DS-4: Solutions — Variability and Spread

Problem 1 — Computing the Range (Ecologist Bird Species)

Dataset: 12, 8, 15, 10, 9, 14, 7, 11 (8 wetland sites).

(a) Find the minimum: Scan all values — the smallest is 7. min = 7 species.

(b) Compute the range: max = 15 (the largest value), min = 7. The bird species counts span 8 species across the 8 sites.

(c) With a 9th site (37 species), new range: The new maximum is 37; the minimum remains 7 (no new low value). The range jumped from 8 to 30 — nearly a 4× increase from a single new observation. This is the range’s fragility in action: one extreme value dominates it completely.

Problem 2 — Sample Variance and Standard Deviation (All 5 Variants)

Every variant follows the same procedure: We use (Bessel’s correction) because we are estimating the population variance from a sample.

Variant 0 — Tomato Plant Fruit Counts (n = 5)

Data: 8, 12, 9, 11, 10.

Step 1 — Mean: .

Step 2 — Deviation table:

8	−2	4
12	+2	4
9	−1	1
11	+1	1
10	0	0

Check: deviations sum to ✓

Step 3 — Sum of squared deviations: .

Step 4 — Sample variance: .

Step 5 — Sample standard deviation: .

Interpretation: The typical plant’s fruit count deviates from the mean of 10 by about 1.58 fruits.

Variant 1 — Runner Sprint Times (n = 6)

Data: 12.1, 11.8, 12.5, 12.0, 11.6, 12.4 (seconds).

Step 1 — Mean: .

Step 2 — Deviation table:

12.1	+0.0333	0.0011
11.8	−0.2667	0.0711
12.5	+0.4333	0.1878
12.0	−0.0667	0.0044
11.6	−0.4667	0.2178
12.4	+0.3333	0.1111

Step 3 — Sum of squared deviations: .

Step 4 — Sample variance: .

Step 5 — Sample standard deviation: seconds.

Interpretation: The typical time deviates from the mean of 12.07 s by about 0.34 s — a small SD relative to the mean indicates a tightly clustered field.

Variant 2 — Cafe Daily Pastry Sales (n = 7)

Data: 24, 30, 22, 28, 26, 20, 25.

Step 1 — Mean: .

Step 2 — Deviation table:

24	−1	1
30	+5	25
22	−3	9
28	+3	9
26	+1	1
20	−5	25
25	0	0

Check: ✓

Step 3 — Sum of squared deviations: .

Step 4 — Sample variance: .

Step 5 — Sample standard deviation: pastries.

Interpretation: Daily sales typically deviate from the mean of 25 by about 3.4 pastries — moderately variable.

Variant 3 — Container Liquid Volumes (n = 8)

Data: 250, 248, 253, 251, 249, 252, 247, 250 (mL).

Step 1 — Mean: .

Step 2 — Deviation table:

250	0	0
248	−2	4
253	+3	9
251	+1	1
249	−1	1
252	+2	4
247	−3	9
250	0	0

Step 3 — Sum of squared deviations: .

Step 4 — Sample variance: .

Step 5 — Sample standard deviation: mL.

Interpretation: The filling process is very consistent — the typical container deviates from 250 mL by only 2.0 mL (under 1% relative variation).

Variant 4 — Package Weights (n = 4)

Data: 3.2, 3.8, 3.5, 3.1 (kg).

Step 1 — Mean: .

Step 2 — Deviation table:

3.2	−0.2	0.04
3.8	+0.4	0.16
3.5	+0.1	0.01
3.1	−0.3	0.09

Step 3 — Sum of squared deviations: .

Step 4 — Sample variance: .

Step 5 — Sample standard deviation: kg.

Interpretation: Packages typically deviate from the mean of 3.4 kg by about 0.32 kg. With n = 4, Bessel’s correction matters: dividing by n = 4 would give , but the corrected is 33% larger.

Problem 3 — Five-Number Summary, IQR, and Outlier Detection (All 5 Variants)

Variant 0 — Statistics Quiz Scores (n = 10)

Sorted: 8, 9, 11, 12, 13, 14, 15, 16, 17, 18 (n = 10, even).

• Q2 (median): positions 5 and 6 → .
• Lower half (positions 1–5): 8, 9, 11, 12, 13. (odd). Q1 = 11 (position 3).
• Upper half (positions 6–10): 14, 15, 16, 17, 18. (odd). Q3 = 16 (position 3).

