Range, Quartiles & Interquartile Range (IQR)
Two datasets can share identical means and medians while having completely different levels of variation. Measures of spread quantify this dispersion.
Quartiles divide an ordered dataset into four equal segments ($25\%$ each):
- Lower Quartile ($Q_1$): The median of the lower half of the data ($25\text{th percentile}$).
- Median ($Q_2$): The middle of the entire dataset ($50\text{th percentile}$).
- Upper Quartile ($Q_3$): The median of the upper half of the data ($75\text{th percentile}$).
- Interquartile Range ($\text{IQR}$): The range of the central $50\%$ of the observations: $$\text{IQR} = Q_3 - Q_1$$
- Semi-Interquartile Range: $\text{SIQR} = \frac{Q_3 - Q_1}{2}$.
Box-and-Whisker Plots & The 1.5×IQR Outlier Fence Rule
A box-and-whisker plot visually encapsulates the Five-Number Summary: $$\{\text{Minimum}, \; Q_1, \; Q_2, \; Q_3, \; \text{Maximum}\}$$
A data point $x$ is formally classified as an outlier if it lies outside the inner fences: $$\text{Lower Fence} = Q_1 - 1.5 \times \text{IQR}$$ $$\text{Upper Fence} = Q_3 + 1.5 \times \text{IQR}$$ $$\text{Outlier Condition: } x < Q_1 - 1.5(\text{IQR}) \quad \text{or} \quad x > Q_3 + 1.5(\text{IQR})$$
Boxplot Drawing Conventions:
- The central rectangular box extends from $Q_1$ to $Q_3$, with a vertical line marking the median $Q_2$.
- Outliers are marked as individual points ($\times$ or $\bullet$).
- The whiskers extend to the smallest and largest non-outlier values in the dataset.
Five-number summary boxplot: showing lower quartile ($Q_1$), median ($Q_2$), upper quartile ($Q_3$), whiskers, and an isolated outlier beyond the fence.
Variance & Standard Deviation (Population vs. Sample)
Standard deviation ($\sigma$) measures the typical average distance of data values from the arithmetic mean ($\bar{x}$).
When estimating the spread of a population from a smaller sample, Bessel's correction divides by $n - 1$: $$s = \sqrt{\frac{\sum (x - \bar{x})^2}{n - 1}}$$
Linear Transformations & Comparative Analysis
When data is converted via $y = ax + b$, measures of spread behave very differently from measures of central tendency.
- Adding a constant ($+b$) shifts every point equally without altering their spacing: $$\text{Range}(y) = |a| \times \text{Range}(x)$$ $$\text{IQR}(y) = |a| \times \text{IQR}(x)$$ $$\sigma(y) = |a| \times \sigma(x)$$ $$\sigma^2(y) = a^2 \times \sigma^2(x)$$
- Note: Adding a constant has zero effect on any measure of spread! Only multiplying by $|a|$ scales the spread.
Worked Example 1: Five-Number Summary & Outlier Test
The resting heart rates (in beats per minute) of 13 athletes are:
$$42, 45, 48, 49, 51, 52, 54, 55, 58, 60, 62, 65, 88$$
a) Determine the five-number summary: Min, $Q_1$, $Q_2$, $Q_3$, Max.
b) Calculate the Interquartile Range (IQR).
c) Apply the $1.5 \times \text{IQR}$ rule to determine whether $88\text{ bpm}$ is an outlier.
Step-by-step Solution:
Part a: Five-Number Summary:
Data is already ordered ($n = 13$).
Median ($Q_2$): Position is $\frac{13+1}{2} = 7$th value $\implies Q_2 = 54$. [M1]
Lower half (first 6 values: $42, 45, 48, 49, 51, 52$):
$Q_1 = \frac{48 + 49}{2} = 48.5$. [A1]
Upper half (last 6 values: $55, 58, 60, 62, 65, 88$):
$Q_3 = \frac{60 + 62}{2} = 61$. [A1]
$\text{Min} = 42$, $\text{Max} = 88$.
Summary: $\{42, \; 48.5, \; 54, \; 61, \; 88\}$.
