Statistics

What this chapter is about

In earlier classes you learnt to organise data, draw bar graphs and calculate basic measures like mean, median and mode. Class 11 Statistics moves a step further: it asks how spread out or scattered the data points are around the central value. Two batches of students may have the same average marks, yet one batch may have marks tightly clustered near that average while the other has marks ranging widely. Measures of dispersion capture this difference.

This chapter introduces range, mean deviation, variance and standard deviation for ungrouped and grouped data. You will also learn how to compare variability across data sets that have different units or different means by using the coefficient of variation. These tools appear throughout science, economics and quality control whenever you need to describe not just the typical value but the consistency or reliability of a set of observations.

By the end of the chapter you should be able to compute each measure from raw or frequency-table data, interpret what a large or small dispersion tells you, and choose the right measure for a given situation.

Key ideas

• Central tendency alone is not enough. Two distributions can share the same mean yet differ greatly in how the individual values scatter around it.
• Range is the simplest dispersion measure: Range = Maximum value − Minimum value. It uses only two data points and ignores the rest.
• Mean deviation about the mean averages the absolute differences of each observation from the mean. Mean deviation about the median is similarly defined and is smallest when taken about the median.
• Variance averages the squared deviations from the mean; squaring removes negative signs and gives extra weight to outliers.
• Standard deviation is the positive square root of variance; it has the same unit as the data, making it easier to interpret.
• For grouped data with frequencies f₁, f₂, …, fₙ and class marks x₁, x₂, …, xₙ, the formulas involve weighted sums using these frequencies.
• Coefficient of variation (CV) = (Standard deviation / Mean) × 100. It is a pure number (no unit) and lets you compare spread across data sets with different scales.
• A smaller standard deviation or CV indicates more consistent or homogeneous data.

Formulas and facts to remember

• Measure: Range · Ungrouped data (n observations x₁, x₂, …, xₙ with mean x̄): Maximum − Minimum
• Measure: Mean deviation about mean · Ungrouped data (n observations x₁, x₂, …, xₙ with mean x̄): M.D.(x̄) = (1/n) × Σ |xᵢ − x̄|
• Measure: Variance · Ungrouped data (n observations x₁, x₂, …, xₙ with mean x̄): σ² = (1/n) × Σ (xᵢ − x̄)²
• Measure: Standard deviation · Ungrouped data (n observations x₁, x₂, …, xₙ with mean x̄): σ = √[(1/n) × Σ (xᵢ − x̄)²]

Short-cut formula for variance (ungrouped): σ² = (1/n) × Σ xᵢ² − (x̄)²

For grouped data with frequencies fᵢ (total N = Σ fᵢ):

• Variance: σ² = (1/N) × Σ fᵢ (xᵢ − x̄)²
• Alternate form: σ² = (1/N) × Σ fᵢ xᵢ² − (x̄)²

Coefficient of variation: CV = (σ / x̄) × 100

Meaning: CV expresses standard deviation as a percentage of the mean, allowing comparison when means or units differ.

Worked examples

### Example 1 – Ungrouped data Problem: Five saplings in a nursery have heights (in cm): 24, 26, 28, 30, 32. Find the mean, variance and standard deviation.

Solution: 1. Mean x̄ = (24 + 26 + 28 + 30 + 32) / 5 = 140 / 5 = 28 cm. 2. Deviations from mean: −4, −2, 0, 2, 4. 3. Squared deviations: 16, 4, 0, 4, 16. Sum = 40. 4. Variance σ² = 40 / 5 = 8 cm². 5. Standard deviation σ = √8 ≈ 2.83 cm.

### Example 2 – Grouped data Problem: The weekly pocket money (in rupees) of 30 students is grouped below. Find the standard deviation.

• Class: Frequency fᵢ · 50–60: 4 · 60–70: 10 · 70–80: 12 · 80–90: 4

Solution: 1. Class marks xᵢ: 55, 65, 75, 85. 2. N = 4 + 10 + 12 + 4 = 30. 3. Σ fᵢ xᵢ = 4×55 + 10×65 + 12×75 + 4×85 = 220 + 650 + 900 + 340 = 2110. 4. Mean x̄ = 2110 / 30 ≈ 70.33 rupees. 5. Σ fᵢ xᵢ² = 4×3025 + 10×4225 + 12×5625 + 4×7225 = 12100 + 42250 + 67500 + 28900 = 150750. 6. Variance σ² = 150750/30 − (70.33)² = 5025 − 4946.3 ≈ 78.7. 7. Standard deviation σ ≈ √78.7 ≈ 8.87 rupees.

### Example 3 – Comparing two data sets using CV Problem: Factory A produces bolts with mean length 5 cm and standard deviation 0.4 cm. Factory B produces bolts with mean length 8 cm and standard deviation 0.5 cm. Which factory shows more consistent production?

Solution: 1. CV for A = (0.4 / 5) × 100 = 8 %. 2. CV for B = (0.5 / 8) × 100 = 6.25 %. 3. Lower CV means less relative variation, so Factory B is more consistent.

Common mistakes

• Using deviations without absolute value or squaring when finding mean deviation or variance → always use |xᵢ − x̄| for mean deviation and (xᵢ − x̄)² for variance.
• Forgetting to multiply by frequency in grouped-data formulas → each term must carry its weight fᵢ.
• Confusing variance and standard deviation units; variance has squared units → take the square root to get σ in the original unit.
• Comparing standard deviations of data sets with very different means → use coefficient of variation instead.
• Calculating range but reporting it as a measure of average → range measures spread, not central tendency.

Quick revision

• Range = Max − Min; quick but ignores most data.
• Mean deviation uses absolute differences: M.D. = (1/n) Σ |xᵢ − x̄|.
• Variance uses squared differences: σ² = (1/n) Σ (xᵢ − x̄)².
• Standard deviation σ = √(variance); same unit as data.
• CV = (σ / x̄) × 100 removes units; use it to compare variability across different scales.
• Lower dispersion → more consistent data.