Five-number summary: min = 8, Q1 = 11, Q2 = 13.5, Q3 = 16, max = 18. IQR = 16 − 11 = 5. Fences: Lower = 11 − 1.5 × 5 = 3.5; Upper = 16 + 1.5 × 5 = 23.5. All values in [3.5, 23.5]. No outliers.

Variant 1 — Delivery Truck Distances (n = 7)

Sorted: 38, 45, 49, 61, 67, 72, 83 (n = 7, odd).

• Q2: position 4 → 61.
• Lower half: 38, 45, 49 → Q1 = 45. Upper half: 67, 72, 83 → Q3 = 72.

Five-number summary: min = 38, Q1 = 45, Q2 = 61, Q3 = 72, max = 83. IQR = 27 km. Fences: Lower = 45 − 1.5 × 27 = 4.5; Upper = 72 + 1.5 × 27 = 112.5. All values in [4.5, 112.5]. No outliers.

Variant 2 — Pharmacy Prescription Counts (n = 7)

Sorted: 88, 98, 115, 142, 160, 175, 205 (n = 7, odd).

• Q2: position 4 → 142.
• Lower half: 88, 98, 115 → Q1 = 98. Upper half: 160, 175, 205 → Q3 = 175.

Five-number summary: min = 88, Q1 = 98, Q2 = 142, Q3 = 175, max = 205. IQR = 77. Fences: Lower = 98 − 1.5 × 77 = −17.5; Upper = 175 + 1.5 × 77 = 290.5. The negative lower fence just means no value can be flagged low (expected for count data). All values in [−17.5, 290.5]. No outliers.

Variant 3 — Car Battery Lifetimes (n = 8)

Sorted: 24, 30, 36, 38, 42, 48, 54, 60 (n = 8, even).

• Q2: positions 4 and 5 → .
• Lower half (positions 1–4): 24, 30, 36, 38. . Q1 = .
• Upper half (positions 5–8): 42, 48, 54, 60. . Q3 = .

Five-number summary: min = 24, Q1 = 33, Q2 = 40, Q3 = 51, max = 60. IQR = 18 months. Fences: Lower = 33 − 1.5 × 18 = 6; Upper = 51 + 1.5 × 18 = 78. All values in [6, 78]. No outliers.

Variant 4 — Apartment Rents (n = 11)

Sorted: 800, 850, 900, 950, 1000, 1050, 1100, 1150, 1250, 1300, 2000 (n = 11, odd).

• Q2: position 6 → 1050.
• Lower half (below Q2): 800, 850, 900, 950, 1000. . Q1 = 900.
• Upper half (above Q2): 1100, 1150, 1250, 1300, 2000. . Q3 = 1250.

Five-number summary: min = 800, Q1 = 900, Q2 = 1050, Q3 = 1250, max = 2000. IQR = 350. Fences: Lower = 900 − 1.5 × 350 = 375; Upper = 1250 + 1.5 × 350 = 1775. Outlier check: 800 > 375 → no low outlier; 2000 > 1775 → 2000 is a potential outlier. All other values (850–1300) are well within the fences. The $2000 apartment should be investigated — luxury unit, data-entry error, or legitimate high-end rental? Do not auto-delete; flag for investigation.

Problem 1 — Sample Variance and Standard Deviation (Generator)

Generated fresh each time. The approach is always the same:

1. Compute the mean: .
2. Compute each deviation and square it.
3. Sum the squared deviations: .
4. Divide by : .
5. Take the square root: .

Sample output 1 (n = 6): Data = 12, 28, 35, 19, 42, 8. . Deviations: −12, +4, +11, −5, +18, −16. Squared: 144, 16, 121, 25, 324, 256. Sum = 886. , .

Sample output 2 (n = 7): Data = 45, 38, 50, 42, 47, 39, 44. . Deviations: +1.43, −5.57, +6.43, −1.57, +3.43, −4.57, +0.43. Squared: 2.04, 31.04, 41.33, 2.47, 11.76, 20.90, 0.18. Sum ≈ 109.71. , .

Sample output 3 (n = 5): Data = 22, 31, 18, 27, 25. . Deviations: −2.6, +6.4, −6.6, +2.4, +0.4. Squared: 6.76, 40.96, 43.56, 5.76, 0.16. Sum = 97.20. , .

Problem 2 — Five-Number Summary and IQR (Generator)

Generated with . Follow the median-of-halves method (Tukey).