Part b: Interquartile Range:
$\text{IQR} = Q_3 - Q_1 = 61 - 48.5 = 12.5\text{ bpm}$. [A1]
Part c: Outlier Fence Test:
$\text{Upper Fence} = Q_3 + 1.5 \times \text{IQR} = 61 + 1.5(12.5) = 61 + 18.75 = 79.75\text{ bpm}$. [M1]
Since $88 > 79.75$, the heart rate of $88\text{ bpm}$ is a mathematically confirmed outlier. [A1][R1: Criterion C Communication]
Worked Example 2: Standard Deviation Calculation
Calculate the exact mean ($\bar{x}$) and population standard deviation ($\sigma$) of the following 5 scores: $$3, 5, 8, 11, 13$$
Step-by-step Solution:
Step 1: Calculate the Mean
$\bar{x} = \frac{3 + 5 + 8 + 11 + 13}{5} = \frac{40}{5} = 8$. [A1]
Step 2: Deviations and Squared Deviations
$(3 - 8)^2 = (-5)^2 = 25$
$(5 - 8)^2 = (-3)^2 = 9$
$(8 - 8)^2 = 0^2 = 0$
$(11 - 8)^2 = 3^2 = 9$
$(13 - 8)^2 = 5^2 = 25$ [M1]
$\sum (x - \bar{x})^2 = 25 + 9 + 0 + 9 + 25 = 68$.
Step 3: Variance & Standard Deviation
$\sigma^2 = \frac{68}{5} = 13.6$. [M1]
$\sigma = \sqrt{13.6} \approx 3.69$ (3 s.f.). [Criterion A: Knowledge & Understanding]
Tiered Practice Exercises
10 problems advancing from range, quartiles, and IQR to standard deviation derivations, boxplot comparisons, and linear scaling transformations.
Find the range and interquartile range (IQR) for the following ordered dataset: $$10, 12, 14, 15, 17, 19, 21, 24, 28$$
View Solution & Markscheme
Method & Marks:
Range $= 28 - 10 = 18$. [A1]
$n = 9$. Median $Q_2 = 17$ (5th value).
Lower half ($10, 12, 14, 15$): $Q_1 = \frac{12 + 14}{2} = 13$. [M1]
Upper half ($19, 21, 24, 28$): $Q_3 = \frac{21 + 24}{2} = 22.5$. [M1]
$\text{IQR} = Q_3 - Q_1 = 22.5 - 13 = 9.5$. [A1]
A dataset has $Q_1 = 34$ and $Q_3 = 58$.
a) Calculate the IQR.
b) Calculate the lower and upper outlier fences.
View Solution & Markscheme
Method & Marks:
a) $\text{IQR} = 58 - 34 = 24$. [A1]
b) Lower Fence $= Q_1 - 1.5(\text{IQR}) = 34 - 1.5(24) = 34 - 36 = -2$. [A1]
Upper Fence $= Q_3 + 1.5(\text{IQR}) = 58 + 1.5(24) = 58 + 36 = 94$. [A1]
The standard deviation of test scores in a class is $8.4$. If the teacher decides to add $5$ bonus marks to every student's score, what is the new standard deviation? Explain.
View Solution & Markscheme
Method & Marks:
The new standard deviation is still $8.4$. [A1]
Explanation: Adding a constant value shifts every data point by $+5$ without changing the distances between scores or their deviations from the new mean. Measures of spread are invariant under addition of a constant. [R1]
The masses (in grams) of six chemistry samples are: $12, 14, 15, 18, 19, 22$.
a) Find the mean mass $\bar{x}$.
b) Calculate the population variance $\sigma^2$ and standard deviation $\sigma$.
View Solution & Markscheme
Method & Marks:
a) $\bar{x} = \frac{12 + 14 + 15 + 18 + 19 + 22}{6} = \frac{100}{6} \approx 16.67\text{ g}$. [M1][A1]
Using $\sigma^2 = \frac{\sum x^2}{n} - \bar{x}^2$:
$\sum x^2 = 144 + 196 + 225 + 324 + 361 + 484 = 1734$.
$\sigma^2 = \frac{1734}{6} - \left(\frac{100}{6}\right)^2 = 289 - 277.78 = 11.22\text{ g}^2$. [A1]
$\sigma = \sqrt{11.22} \approx 3.35\text{ g}$. [A1]
Two runners record their race times (in seconds) over 8 training runs:
Runner A: $\text{Mean} = 12.2\text{ s}, \; \text{SD} = 0.15\text{ s}$.
Runner B: $\text{Mean} = 12.2\text{ s}, \; \text{SD} = 0.65\text{ s}$.
a) Which runner is faster?
b) Which runner is more consistent? Justify using statistical concepts.
