Statistics ISC Class 11: Measures of Dispersion

What Statistics is about

Weightage: Statistics belongs to the unit Statistics & Probability, which CISCE weights at 8 of the 80 marks in the ISC Class XI Mathematics Paper I (Theory). Class XI is examined by the school, so your school sets the paper within this weightage.Source: CISCE ISC Mathematics syllabus, examination year 2028, Class XI unit weightage

Key concepts

Dispersion and range

Dispersion is the extent to which observations are spread out about a central value. The simplest measure is the range = maximum − minimum. It is easy to compute but uses only the two extreme values, so a single unusual value changes it a lot, and it says nothing about how the rest of the data are spread. For grouped data, the range is usually taken as upper limit of the last class minus lower limit of the first.

Worked example

Find the range of the marks 23, 45, 31, 50, 12.

Try it first, then show the solutionSolution

1. Largest 50, smallest 12
2. Range = 50 − 12 = 38

Answer: 38

Mean deviation about the mean: raw data

The mean deviation about the mean is MD = Σ|xᵢ − x̄|/n. Absolute values are used because the ordinary deviations xᵢ − x̄ always add to zero. Steps: find x̄; find each |xᵢ − x̄|; add; divide by n. MD has the same units as the data.

Worked example

Find the mean deviation about the mean of 3, 5, 8, 10, 14.

Try it first, then show the solutionSolution

1. x̄ = 40/5 = 8
2. |xᵢ − 8|: 5, 3, 0, 2, 6; total 16
3. MD = 16/5

Answer: 3.2

Mean deviation about the mean: frequency and grouped data

For a frequency distribution, MD = Σfᵢ|xᵢ − x̄|/N, with N = Σfᵢ and x̄ = Σfᵢxᵢ/N. For grouped data, take xᵢ as the class mark (mid-value) of each class, assuming the observations in a class are centred there. A working table with columns x, f, fx, |x − x̄| and f|x − x̄| keeps the calculation organised.

Worked example

Find the mean deviation about the mean for classes 0–10, 10–20, 20–30, 30–40, 40–50 with frequencies 5, 8, 15, 16, 6.

Try it first, then show the solutionSolution

1. Class marks 5, 15, 25, 35, 45; N = 50; Σfx = 25 + 120 + 375 + 560 + 270 = 1350; x̄ = 27
2. |x − 27|: 22, 12, 2, 8, 18; f|x − 27|: 110, 96, 30, 128, 108; total 472
3. MD = 472/50

Answer: 9.44

Variance and standard deviation: direct method

Variance σ² = Σ(xᵢ − x̄)²/n (or Σfᵢ(xᵢ − x̄)²/N for frequency data) and standard deviation σ = √variance. Squaring removes signs and gives more weight to large deviations. An equivalent form, often quicker, is σ² = Σx²/n − x̄² (mean of squares minus square of mean). Adding a constant to every observation leaves σ unchanged; multiplying every observation by k multiplies σ by |k|.

Worked example

Find the variance and SD of 3, 5, 8, 10, 14.

Try it first, then show the solutionSolution

1. x̄ = 8; deviations −5, −3, 0, 2, 6; squares 25, 9, 0, 4, 36; total 74
2. σ² = 74/5 = 14.8
3. σ = √14.8 ≈ 3.85

Answer: Variance 14.8; SD ≈ 3.85

Short-cut (assumed mean) method

Choose an assumed mean A near the middle and write d = x − A. Then x̄ = A + Σfd/N and σ² = Σfd²/N − (Σfd/N)². Because variance does not depend on where you measure from, this gives exactly the same σ as the direct method, with smaller numbers. It is especially useful when the true mean is not a whole number.

Worked example

Find the mean and SD of x: 10, 20, 30, 40, 50 with f: 3, 7, 12, 5, 3.

Try it first, then show the solutionSolution

1. A = 30; d: −20, −10, 0, 10, 20; N = 30
2. Σfd = −60 − 70 + 0 + 50 + 60 = −20; Σfd² = 1200 + 700 + 0 + 500 + 1200 = 3600
3. x̄ = 30 − 20/30 ≈ 29.33; σ² = 3600/30 − (2/3)² = 120 − 0.444 = 119.56; σ ≈ 10.93

Answer: Mean ≈ 29.33; SD ≈ 10.93

Step-deviation method for grouped data

When class widths are equal to h, put u = (x − A)/h, with x the class mark. Then x̄ = A + h·Σfu/N and σ² = h²[Σfu²/N − (Σfu/N)²]. The factor h² appears because variance is in squared units. This keeps the u-values small integers (−2, −1, 0, 1, 2, …).