Sample output 1 (n = 9, odd): Sorted 12, 23, 28, 35, 47, 52, 61, 74, 88. Q2 = position 5 = 47. Lower half: 12, 23, 28, 35 → Q1 = (23+28)/2 = 25.5. Upper half: 52, 61, 74, 88 → Q3 = (61+74)/2 = 67.5. Summary: min = 12, Q1 = 25.5, Q2 = 47, Q3 = 67.5, max = 88. IQR = 42.

Sample output 2 (n = 10, even): Sorted 5, 18, 22, 31, 40, 49, 55, 63, 77, 91. Q2 = (40+49)/2 = 44.5. Lower half (1–5): 5, 18, 22, 31, 40 → Q1 = 22. Upper half (6–10): 49, 55, 63, 77, 91 → Q3 = 63. Summary: min = 5, Q1 = 22, Q2 = 44.5, Q3 = 63, max = 91. IQR = 41.

Sample output 3 (n = 11, odd): Sorted 8, 15, 21, 30, 36, 44, 52, 58, 67, 75, 82. Q2 = position 6 = 44. Lower half: 8, 15, 21, 30, 36 → Q1 = 21. Upper half: 52, 58, 67, 75, 82 → Q3 = 67. Summary: min = 8, Q1 = 21, Q2 = 44, Q3 = 67, max = 82. IQR = 46.

Problem 3 — Outlier Detection via Fences (Generator)

Datasets may contain 0, 1, or 2 potential outliers. Always use 1.5 × IQR fences.

Sample 1 — No outliers (n = 12): Sorted 14, 18, 22, 27, 31, 35, 39, 43, 48, 52, 58, 72. Q2 = (35+39)/2 = 37. Lower half (1–6): Q1 = (22+27)/2 = 24.5. Upper half (7–12): Q3 = (48+52)/2 = 50. IQR = 25.5. Lower fence = 24.5 − 1.5 × 25.5 = −13.75; Upper fence = 50 + 1.5 × 25.5 = 88.25. All values in [−13.75, 88.25]. No outliers.

Sample 2 — Single outlier (n = 10): Sorted 11, 15, 19, 24, 28, 33, 37, 42, 48, 112. Q2 = (28+33)/2 = 30.5. Lower half (1–5): Q1 = 19. Upper half (6–10): Q3 = 42. IQR = 23. Lower fence = 19 − 34.5 = −15.5; Upper fence = 42 + 34.5 = 76.5. 112 > 76.5 → 112 is a potential outlier. Could be a data-entry error (e.g., 11.2 typed as 112) or a genuine extreme.

Sample 3 — Larger sample (n = 15): Sorted 8, 12, 16, 20, 23, 27, 31, 35, 40, 44, 49, 55, 62, 70, 145. Q2 = position 8 = 35. Lower half (below Q2): 8, 12, 16, 20, 23, 27, 31 → Q1 = position 4 = 20. Upper half (above Q2): 40, 44, 49, 55, 62, 70, 145 → Q3 = position 4 = 55. IQR = 35. Lower fence = 20 − 52.5 = −32.5; Upper fence = 55 + 52.5 = 107.5. 145 > 107.5 → 145 is a potential outlier; no low outliers.

Problem 4 — Range vs. Variance: Same Range, Different Spread (Generator)

Two datasets can share the same range yet differ greatly in variance — because the range uses only the two endpoints, while the variance uses every value.

Clustered dataset: min = 20, max = 27, Range = 7. Values bunched near the centre: 22, 23, 24, 24, 25, 25, 26. (tight around ).

Dispersed dataset: min = 20, max = 27, Range = 7 (identical). Values spread across the range: 20, 21, 23, 24, 25, 26, 27. (about 4× larger).

Why range and variance diverge: the range is blind to everything between the endpoints; the variance captures internal clustering or dispersion. Two datasets can have identical bookends but completely different internal structure — the variance sees it, the range does not. (The generator constructs the dispersed set so that ; the specific values vary per generation.)

Problem 5 — Find the Error in Spread-Measure Computations (All 5 Variants)

Variant 0 — Student’s SD Computation (Study Hours)

Data: 8, 6, 10, 4, 12. Student computed , , .

Error: divided by n = 5 instead of n−1 = 4. The squared deviations are correct: . Correct: (not 8), hours (not 2.83). The error understates both — exactly the bias Bessel’s correction fixes.

Variant 1 — Analyst’s Five-Number Summary (Customer Counts)

Sorted: 23, 26, 28, 30, 34, 37, 39, 41, 42, 45.

