 = 20; Σf = 25; mean = 50 + 20 ÷ 25 = 50.8.
  • Exam-type: x = 105, 115, 125, 135; f = 4, 6, 6, 4; A = 120 ⇒ d = −15, −5, 5, 15; Σfd = −60 − 30 + 30 + 60 = 0; mean = 120 + 0 ÷ 20 = 120.

Rule 3: Step deviation method — If all classes have the same width h, take u = (x − A) ÷ h. Mean = A + h × Σfu ÷ Σf.

  • Easy: x = 5, 15, 25; f = 3, 4, 3; A = 15; h = 10 ⇒ u = −1, 0, 1; Σfu = −3 + 0 + 3 = 0; mean = 15 + 10 × 0 ÷ 10 = 15.
  • Medium: x = 15, 25, 35, 45, 55; f = 4, 6, 10, 8, 2; A = 35; h = 10 ⇒ u = −2, −1, 0, 1, 2; Σfu = −8 − 6 + 0 + 8 + 4 = −2; Σf = 30; mean = 35 + 10 × (−2) ÷ 30 ≈ 34.33.
  • Exam-type: x = 50, 150, 250, 350; f = 5, 10, 15, 10; A = 250; h = 100 ⇒ u = −2, −1, 0, 1; Σfu = −10 − 10 + 0 + 10 = −10; Σf = 40; mean = 250 + 100 × (−10) ÷ 40 = 225.
Remember

All three methods give the same answer. Pick the easy one: direct if numbers are small, assumed mean or step deviation if they are big.

Common mistake

Do not use only the lower limit as x. Always use the class mark: (lower + upper) ÷ 2.

Shortcut

The mean always lies between the smallest and the largest class mark; if your answer is outside, there is a mistake. Memory clue: "f times x, add up, divide by N".

Spot the difference

PointDirect methodStep deviation
Use whenSmall numbersEqual widths, big numbers
Value usedxu = (x − A) ÷ h
SumΣfxΣfu
FormulaΣfx ÷ ΣfA + h × Σfu ÷ Σf

Practice questions

Easy
  1. 1What are the class marks of the classes 0–20 and 20–40?
  2. 2If f = 2, 3 and x = 10, 20, what is Σfx?
  3. 3If the frequencies are 4, 6, 10, what is N?
Medium
  1. 4x = 5, 15; f = 6, 4; find the mean.
  2. 5x = 20, 30, 40; f = 5, 10, 5; find the mean with A = 30.
  3. 6x = 12, 16, 20, 24; f = 3, 7, 6, 4; find the mean with A = 18.
Challenging
  1. 7x = 25, 35, 45, 55; f = 2, 5, 8, 5; find the mean by step deviation with A = 45, h = 10.
  2. 8Classes 0–10 to 40–50; f = 4, 6, 10, 6, 4: what is the mean? Give an easy way to check the answer.
Show answers (try them yourself first)
  1. 110 and 30.
  2. 220 + 60 = 80.
  3. 3N = 20.
  4. 4Σfx = 30 + 60 = 90; Σf = 10; mean = 9.
  5. 5d = −10, 0, 10; Σfd = 0; mean = 30.
  6. 6d = −6, −2, 2, 6; Σfd = −18 − 14 + 12 + 24 = 4; Σf = 20; mean = 18 + 0.2 = 18.2.
  7. 7u = −2, −1, 0, 1; Σfu = −4 − 5 + 0 + 5 = −4; Σf = 20; mean = 45 + 10 × (−4) ÷ 20 = 43.
  8. 8The frequencies are symmetric about the middle class, so the mean = the middle class mark = 25.

Median and mode of grouped data

From daily life

If the students of a class stand in a line by height, the middle student shows the median; the height shared by most students points to the mode.

Simple definition

The median is the middle value of the data. The mode is the value that occurs most often. For grouped data we find the median class (the class in which N ÷ 2 falls in the cumulative frequency) and the modal class (the class with the biggest f).

Formula: median = L + (N ÷ 2 − cf) ÷ f × h; mode = L + (f₁ − f₀) ÷ (2f₁ − f₀ − f₂) × h

3584381620N = 20N/2 = 10cf

In the figure: Classes 0–10 to 30–40; f = 3, 5, 8, 4; cumulative frequency cf = 3, 8, 16, 20. N ÷ 2 = 10 falls in cf = 16, so the median class is 20–30.

Rule 1: Median: cumulative frequency — Write the cf, find N ÷ 2, and the class where cf first becomes at least N ÷ 2 is the median class. L = lower limit of that class, cf = cumulative frequency before it.