Worked example

Using the step-deviation method, find the mean and SD for classes 0–10, …, 40–50 with frequencies 5, 8, 15, 16, 6.

Try it first, then show the solutionSolution

1. A = 25, h = 10; u: −2, −1, 0, 1, 2; N = 50
2. Σfu = −10 − 8 + 0 + 16 + 12 = 10; Σfu² = 20 + 8 + 0 + 16 + 24 = 68
3. x̄ = 25 + 10 × 10/50 = 27; σ² = 100(68/50 − (10/50)²) = 100(1.36 − 0.04) = 132
4. σ = √132 ≈ 11.49

Answer: Mean 27; SD ≈ 11.49

Combined mean and standard deviation

For two groups (n₁, x̄₁, σ₁) and (n₂, x̄₂, σ₂): combined mean x̄ = (n₁x̄₁ + n₂x̄₂)/(n₁ + n₂). With d₁ = x̄₁ − x̄ and d₂ = x̄₂ − x̄, the combined variance is σ² = [n₁(σ₁² + d₁²) + n₂(σ₂² + d₂²)]/(n₁ + n₂). The d² terms account for the gap between the group means: even if both groups have small spreads, the combined spread is large if their means are far apart. Equivalently, you can rebuild Σx and Σx² for each group and combine.

Worked example

Group 1: 40 values, mean 50, SD 6. Group 2: 60 values, mean 55, SD 8. Find the combined mean and SD.

Try it first, then show the solutionSolution

1. x̄ = (2000 + 3300)/100 = 53
2. d₁ = −3, d₂ = 2
3. σ² = [40(36 + 9) + 60(64 + 4)]/100 = (1800 + 4080)/100 = 58.8
4. σ = √58.8 ≈ 7.67

Answer: Mean 53; SD ≈ 7.67

Maths in real life: Statistics

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.

Topics in this chapter

• Range and Mean Deviation about the Mean and Median
• Variance and Standard Deviation

Statistics formula sheet

Range

$\text{Range}=x_{\max }-x_{\min }$

Uses only the extremes.

Mean deviation about the mean

$\text{MD}=\frac{\sum f_i|x_i-\bar{x}|}{N}$

f = 1 for raw data; class marks for grouped data.

Variance (direct)

${\sigma }^2=\frac{\sum f_i(x_i-\bar{x}{)}^2}{N}=\frac{\sum f_i{x}_i^2}{N}-{\bar{x}}^2$

SD σ = √variance.

Short-cut method

${\sigma }^2=\frac{\sum f{d}^2}{N}-{\left(\frac{\sum fd}{N}\right)}^2,d=x-A$

x̄ = A + Σfd/N.

Step-deviation method

${\sigma }^2=h^2\left[\frac{\sum f{u}^2}{N}-{\left(\frac{\sum fu}{N}\right)}^2\right],u=\frac{x-A}{h}$

x̄ = A + h·Σfu/N.

Combined mean

$\bar{x}=\frac{n_1{\bar{x}}_1+n_2{\bar{x}}_2}{n_1+n_2}$

Weighted by group sizes.

Combined variance

${\sigma }^2=\frac{n_1({\sigma }_1^2+d_1^2)+n_2({\sigma }_2^2+d_2^2)}{n_1+n_2},d_i={\bar{x}}_i-\bar{x}$

Do not simply average the SDs.

Statistics: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the school exam asks Statistics

The question patterns to expect, and the method that works for each.

Tricks to remember Statistics

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

• Mean of squares minus square of mean

  σ² = (Σx²/n) − (x̄)².

  Use it when: Quick variance.

• Shift doesn't matter, scale does

  Adding a constant leaves σ unchanged; multiplying by k multiplies σ by |k| (variance by k²).

  Use it when: Effects of changing data; step-deviation method.

• Square the step

  In step deviation, variance gets h², SD gets h.