Error: approximated Q1 and Q3 by raw position (“3rd value,” “8th value”) instead of the median-of-halves method. Correct: n = 10 (even); Q2 = (34+37)/2 = 35.5; lower half 23, 26, 28, 30, 34 → Q1 = 28; upper half 37, 39, 41, 42, 45 → Q3 = 41. IQR = 13. The analyst’s IQR happened to match by coincidence, but the method was wrong and fails on other datasets.

Variant 2 — Lab Technician’s Range-Only Report (Chemical Samples)

Data: 5.12, 5.08, 5.15, 5.11, 5.09, 5.14 g. Technician reported range = 0.07 g and stopped.

Error: relying solely on the range, which uses only min and max and reveals nothing about internal consistency. Should have reported the SD (or IQR) too: , sum of squared deviations ≈ 0.00335, , g. The SD of 0.026 g confirms excellent consistency (CV ≈ 0.5%) — but only computing it proves this.

Variant 3 — Student Forgets to Square Deviations (Exam Scores)

Data: 65, 70, 75, 80, 85, 90, 95. Student: , .

Error: summed the absolute deviations (15+10+5+0+5+10+15 = 60) and divided by n−1 = 6 — that computes the mean absolute deviation, not the SD. Correct: deviations −15, −10, −5, 0, +5, +10, +15; squared 225, 100, 25, 0, 25, 100, 225; sum = 700; ; (not 10). The answer is close here only because the data are evenly spaced.

Variant 4 — HR Analyst’s Range Comparison with an Outlier (Salaries)

Dept A: 42, 48, 45, 52, 47, 44, 50, 46. Dept B: 38, 42, 55, 40, 95, 44, 41, 39. Analyst: Range A = 10, Range B = 57 → “Dept B has much more pay inequity.”

Error: the range is dominated by Dept B’s single $95K outlier. Better — compare IQRs: Dept A sorted 42, 44, 45, 46, 47, 48, 50, 52 → Q1 = 44.5, Q3 = 49, IQR(A) = 4.5. Dept B sorted 38, 39, 40, 41, 42, 44, 55, 95 → Q1 = 39.5, Q3 = 49.5, IQR(B) = 10.0. The IQRs are much closer (about 2.2×, not 5.7×). The dramatic ratio came almost entirely from the $95K outlier, not genuine dispersion — investigate it before drawing equity conclusions.

Problem 6 — Multi-Step Synthesis: Forestry Researcher’s Maple Tree Data

Data: 28, 34, 22, 31, 45, 26, 38, 24, 29 (cm, diameter at breast height). Sorted: 22, 24, 26, 28, 29, 31, 34, 38, 45.

(a) Mean, Variance, and Standard Deviation

Mean: , , cm.

22	−8.78	77.09
24	−6.78	45.97
26	−4.78	22.85
28	−2.78	7.73
29	−1.78	3.17
31	+0.22	0.05
34	+3.22	10.37
38	+7.22	52.13
45	+14.22	202.21

. cm². cm.

(b) Five-Number Summary, IQR, and Outliers

n = 9 (odd). Q2: position 5 → 29 cm. Lower half: 22, 24, 26, 28 → Q1 = (24+26)/2 = 25 cm. Upper half: 31, 34, 38, 45 → Q3 = (34+38)/2 = 36 cm.

Five-number summary: min = 22, Q1 = 25, Q2 = 29, Q3 = 36, max = 45. IQR = 11 cm. Fences: Lower = 25 − 1.5 × 11 = 8.5; Upper = 36 + 1.5 × 11 = 52.5. All 9 values in [8.5, 52.5]. No outliers.

(c) Adding a 10th Tree (68 cm) — Resistance Comparison

New sorted data: 22, 24, 26, 28, 29, 31, 34, 38, 45, 68 (n = 10). , cm.

22	−12.5	156.25
24	−10.5	110.25
26	−8.5	72.25
28	−6.5	42.25
29	−5.5	30.25
31	−3.5	12.25
34	−0.5	0.25
38	+3.5	12.25
45	+10.5	110.25
68	+33.5	1122.25

. , cm. Change in SD: .

New five-number summary (n = 10): Q2 = (29+31)/2 = 30. Lower half (1–5): 22, 24, 26, 28, 29 → Q1 = 26. Upper half (6–10): 31, 34, 38, 45, 68 → Q3 = 38. IQR = 12 cm. Change in IQR: .