View Solution & Markscheme
Method & Marks:
a) Both runners have the exact same average speed since their means are identical ($12.2\text{ s}$). [A1]
b) Runner A is significantly more consistent. A lower standard deviation ($0.15 < 0.65$) indicates that Runner A's race times cluster much more tightly around the mean, with minimal variation between runs. Runner B has greater variability. [R1: Criterion C Communication]
For a dataset, the five-number summary is: $\text{Min} = 12, Q_1 = 28, Q_2 = 45, Q_3 = 60, \text{Max} = 115$.
a) Determine whether the maximum value $115$ is an outlier.
b) Determine whether the minimum value $12$ is an outlier.
View Solution & Markscheme
Method & Marks:
$\text{IQR} = Q_3 - Q_1 = 60 - 28 = 32$. [M1]
$1.5 \times \text{IQR} = 1.5 \times 32 = 48$.
$\text{Upper Fence} = 60 + 48 = 108$.
$\text{Lower Fence} = 28 - 48 = -20$.
a) Since $\text{Max} = 115 > 108$, $115$ is an outlier. [A1]
b) Since $\text{Min} = 12 > -20$, $12$ is within the lower fence and is NOT an outlier. [A1]
A set of measurements has a range of $40$ and standard deviation of $6.5$. If every score is converted using $y = 3x - 10$, find the new range, new standard deviation, and new variance.
View Solution & Markscheme
Method & Marks:
Under $y = ax + b$, multiplier $|a| = 3$: [M1]
New Range $= 3 \times 40 = 120$. [A1]
New Standard Deviation $= 3 \times 6.5 = 19.5$. [A1]
New Variance $= 19.5^2 = 380.25$ (or $3^2 \times 6.5^2 = 9 \times 42.25 = 380.25$). [A1]
Prove algebraically that for any set of numbers $x_1, x_2, \dots, x_n$ with mean $\bar{x}$, the variance can be calculated via the alternate formula: $$\sigma^2 = \frac{\sum x^2}{n} - \bar{x}^2$$
View Solution & Markscheme
Method & Marks:
Start from the definition: $\sigma^2 = \frac{1}{n}\sum_{i=1}^n (x_i - \bar{x})^2$. [M1]
Expand the squared binomial: $(x_i - \bar{x})^2 = x_i^2 - 2\bar{x}x_i + \bar{x}^2$. [M1]
Distribute summation: $\sigma^2 = \frac{1}{n}\left[\sum x_i^2 - 2\bar{x}\sum x_i + \sum \bar{x}^2\right]$.
Note that $\sum \bar{x}^2 = n\bar{x}^2$ and $\sum x_i = n\bar{x}$: [M1]
$\sigma^2 = \frac{1}{n}\left[\sum x_i^2 - 2\bar{x}(n\bar{x}) + n\bar{x}^2\right] = \frac{1}{n}\left[\sum x_i^2 - 2n\bar{x}^2 + n\bar{x}^2\right] = \frac{1}{n}\left[\sum x_i^2 - n\bar{x}^2\right]$.
$\sigma^2 = \frac{\sum x^2}{n} - \bar{x}^2$. $\blacksquare$ [A1][R1: Rigorous proof]
The heights of 50 male students have mean $175\text{ cm}$ and standard deviation $6\text{ cm}$. The heights of 50 female students have mean $165\text{ cm}$ and standard deviation $5\text{ cm}$. Calculate the overall standard deviation of all 100 students combined.
View Solution & Markscheme
Method & Marks:
Combined mean: $\bar{x}_c = \frac{50(175) + 50(165)}{100} = 170\text{ cm}$. [M1]
For males: $\sigma_m^2 = \frac{\sum x_m^2}{50} - 175^2 \implies \frac{\sum x_m^2}{50} = 6^2 + 175^2 = 36 + 30625 = 30661 \implies \sum x_m^2 = 1{,}533{,}050$. [M1]
For females: $\frac{\sum x_f^2}{50} = 5^2 + 165^2 = 25 + 27225 = 27250 \implies \sum x_f^2 = 1{,}362{,}500$. [M1]
Combined sum of squares: $\sum x_c^2 = 1{,}533{,}050 + 1{,}362{,}500 = 2{,}895{,}550$.
Combined variance: $\sigma_c^2 = \frac{2{,}895{,}550}{100} - 170^2 = 28955.5 - 28900 = 55.5$. [A1]
Combined standard deviation: $\sigma_c = \sqrt{55.5} \approx 7.45\text{ cm}$. [A1]
Criterion B & D Quality Control Investigation: An automated factory produces ball bearings with a target diameter of $10.00\text{ mm}$. The factory operates two machines (Machine X and Machine Y). Random samples of 100 bearings from each machine yield:
Machine X: $\bar{x} = 10.02\text{ mm}, \; \sigma = 0.08\text{ mm}$, with no outliers.
