📏 Measures of Dispersion
Two datasets can share the same mean yet behave completely differently. Dispersion measures — range, variance, standard deviation — capture how spread out or consistent the data really is, which is often what matters most in quality control.
2Topics
8Worked Examples
4Eng. Cases
12Practice Qs
↔️Range & DeviationSimple spread measures
√Variance & SDThe standard engineering measure
⚙️ApplicationsReal engineering
🎮Quick-Fire GameTest your speed
📝PracticeGraded questions
🎯What you will learn
Calculate range and mean deviation as simple spread indicators; understand why variance (mean squared deviation) is the mathematically preferred measure; and compute standard deviation for both raw and grouped data.
💡 Standard deviation is just √variance — it's preferred over variance itself because it's back in the original units (mm, not mm²), making it directly comparable to the mean.
🧭How to use this resource
1. Read the rules banner at the top of each tab
2. Experiment with the interactive tool
3. Check the key vocabulary panel
4. Study the visual diagram, then the worked examples
↔️
Range & Deviation
The simplest ways to describe how spread out a dataset is.
🔧Range & Deviation CalculatorEnter data separated by commas
Enter a list of numbers, then press Calculate to see the range and each value's deviation from the mean…
📖 Key Vocabulary
Range — the difference between the largest and smallest values; sensitive to outliers.
Deviation — the signed distance of a value from the mean.
Mean absolute deviation — the average of |deviations|, avoiding the cancellation problem.
Interquartile range — Q3−Q1, a range measure that ignores the extreme 25% at each end.
👁️ Visual aid: values below the mean give negative deviations, values above give positive ones — they always balance to zero overall.
✏️Worked Example 1 — Finding the rangeType: Range
Question Find the range of 23, 45, 12, 38, 51, 19.
Identify max & min
max=51, min=12Subtract
51 − 12Answer Range = 39
📐Worked Example 2 — Calculating individual deviationsType: Deviation
Question Data: 8, 12, 10, 14, 6 (mean=10). Find each deviation and confirm they sum to zero.
Deviations
8−10=−2, 12−10=+2, 10−10=0, 14−10=+4, 6−10=−4Sum
−2+2+0+4−4 = 0Answer Deviations sum to zero exactly, as expected for any dataset.
📊Worked Example 3 — Mean absolute deviationType: MAD
Question Using the data from Example 2, find the mean absolute deviation.
Take absolute values
|−2|,|2|,|0|,|4|,|−4| = 2,2,0,4,4Average them
(2+2+0+4+4)/5 = 12/5Answer MAD = 2.4 — a typical value differs from the mean by about 2.4 units.
⚙️Worked Example 4 — Engineering context (range as a quick QC check)Type: Engineering
Question Ten shaft diameters (mm) range from 24.97 to 25.04. The tolerance band is ±0.05 mm (0.10 mm total). Is the range within tolerance?
Range
25.04 − 24.97 = 0.07 mmCompare
0.07 mm < 0.10 mm tolerance bandAnswer Yes — the full spread of measured diameters fits comfortably inside the tolerance, a quick first check before deeper statistical analysis.
🧠 Common slip: relying on range alone to judge consistency — two datasets can have identical ranges but very different spreads everywhere else. Range only looks at the two extremes.
√
Variance & Standard Deviation
Squaring the deviations fixes the cancellation problem — and taking the square root brings the units back to normal.
🔧Variance & Standard Deviation CalculatorEnter data separated by commas
Enter a list of numbers, then press Calculate to see the mean, squared deviations, variance and standard deviation…
📖 Key Vocabulary
Variance (σ²) — the average squared deviation from the mean; in squared units.
Standard deviation (σ) — √variance; back in the original units, so directly comparable to the mean.
Population vs sample SD — sample SD divides by (n−1), not n, to correct for bias when estimating from a subset.
Low vs high SD — low SD means tightly clustered data; high SD means widely spread data.
👁️ Visual aid: both curves are centred on the same mean, but the narrow green curve represents far more consistent (lower variance) data.
✏️Worked Example 1 — Full variance and SD calculationType: Full calculation
Question Find the variance and SD of 8, 12, 10, 14, 6.
Mean
(8+12+10+14+6)/5 = 10Squared deviations
(−2)²+2²+0²+4²+(−4)² = 4+4+0+16+16 = 40Variance
40 ÷ 5 = 8Answer Variance = 8, SD = √8 ≈ 2.83
⚖️Worked Example 2 — Comparing two machines by SDType: Comparison
Question Machine A output SD = 1.2 kg, Machine B output SD = 3.8 kg, both with the same mean output. Which machine is more consistent?
Compare Lower SD means less variation around the mean
Answer Machine A (SD=1.2 kg) is far more consistent than Machine B (SD=3.8 kg), even though both produce the same average output.
📊Worked Example 3 — Grouped standard deviationType: Grouped SD
Question Midpoints 5,15,25 with frequencies 4,10,6 (mean=16.5, found separately). Estimate the variance.
Deviations
5−16.5=−11.5, 15−16.5=−1.5, 25−16.5=8.5f×deviation²
4(132.25)+10(2.25)+6(72.25) = 529+22.5+433.5 = 985Answer Variance ≈ 985 ÷ 20 = 49.25, so SD ≈ √49.25 ≈ 7.02
⚙️Worked Example 4 — Engineering context (process capability)Type: Engineering
Question A shaft's target diameter is 20.00 mm with tolerance ±0.06 mm. Ten samples give SD = 0.018 mm. Roughly how many SDs fit inside the tolerance band?