Interpretation: One extreme observation (68 cm) nearly doubled the SD (+87.6%) but barely moved the IQR (+9.1%). The SD squares deviations, so the 68 cm value alone contributed 1122.25 — 67% of the new sum of squares. The IQR ignores the tails, so it was largely unaffected. Principle: SD is sensitive to outliers; IQR is resistant — mirroring mean (sensitive) vs. median (resistant) from DS-3. Use mean + SD for symmetric data without outliers; median + IQR for skewed data or data with outliers.

Problem 1 — Feynman Test: Why n−1? (Model Answer)

When you have the entire population you know the true centre, so dividing by N gives the actual average squared deviation. But with a sample you estimate the centre using the sample mean, which is always pulled slightly toward your sample values. This makes the sample deviations a bit smaller than the true deviations would be. Dividing by a slightly smaller number (n−1 instead of n) inflates the variance just enough to correct for this built-in underestimation.

Think of it as a statistical honesty tax: since your sample is almost certainly less variable than the population it came from, you compensate by making the variance a touch larger. The smaller the sample, the bigger the correction — with 5 values, dividing by 4 really matters; with 500 values, dividing by 499 barely changes anything.

There is also a mathematical reason: once you compute and use it for deviations, the n deviations are not independent — they must sum to zero, so once you know n−1 of them, the last is forced. You have n−1 genuine pieces of information — n−1 degrees of freedom — so you divide by n−1.

Problem 2 — Apply: Choosing the Right Spread Measure (Two Cities)

(a) City A — symmetric, bell-shaped, no outliers → standard deviation. It uses every observation, connects directly to the mean, underpins most inferential statistics (CIs, hypothesis tests), and supports the 68–95–99.7 rule (≈68% of values within ±1 SD).

(b) City B — right-skewed with luxury outliers → IQR. It is resistant (luxury prices sit in the upper tail and don’t affect Q1/Q3), describes the middle 50% — the “typical” market — and avoids the SD’s massive inflation from large squared deviations. For City B the SD might be $300K+ while the IQR is only $50K–$80K; the SD would describe nobody’s actual experience.

(c) Principle: the choice of spread measure mirrors the choice of centre from DS-3. Symmetric, outlier-free → mean + SD (uses all data). Skewed or outliers → median + IQR (resistant; ignores the tails). Always pair the spread measure with the matching centre.

Problem 3 — Error Analysis: Variance vs. Standard Deviation Confusion

Data: 18, 22, 20, 24, 19, 23, 21 (°C). Both students report °C.

The tested error (Student 1): claiming SD = 4.33°C because the variance = 4.33°C² — treating variance and SD as the same number. They differ: . Variance is in squared units (°C²); SD is in original units (°C). This variance/SD confusion is one of the most common and consequential errors in introductory statistics.

Verifying the computation: . Deviations: −3, +1, −1, +3, −2, +2, 0. Squared: 9, 1, 1, 9, 4, 4, 0. Sum = 28. Sample variance (n−1 = 6): °C². SD: °C. (Student 2 is correct in concept — .)

Path A — The Analyst: Departmental Salary Equity

Engineering (12): 62, 58, 71, 65, 68, 60, 64, 70, 67, 63, 66, 59 ($K). Marketing (10): 48, 52, 55, 50, 95, 53, 47, 51, 54, 49 ($K).

Task A1 — Mean Salary

Engineering: , ($64,420). Marketing: , ($55,400). Engineering’s mean is about $9,000 higher.

Task A2 — Range and Standard Deviation

Engineering: Range = 71 − 58 = 13 ($13,000). Deviations from :

62	−2.42	5.86
58	−6.42	41.22
71	+6.58	43.30
65	+0.58	0.34
68	+3.58	12.82
60	−4.42	19.54
64	−0.42	0.18
70	+5.58	31.14
67	+2.58	6.66
63	−1.42	2.02
66	+1.58	2.50
59	−5.42	29.38

. , ($4,210).

Marketing: Range = 95 − 47 = 48 ($48,000). Deviations from :

48	−7.40	54.76
52	−3.40	11.56
55	−0.40	0.16
50	−5.40	29.16
95	+39.60	1568.16
53	−2.40	5.76
47	−8.40	70.56
51	−4.40	19.36
54	−1.40	1.96
49	−6.40	40.96

. , ($14,150).

Marketing’s SD ($14,150) is about 3.4× Engineering’s ($4,210) — but the single $95K salary contributes 1568.16 of 1802.40 to the sum of squares (87% from one data point). The SD is mostly measuring how far $95K sits from the mean, not typical pay dispersion.

