Measures of Spread — Solutions

Fluency
Find the range and IQR for the dataset: 3, 7, 8, 12, 14, 15, 18, 22, 25

Ordered data: 3, 7, 8, 12, 14, 15, 18, 22, 25 (n = 9)

Range = max − min = 25 − 3 = 22

Median (Q2): middle value (5th) = 14

Lower half (values below 14): 3, 7, 8, 12 → Q1 = (7 + 8) ÷ 2 = 7.5

Upper half (values above 14): 15, 18, 22, 25 → Q3 = (18 + 22) ÷ 2 = 20

IQR = Q3 − Q1 = 20 − 7.5 = 12.5
Fluency
Find Q1, Q2 (median), Q3 and IQR for: 5, 9, 11, 13, 15, 17, 19, 23

Ordered data: 5, 9, 11, 13, 15, 17, 19, 23 (n = 8, even)

Q2 (Median): average of 4th and 5th values = (13 + 15) ÷ 2 = 14

Lower half: 5, 9, 11, 13 → Q1 = (9 + 11) ÷ 2 = 10

Upper half: 15, 17, 19, 23 → Q3 = (17 + 19) ÷ 2 = 18

IQR = 18 − 10 = 8
Fluency
Calculate the standard deviation (to 2 d.p.) of: 2, 4, 4, 4, 5, 5, 7, 9
Hint: the mean is 5.

Deviations from mean (x̅ = 5): −3, −1, −1, −1, 0, 0, 2, 4

Squared deviations: 9, 1, 1, 1, 0, 0, 4, 16 → sum = 32

Sample variance: s² = 32 ÷ (8 − 1) = 32 ÷ 7 ≈ 4.571

Sample standard deviation: s = √4.571 ≈ 2.14
Fluency
A dataset has Q1 = 20 and Q3 = 35. (a) Find the IQR. (b) Calculate the lower and upper outlier fences.

(a) IQR = Q3 − Q1 = 35 − 20 = 15

(b) Outlier fences (1.5 × IQR rule):

Lower fence = Q1 − 1.5 × IQR = 20 − 1.5 × 15 = 20 − 22.5 = −2.5

Upper fence = Q3 + 1.5 × IQR = 35 + 1.5 × 15 = 35 + 22.5 = 57.5

Any value below −2.5 or above 57.5 is classified as an outlier.
Understanding
Heights (cm) of 9 basketball players: 172, 175, 178, 180, 182, 183, 185, 188, 220. Use the IQR outlier method to determine whether 220 cm is an outlier. Show all working.

Ordered data: 172, 175, 178, 180, 182, 183, 185, 188, 220 (n = 9)

Median: 5th value = 182

Q1: lower half = 172, 175, 178, 180 → Q1 = (175 + 178) ÷ 2 = 176.5

Q3: upper half = 183, 185, 188, 220 → Q3 = (185 + 188) ÷ 2 = 186.5

IQR = 186.5 − 176.5 = 10

Upper fence = Q3 + 1.5 × 10 = 186.5 + 15 = 201.5

Conclusion: 220 > 201.5, so 220 cm IS an outlier.
Understanding
Two classes sat the same test. Class A has mean 70 and SD = 5. Class B has mean 70 and SD = 15. Interpret what each standard deviation tells you about the students in each class.

Class A (SD = 5): Scores are closely clustered around the mean of 70. Most students performed consistently — the typical score was within about 5 marks of 70. Low spread indicates uniform performance.

Class B (SD = 15): Scores are widely spread around the mean of 70. Some students scored much higher and some much lower than 70. High spread indicates a diverse range of abilities or inconsistent preparation.

Key point: Both classes have the same mean, but Class A’s performance was far more consistent. SD measures the typical distance from the mean — it does not measure the average itself.
Understanding
Dataset: 10, 12, 13, 15, 16, 18, 45. Calculate the range and IQR. Which measure better represents the spread of the central data? Justify your answer.

Ordered data: 10, 12, 13, 15, 16, 18, 45 (n = 7)

Range = 45 − 10 = 35

Median: 4th value = 15

Q1: lower half = 10, 12, 13 → Q1 = 12

Q3: upper half = 16, 18, 45 → Q3 = 18

IQR = 18 − 12 = 6

Which is better? The IQR better represents the spread of the central data. The value 45 is an outlier that inflates the range to 35, making it appear the data is far more spread than the central values actually are. The IQR of 6 captures the spread of the middle 50% and is unaffected by the outlier.
Understanding
The five-number summary for a dataset is: Min = 8, Q1 = 15, Median = 22, Q3 = 30, Max = 45. (a) Find the IQR. (b) Use the fence test to check if there are any outliers. (c) Describe the shape of the distribution.

(a) IQR = Q3 − Q1 = 30 − 15 = 15

(b) Outlier fences:

Lower = 15 − 1.5 × 15 = 15 − 22.5 = −7.5 Min = 8 > −7.5 → no lower outlier

Upper = 30 + 1.5 × 15 = 30 + 22.5 = 52.5 Max = 45 < 52.5 → no upper outlier

No outliers.

(c) Shape: The median (22) is approximately equidistant from Q1 (22 − 15 = 7) and Q3 (30 − 22 = 8), suggesting roughly symmetric data in the middle 50%. However, the upper whisker (45 − 30 = 15) is longer than the lower whisker (15 − 8 = 7), indicating a slight positive (right) skew overall.
Problem Solving
Two investment portfolios both have a mean annual return of 8%. Portfolio A has SD = 2%, Portfolio B has SD = 9%. Explain which portfolio is more suitable for: (a) a retiree needing stable income, (b) a young investor comfortable with risk.

(a) Retiree (stable income): Portfolio A (SD = 2%) is more suitable. The low standard deviation means returns are predictable and stay close to the 8% average each year. A retiree cannot afford large losses and needs reliable income — low volatility is essential.

(b) Young investor (comfortable with risk): Portfolio B (SD = 9%) can be suitable. High standard deviation means returns vary widely from year to year — some years may produce large gains and others losses. A young investor has a long time horizon to recover from poor years and can benefit from the potentially high returns that come with high volatility.

Key concept: Standard deviation measures risk in finance — higher SD means higher risk and higher potential reward.
Problem Solving
A quality control inspector measures 8 bolts (lengths in mm): 50.1, 49.8, 50.2, 50.0, 49.9, 50.3, 49.7, 50.0.
1. Calculate the mean and standard deviation.
2. Bolts outside x̅ ± 2s are rejected. Identify any rejected bolts.
3. If the acceptable tolerance were tightened to x̅ ± 1s, how many bolts would be rejected?
(a) Mean: x̅ = (50.1 + 49.8 + 50.2 + 50.0 + 49.9 + 50.3 + 49.7 + 50.0) ÷ 8 = 400.0 ÷ 8 = 50.0 mm

Deviations: 0.1, −0.2, 0.2, 0, −0.1, 0.3, −0.3, 0

Squared deviations: 0.01, 0.04, 0.04, 0, 0.01, 0.09, 0.09, 0 → sum = 0.28

Sample SD: s = √(0.28 ÷ 7) = √0.04 = 0.2 mm

(b) x̅ ± 2s = 50.0 ± 0.4 → acceptable range: [49.6, 50.4]

All bolts (49.7 to 50.3) fall within [49.6, 50.4]. No bolts rejected.

(c) x̅ ± 1s = 50.0 ± 0.2 → acceptable range: [49.8, 50.2]

Bolts outside this range: 50.3 (above 50.2) and 49.7 (below 49.8). 2 bolts rejected.