  Use it when: Avoiding a missing factor.

• Combined needs the gaps

  Combined variance adds each group's σ² plus its squared gap d² from the overall mean.

  Use it when: Combined SD.

• Absolute for MD, square for SD

  Deviations sum to zero, so remove signs: by |…| for MD, by squaring for SD.

  Use it when: Remembering the definitions.

Statistics: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section D style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 5-mark question split into 1 + 1 + 1 + 2 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Statistics

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

• ✗Mistake: Forgetting h² in the step-deviation method

  Students compute the bracket and stop.

  Wrong working

  σ² = 68/50 − (10/50)² = 1.32.

  Right working

  σ² = 10² × 1.32 = 132.

  ✓Do this: Write h² outside the bracket before calculating.

  Usually costs: 1-2 marks.

• ✗Mistake: Using deviations without absolute values for MD

  Signed deviations cancel.

  Wrong working

  MD of 3, 5, 8, 10, 14 = (−5 − 3 + 0 + 2 + 6)/5 = 0.

  Right working

  (5 + 3 + 0 + 2 + 6)/5 = 3.2.

  ✓Do this: Write |x − x̄| as a column heading.

  Usually costs: Full marks for the part.

• ✗Mistake: Averaging SDs for combined groups

  Students take (σ₁ + σ₂)/2 or a weighted average of σ.

  Wrong working

  Combined SD = (30 × 8 + 20 × 6)/50 = 7.2.

  Right working

  Use [n₁(σ₁² + d₁²) + n₂(σ₂² + d₂²)]/N: ≈ 8.76.

  ✓Do this: Combine variances with the d² terms, then take the root.

  Usually costs: 2 marks.

• ✗Mistake: Using class limits instead of class marks

  Students multiply f by the lower limit.

  Wrong working

  For 20–30, x = 20.

  Right working

  x = 25, the mid-value.

  ✓Do this: Add a 'class mark' column first.

  Usually costs: Every later value.

• ✗Mistake: Reporting variance as SD

  Students forget the final square root.

  Wrong working

  SD of 3, 5, 6, 6, 7, 8, 9, 12 = 6.5.

  Right working

  Variance 6.5, SD = √6.5 ≈ 2.55.

  ✓Do this: Check units: SD has the units of the data.

  Usually costs: 1 mark.

Practice questions on Statistics

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

1. Q1 · 1 mark · MCQExam level

   The range of 5, 9, 2, 14, 8 is:
   (a) 9
   (b) 12
   (c) 14
   (d) 7.6

   Hint

   Max − min.

   Answer

   (b)

2. Q2 · 1 mark · MCQExam level

   If the variance of a data set is 16, its standard deviation is:
   (a) 2
   (b) 4
   (c) 8
   (d) 256

   Hint

   Square root.

   Answer

   (b)

3. Q3 · 1 mark · Very short answerExam level

   Write the standard deviation of 6, 6, 6, 6.

   Hint

   No spread.

   Answer

   0

4. Q4 · 1 mark · Very short answerExam level

   If every observation is multiplied by 3, what happens to the SD and the variance?

   Hint

   Scale.

   Answer

   SD is multiplied by 3; variance by 9.

5. Q5 · 2 marks · Short answerExam level

   Find the mean deviation about the mean of 10, 20, 30, 40.

   Hint

   Mean 25.

   Answer

   (15 + 5 + 5 + 15)/4 = 10

6. Q6 · 2 marks · Short answerHarder exam level

   Find the variance of 4, 8, 12, 16, 20.

   Hint

   Mean 12.

   Answer

   (64 + 16 + 0 + 16 + 64)/5 = 32

7. Q7 · 2 marks · Short answerHarder exam level

   For 10 observations, Σx = 60 and Σx² = 400. Find the mean and SD.

   Hint

   σ² = Σx²/n − x̄².

   Answer

   Mean 6; σ² = 40 − 36 = 4; SD 2

8. Q8 · 3 marks · Short answerHarder exam level

   Find the mean deviation about the mean for classes 0–4, 4–8, 8–12, 12–16 with frequencies 2, 6, 8, 4.

   Hint

   Class marks 2, 6, 10, 14.