Task A3 — Five-Number Summary and IQR

Engineering (sorted): 58, 59, 60, 62, 63, 64, 65, 66, 67, 68, 70, 71. Q2 = (64+65)/2 = 64.5. Q1 = (60+62)/2 = 61. Q3 = (67+68)/2 = 67.5. IQR(E) = 6.5 ($6,500).

Marketing (sorted): 47, 48, 49, 50, 51, 52, 53, 54, 55, 95. Q2 = (51+52)/2 = 51.5. Q1 = 49. Q3 = 54. IQR(M) = 5 ($5,000).

Key observation: Marketing’s IQR ($5K) is actually smaller than Engineering’s ($6.5K) — the middle 50% of Marketing salaries are more tightly clustered. The range and SD gave the opposite impression because one outlier dominated them.

Task A4 — Outlier Detection

Engineering: fences [61 − 9.75, 67.5 + 9.75] = [51.25, 77.25]. All values in [58, 71]. No outliers. Marketing: fences [49 − 7.5, 54 + 7.5] = [41.5, 61.5]. 95 > 61.5 → $95K is flagged. All other values (47–55) are inside. Do not auto-delete — investigate (data-entry error, misclassification, legacy contract, or a legitimate specialist), then decide.

Task A5 — Synthesize and Advise

Path B — The Architect: Quality-Control Study Design

Task B1 — Variable Type

“Tablet mass in mg” is quantitative continuous — mass can take any value within the scale’s precision (e.g., 49.87 mg). Continuous data support a richer toolset (normal distribution, SD, CV) than discrete data.

Task B2 — Primary Spread Measure

Standard deviation. The data are symmetric, bell-shaped, and outlier-free (n = 30, ≈ normal). The SD connects to the normal distribution for action limits (±2s, ±3s), underpins SPC control charts, and supports the 68–95–99.7 rule. IQR isn’t wrong but discards information when the data are normal.

Task B3 — Outlier Detection Thresholds

Required: Q1, Q3, IQR from the batch. Multiplier: 1.5 (Tukey). Fences: [Q1 − 1.5 × IQR, Q3 + 1.5 × IQR]. In QC you can supplement with an SD-based rule (flag beyond ±3s — Western Electric rules); the two methods catch different patterns and can coexist.

Task B4 — Comparative Spread Analysis (50 mg vs. 100 mg Tablets)

(a) Fair-comparison measure: Coefficient of Variation (CV). CV is unitless (SD as a percentage of the mean), so it compares across different target masses. An SD of 2 mg is 4% of a 50 mg target but 2% of a 100 mg target.

(b) Computation:

• 50 mg line: mg, mg. .
• 100 mg line: mg, mg. .

Interpretation: the 100 mg line has the lower CV (2.39% vs. 3.59%) — smaller relative variability — even though its absolute SD is larger. Bottom line: the 100 mg process is more consistent. Both CVs are well under 5%, excellent for pharmaceutical content uniformity.

Challenge 1 — Proving and Connecting to n−1

Claim: For any numbers with , we have .

Proof: Since is constant across the sum, . Therefore

Connection to n−1 (degrees of freedom): because always holds, the n deviations are not independent — once you know n−1 of them, the last is forced. Example: Data 2, 5, 8, , deviations −3, 0, +3. Knowing the first two (−3, 0) forces the third (+3). Three values, but only 2 independent deviations → n−1 = 2 degrees of freedom.

Minimum sample size for variance: . If , then and the formula divides by zero. You cannot measure spread from a single observation — variability inherently requires at least two values.

Challenge 2 — Comparing Spread Across Different Units with CV

Farm A (baseline): Height cm, cm → . Ear mass g, g → . Ear mass is relatively more variable.

Variant 0 — Farm B: Height ; Mass . Mass remains relatively more variable; both CVs slightly exceed Farm A’s.

Variant 1 — Farm C: Height ; Mass .

Farm	Height CV
Farm A	15.33%
Farm B	16.31%
Farm C	8.98%

Farm C has the most consistent plant height — a CV of 8.98%, nearly half of A’s and B’s. Even though its plants are taller on average, they are far more uniform relative to their mean.

Variant 2 — Farm D (unit-invariance of CV): Height in metres: m, m → . Converting Farm A to metres ( m, m) gives CV = 15.33% — exactly the same as in cm. Why: converting cm → m divides every value (and thus both and ) by 100, leaving the ratio unchanged: CV does not depend on the measurement units — which is what makes it valid for cross-variable comparisons.