Half-tolerance
0.06 mmDivide by SD
0.06 ÷ 0.018 ≈ 3.33Answer The tolerance band spans about ±3.3 standard deviations — comfortably capturing nearly all production under a normal distribution (a process capability check engineers run routinely).
🧠 Common slip: forgetting to square-root the variance — variance and standard deviation are often confused, but they are in different units (mm² vs mm) and mean different things physically.
⚙️
Engineering Applications
Four live scenarios — each with its own calculator. Enter your own numbers.
🔩 Process Capability — Tolerance vs SD
Comparing tolerance width to standard deviation gives an instant sense of how capable a process is of staying within spec.
Capability ratio = tolerance half-width ÷ SD
📟 Try it — Capability Ratio
Enter tolerance and SD, then press Calculate…
🏭 Batch Consistency — Standard Deviation from Data
Direct SD calculation on a batch of readings reveals whether a process is running consistently.
σ = √[Σ(x−mean)² ÷ n]
📟 Try it — Batch SD Calculator
Enter the readings, then press Calculate…
📶 Signal Noise — Standard Deviation as Noise Level
In signal processing, the standard deviation of a noisy signal (after removing the mean/trend) quantifies the noise level directly.
Noise level ≈ σ of the residual signal
📟 Try it — Noise Level Calculator
Enter the residual values, then press Calculate…
⚖️ Comparing Two Batches — Coefficient of Variation
Comparing SD alone can mislead if means differ greatly; the coefficient of variation (SD/mean ×100%) normalises for scale.
CV (%) = (SD ÷ mean) × 100
📟 Try it — Coefficient of Variation
Enter mean and SD, then press Calculate…
🎮
Quick-Fire Dispersion Game
12 questions on range, deviation, variance and SD. How many can you get?
⚡ Dispersion Challenge
Score: 0Question: 1/12Streak: 0🔥
Press Start to begin!
📝
Practice Questions
Graded from warm-up to challenge. Attempt fully before revealing the working.
Question 1 🥉2 marks
Find the range of 14, 8, 22, 17, 5.
22 − 5 = 17
Question 2 🥉2 marks
Data 4,6,8,10,12 has mean 8. Find the deviation of each value.
−4, −2, 0, +2, +4 (sum = 0)
Question 3 🥉2 marks
Find the variance of 2, 4, 6 (mean=4).
Deviations: −2,0,2 → squares: 4,0,4, sum=8
Variance = 8/3 = 2.67
Variance = 8/3 = 2.67
Question 4 🥉2 marks
If variance = 25, find the standard deviation.
√25 = 5
Question 5 🥈3 marks
Find the variance and SD of 10, 14, 18, 12, 16 (mean=14).
Deviations: −4,0,4,−2,2 → squares: 16,0,16,4,4, sum=40
Variance = 40/5 = 8
SD = √8 ≈ 2.83
Variance = 40/5 = 8
SD = √8 ≈ 2.83
Question 6 🥈4 marks
Machine X has mean 50, SD 2; Machine Y has mean 50, SD 6. Which is more consistent and why?
Same mean, but X has lower SD
Machine X is more consistent — less spread around the mean
Machine X is more consistent — less spread around the mean
Question 7 🥈3 marks
A dataset has mean 40 and SD 5. Find the coefficient of variation.
CV = (5/40)×100 = 12.5%
Question 8 🥈4 marks
Estimate variance from grouped data: midpoints 10,20,30 with frequencies 3,5,2 (mean=18.4).
Deviations: −8.4,1.6,11.6 → squares:70.56,2.56,134.56
f×sq: 3(70.56)+5(2.56)+2(134.56)=211.68+12.8+269.12=493.6
Variance=493.6/10=49.36
f×sq: 3(70.56)+5(2.56)+2(134.56)=211.68+12.8+269.12=493.6
Variance=493.6/10=49.36
Question 9 🥇4 marks
A shaft tolerance is ±0.04 mm; sample SD is 0.012 mm. Find how many SDs fit in the half-tolerance.
0.04 ÷ 0.012 = ≈ 3.33 SDs
Question 10 🥇5 marks
Two production batches: Batch A mean=200,SD=8; Batch B mean=50,SD=4. Using CV, determine which batch is relatively more variable.
CV_A = (8/200)×100 = 4%
CV_B = (4/50)×100 = 8%
Batch B is relatively more variable despite its smaller absolute SD
CV_B = (4/50)×100 = 8%
Batch B is relatively more variable despite its smaller absolute SD
Question 11 🥇5 marks
Find the variance and SD of 5, 9, 5, 13, 8 (calculate the mean first).
Mean = (5+9+5+13+8)/5 = 40/5 = 8
Deviations: −3,1,−3,5,0 → squares: 9,1,9,25,0, sum=44
Variance = 44/5 = 8.8
SD = √8.8 ≈ 2.97
Deviations: −3,1,−3,5,0 → squares: 9,1,9,25,0, sum=44
Variance = 44/5 = 8.8
SD = √8.8 ≈ 2.97
Question 12 🥇5 marks
Grouped data: midpoints 100,110,120,130 with frequencies 2,7,8,3 (mean=116). Estimate the variance and SD.
Deviations: −16,−6,4,14 → squares: 256,36,16,196
f×sq: 2(256)+7(36)+8(16)+3(196) = 512+252+128+588 = 1480
Σf=20, Variance = 1480/20 = 74
SD = √74 ≈ 8.60
f×sq: 2(256)+7(36)+8(16)+3(196) = 512+252+128+588 = 1480
Σf=20, Variance = 1480/20 = 74
SD = √74 ≈ 8.60