   Answer

   N = 20; Σfx = 176, x̄ = 8.8. f|x − x̄|: 13.6, 16.8, 9.6, 20.8; total 60.8; MD = 3.04

9. Q9 · 3 marks · Short answerHarder exam level

   Find the SD of x: 1, 2, 3, 4, 5 with f: 4, 6, 10, 6, 4.

   Hint

   The distribution is symmetric about 3.

   Answer

   Mean 3; Σf(x − 3)² = 16 + 6 + 0 + 6 + 16 = 44; σ² = 44/30 ≈ 1.467; σ ≈ 1.21

10. Q10 · 3 marks · Short answerHarder exam level

    Group 1 has 50 values with mean 20 and SD 4; Group 2 has 30 values with mean 28 and SD 6. Find the combined mean and SD.

    Hint

    d₁ = −3, d₂ = 5.

    Answer

    Mean (1000 + 840)/80 = 23; σ² = [50(16 + 9) + 30(36 + 25)]/80 = 3080/80 = 38.5; σ ≈ 6.20

11. Q11 · 2 marks · Short answerHarder exam level

    Find the mean and variance of 1, 2, 3, …, 20.

    Hint

    Use (n + 1)/2 and (n² − 1)/12.

    Answer

    Mean 10.5; variance 399/12 = 33.25

12. Q12 · 2 marks · Short answerHarder exam level

    The mean of 6 observations is 8 and their SD is 3. Find the sum of their squares.

    Hint

    Σx² = n(σ² + x̄²).

    Answer

    6(9 + 64) = 438

13. Q13 · 5 marks · Long answerHarder exam level

    Classes 0–20, 20–40, 40–60, 60–80, 80–100 have frequencies 6, 10, 14, 7, 3.
    (a) Find the mean by the step-deviation method. [2]
    (b) Find the variance. [2]
    (c) Find the SD. [1]

    Hint

    A = 50, h = 20.

    Answer

    u: −2, −1, 0, 1, 2; N = 40; Σfu = −9; Σfu² = 53. (a) x̄ = 50 + 20(−9/40) = 45.5. (b) σ² = 400[53/40 − (9/40)²] = 400 × 1.274375 = 509.75. (c) σ ≈ 22.58.

14. Q14 · 3 marks · Short answerHOTS / competency

    Show that the variance of the 2n + 1 numbers a, a + d, a + 2d, …, a + 2nd is n(n + 1)d²/3.

    Hint

    The mean is the middle term a + nd.

    Answer

    Deviations from a + nd are kd for k = −n, …, n. Σ(kd)² = 2d² × n(n + 1)(2n + 1)/6. Dividing by 2n + 1: σ² = n(n + 1)d²/3.

15. Q15 · 3 marks · Short answerHOTS / competency

    The mean and SD of 25 observations were 60 and 4. One reading, 72, had been copied as 52. Find the correct mean and SD.

    Hint

    Correct Σx and Σx².

    Answer

    Σx = 1500 → 1520, mean 60.8. Σx² = 25(16 + 3600) = 90 400 → 90 400 − 2704 + 5184 = 92 880. σ² = 3715.2 − 3696.64 = 18.56; σ ≈ 4.31.

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. 1.Range?Show answer

   Largest − smallest value.

2. 2.Mean deviation about the mean?Show answer

   Σf|x − x̄|/N.

3. 3.Variance short form?Show answer

   Σx²/n − x̄².

4. 4.Step-deviation variance?Show answer

   h²[Σfu²/N − (Σfu/N)²].

5. 5.Effect of adding k to every value on σ?Show answer

   None.

6. 6.Effect of multiplying every value by k on σ?Show answer

   σ becomes |k|σ.

7. 7.Combined mean of two groups?Show answer

   (n₁x̄₁ + n₂x̄₂)/(n₁ + n₂).

8. 8.Variance of the first n natural numbers?Show answer

   (n² − 1)/12.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you find the range and say why it is a weak measure?
• ☐Can you find the MD about the mean for grouped data?
• ☐Can you find the SD of raw data quickly?
• ☐Can you use an assumed mean correctly?
• ☐Can you apply step deviation without forgetting h²?
• ☐Can you find the combined mean and SD of two groups?
• ☐Can you recover missing observations from a mean and variance?
• ☐Can you correct the mean and SD after a misread value?

Statistics: frequently asked questions