Machine Y: $\bar{x} = 10.00\text{ mm}, \; \sigma = 0.22\text{ mm}$, with 4 outliers beyond $3\sigma$.
a) Compare the accuracy and precision of both machines.
b) If quality specifications require all bearings to be within $10.00 \pm 0.15\text{ mm}$, recommend which machine should be prioritized for high-precision aerospace contracts. Justify using measures of spread.
View Solution & Markscheme
Method & Marks:
Part a: Accuracy vs. Precision:
- Accuracy is proximity of the mean to the target: Machine Y has higher accuracy ($\bar{x} = 10.00\text{ mm}$ exactly on target vs. $10.02\text{ mm}$).
- Precision is repeatability and closeness of data: Machine X has far higher precision due to a significantly smaller standard deviation ($\sigma = 0.08\text{ mm}$ vs $0.22\text{ mm}$) and zero outliers. [R1: Criterion C]
Part b: Industrial Recommendation:
Machine X should be prioritized. For Machine X, the entire $2\sigma$ spread ($10.02 \pm 0.16\text{ mm}$) lies comfortably within or near tolerance, and a simple calibration adjustment can shift the mean down by $0.02\text{ mm}$. Machine Y's large spread ($\sigma = 0.22\text{ mm}$) means a substantial fraction of bearings will fail the $\pm 0.15\text{ mm}$ specification, causing high scrap rates and safety hazards in aerospace applications. [A1][R1: Criterion D Evaluation]
Measures of Spread Quick Reference
Range
$$\text{Range} = \text{Max} - \text{Min}$$
Full span of dataset.Interquartile Range
$$\text{IQR} = Q_3 - Q_1$$
Spread of middle 50%.Outlier Rule
$$x < Q_1 - 1.5\text{IQR} \text{ or } x > Q_3 + 1.5\text{IQR}$$
Inner fence boundaries.Standard Deviation
$$\sigma = \sqrt{\frac{\sum x^2}{N} - \bar{x}^2}$$
Root mean square deviation.Sample SD
$$s = \sqrt{\frac{\sum(x - \bar{x})^2}{n - 1}}$$
Bessel's correction factor.Linear Shift Invariance
$$y = ax + b \implies \sigma(y) = |a|\sigma(x)$$
Constant $+b$ does NOT alter spread.Top 5 Common MYP Spread Pitfalls
- Changing Spread When Adding a Constant: Students often think adding 10 to every score increases the standard deviation by 10. Spread measures the distances between values, which stay completely identical when shifted!
- Squaring the Multiplier for Standard Deviation: Under $y = ax + b$, the variance multiplies by $a^2$, but the standard deviation multiplies by $|a|$. Do NOT square $a$ when finding the new standard deviation.
- Extending Whiskers to Outliers: Whiskers on a modified boxplot should NOT extend to outliers. Outliers are plotted as individual detached symbols, and whiskers stop at the nearest non-outlier value.
- Using Range for Outlier-Prone Data: Range relies on the two extreme endpoints. A single data entry error of 1000 ruins the range. Quote the IQR instead.
- Confusing Population ($\sigma$) with Sample ($s$): Pay attention to whether the problem involves the entire population ($N$) or a small sample survey ($n - 1$).
Self-Grading Mini Test
5 quiz questions on IQR, outlier rules, standard deviation, and boxplots.
In a boxplot, $Q_1 = 22$ and $Q_3 = 38$. What is the IQR?
Explanation: $\text{IQR} = Q_3 - Q_1 = 38 - 22 = 16$.
Given $Q_1 = 20$ and $Q_3 = 40$, which of the following scores is an outlier?
Explanation: $\text{IQR} = 20$. Upper fence $= 40 + 1.5(20) = 70$. Lower fence $= 20 - 1.5(20) = -10$. Since $75 > 70$, $75$ is an outlier.
A dataset has standard deviation $\sigma = 4$. If each value is multiplied by $3$ and then has $7$ added to it, what is the new standard deviation?
Explanation: For $y = 3x + 7$, $\sigma_y = |3|\sigma_x = 3 \times 4 = 12$. The addition of $7$ does not affect standard deviation.
Which statistic is the best indicator of consistency in experimental trials?
Explanation: Consistency is measured by dispersion/spread. A smaller standard deviation indicates higher consistency.
If the variance of a dataset is $25$, what is its standard deviation?
Explanation: Standard deviation is the square root of variance: $\sigma = \sqrt{25} = 5$.