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Median and mode of grouped data

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  • Easy: f = 3, 5, 8, 4 (classes 0–10 to 30–40) ⇒ N = 20, N ÷ 2 = 10; median class 20–30; L = 20, cf = 8, f = 8, h = 10 ⇒ median = 20 + (10 − 8) ÷ 8 × 10 = 22.5.
  • Medium: f = 5, 10, 15, 12, 8 (classes 100–120 to 180–200) ⇒ N = 50, N ÷ 2 = 25; cf = 5, 15, 30, 42, 50; class 140–160; median = 140 + (25 − 15) ÷ 15 × 20 ≈ 153.33.
  • Exam-type: f = 6, 10, 14, 12, 8 (classes 0–5 to 20–25) ⇒ N = 50, N ÷ 2 = 25; cf = 6, 16, 30, 42, 50; class 10–15; median = 10 + (25 − 16) ÷ 14 × 5 ≈ 13.21.

Rule 2: Mode: the tallest bar — The class with the biggest f is the modal class. f₁ = its f, f₀ = f of the class before it, f₂ = f of the class after it.

  • Easy: Classes 0–10 to 30–40; f = 3, 5, 9, 4 ⇒ modal class 20–30; f₁ = 9, f₀ = 5, f₂ = 4; mode = 20 + (9 − 5) ÷ (18 − 5 − 4) × 10 ≈ 24.44.
  • Medium: Classes 10–20 to 50–60; f = 2, 6, 12, 8, 2 ⇒ class 30–40; f₁ = 12, f₀ = 6, f₂ = 8; mode = 30 + (12 − 6) ÷ (24 − 6 − 8) × 10 = 36.
  • Exam-type: Classes 0–50 to 150–200; f = 8, 15, 24, 13 ⇒ class 100–150; f₁ = 24, f₀ = 15, f₂ = 13; mode = 100 + (24 − 15) ÷ (48 − 15 − 13) × 50 = 122.5.

Rule 3: The relation between the three — Mode = 3 × median − 2 × mean. If two values are known the third can be found.

  • Easy: Median 20, mean 21 ⇒ mode = 60 − 42 = 18.
  • Medium: Mean 45, median 42 ⇒ mode = 126 − 90 = 36.
  • Exam-type: Mode 12, mean 15 ⇒ 12 = 3 × median − 30, so median = 42 ÷ 3 = 14.
Remember

Write the cf in order by adding up; the last cf must equal N.

Common mistake

The cf in the formula is the cumulative frequency before the median class, not of that class.

Shortcut

In the mode formula f₁ − f₀ goes on top and 2f₁ − f₀ − f₂ at the bottom; then multiply by the class width h. Memory clue: median = middle (N ÷ 2), mode = most (the tallest bar).

Spot the difference

PointMedianMode
MeaningMiddle valueMost frequent
ClassWhere N ÷ 2 falls in cfBiggest f
NeedsL, cf, f, hL, f₀, f₁, f₂, h
FormulaL + (N ÷ 2 − cf) ÷ f × hL + (f₁ − f₀) ÷ (2f₁ − f₀ − f₂) × h

Practice questions

Easy
  1. 1If f = 4, 6, 10, what is the cf?
  2. 2If N = 40, what is N ÷ 2?
  3. 3f = 2, 7, 4; which class is the modal class (first, second or third)?
Medium
  1. 4If mean = 30 and median = 28, what is the mode?
  2. 5Classes 0–10 to 30–40; f = 2, 6, 8, 4: find the median.
  3. 6Classes 10–20 to 40–50; f = 3, 9, 15, 3: find the mode.
Challenging
  1. 7Classes 0–20 to 80–100; f = 6, 10, 18, 10, 6: find the median.
  2. 8If the mean is 40 and the mode is 34, what is the median?
Show answers (try them yourself first)
  1. 14, 10, 20.
  2. 220.
  3. 3The second, because f = 7 is the biggest.
  4. 43 × 28 − 2 × 30 = 84 − 60 = 24.
  5. 5N = 20, N ÷ 2 = 10; cf = 2, 8, 16, 20; class 20–30; median = 20 + (10 − 8) ÷ 8 × 10 = 22.5.
  6. 6Class 30–40; mode = 30 + (15 − 9) ÷ (30 − 9 − 3) × 10 = 30 + 60 ÷ 18 ≈ 33.33.
  7. 7N = 50, N ÷ 2 = 25; cf = 6, 16, 34, 44, 50; class 40–60; median = 40 + (25 − 16) ÷ 18 × 20 = 50.
  8. 834 = 3 × median − 80 ⇒ median = 114 ÷ 3 = 38.

Graphs: histogram, polygon and pie diagram

From daily life

In a cricket match, if the runs in each over are shown by the height of bars, we see at once which over went best.

Simple definition

A histogram: classes on the horizontal axis, frequency on the vertical axis; bars touch each other. A frequency polygon: a graph joining the midpoints of the tops of the bars. A pie diagram: the whole circle is 360° and the angle of each part = (part ÷ total) × 360°.

90°120°60°90°360°

In the figure: The 360° is split into four parts: 90°, 120°, 60°, 90°; sum 90 + 120 + 60 + 90 = 360.

Rule 1: Draw a histogram — Draw a bar of the class width for every class with height f. There is no gap between bars. The tallest bar shows the modal class.

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Graphs: histogram, polygon and pie diagram

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  • Easy: Classes 0–10, 10–20, 20–30; f = 4, 7, 5 ⇒ bar heights 4, 7, 5; the tallest bar is 10–20.
  • Medium: f = 3, 5, 9, 6, 2 (classes 20–30 to 60–70) ⇒ 5 bars; modal class 40–50; the heights add up to N = 3 + 5 + 9 + 6 + 2 = 25.
  • Exam-type: Classes 0–5, 5–10, 10–15, 15–20; f = 2, 8, 6, 4 ⇒ 4 bars, each of width 5, heights 2, 8, 6, 4; N = 20.

Rule 2: Frequency polygon — Take the midpoint of the top of each bar. Add one class with f = 0 before and after, and join the points by line segments.

  • Easy: Classes 10–20, 20–30, 30–40; f = 4, 7, 5 ⇒ points (15, 4), (25, 7), (35, 5); plus (5, 0) before and (45, 0) after.
  • Medium: Classes 0–20, 20–40, 40–60; f = 6, 9, 3 ⇒ class marks 10, 30, 50; points (10, 6), (30, 9), (50, 3); end points (−10, 0) and (70, 0).
  • Exam-type: With 4 classes the polygon has 4 + 2 = 6 points.

Rule 3: Angles of a pie diagram — Angle = (part ÷ total) × 360°. All the angles always add up to 360°.

  • Easy: Out of a total of 40, the part 10 ⇒ angle = 10 ÷ 40 × 360° = 90°.
  • Medium: Out of a total of 60, the part 20 ⇒ angle = 20 ÷ 60 × 360° = 120°.
  • Exam-type: Income 20000 rupees, 6000 spent on food ⇒ angle = 6000 ÷ 20000 × 360° = 108°.
Remember

In a histogram the classes should be continuous (for example 10–20, 20–30) so that the bars touch.

Common mistake

Do not forget to close the polygon on the x-axis at the class marks before and after.

Shortcut

After finding the angles, check at once that they add up to 360°. Memory clue: first find the angle of 1 part = 360° ÷ total, then multiply by the part.

Spot the difference

PointHistogramPie diagram
ShowsFrequency (height)Share of parts (angle)
Axesx: class, y: fNo axes, a circle
TotalN = Σf360°
FormulaHeight = f(part ÷ total) × 360°

Practice questions

Easy
  1. 1What does the height of a bar in a histogram show?
  2. 2Out of a total of 36 the part is 9; what is the angle in the pie diagram?
  3. 3Classes 0–10 to 30–40; f = 3, 5, 8, 4: which bar is the tallest?
Medium
  1. 4Out of a total of 50 the part is 15; find the angle.
  2. 5Classes 20–30 to 40–50; f = 5, 8, 3: write all the points of the polygon.
  3. 6In a school 45 of 120 children play cricket. What is the cricket angle in the pie diagram?
Challenging
  1. 7Monthly spending is 12000 rupees, of which 4500 is rent. What is the rent angle?
  2. 8Three angles 100°, 80° and 70° are known. What is the fourth angle?
Show answers (try them yourself first)
  1. 1The frequency of that class.
  2. 29 ÷ 36 × 360° = 90°.
  3. 320–30 (f = 8).
  4. 415 ÷ 50 × 360° = 108°.
  5. 5(15, 0), (25, 5), (35, 8), (45, 3), (55, 0).
  6. 645 ÷ 120 × 360° = 135°.
  7. 74500 ÷ 12000 × 360° = 135°.
  8. 8360° − (100° + 80° + 70°) = 110°.