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DS-5: Position and Distribution Shape

Module 1 · Descriptive Statistics

Section 1: Introduction

A student scores 78 on an exam. Should she be pleased? Worried? Relieved? The raw number is almost useless in isolation. If the class average was 60 and the standard deviation was 8, a 78 is outstanding — more than two standard deviations above the mean. But if the average was 85 and nearly everyone scored between 80 and 90, a 78 is below average and sits in the lower portion of the class.

Raw numbers only acquire meaning when placed in context. This lesson gives you the tools to locate any value precisely within its distribution — and to understand what the shape of that distribution tells you about where the data tend to cluster, how they spread, and whether standard benchmarks like the 68–95–99.7 rule can be applied.

After this lesson, you will be able to:

  • Locate and interpret percentiles (), quartiles (Q1, Q2, Q3), and deciles for a dataset
  • Compute a z-score for a data value using both population (, ) and sample (, ) parameters, and interpret its sign and magnitude
  • Classify the shape of a distribution as right-skewed, left-skewed, or symmetric by comparing the mean and median
  • Apply the Empirical Rule (68–95–99.7 rule) to approximately normal distributions and identify when it cannot be applied

Section 2: Prerequisites

The tools introduced in this lesson are built directly on top of the spread measures you learned in DS-4. Before continuing, confirm that you can retrieve the key quantities and vocabulary that will appear throughout every worked example and practice problem.

  • From DS-4: Standard deviation and . is the population standard deviation; is the sample standard deviation. Both measure spread in the original units. (You will divide by one of these to compute every z-score in this lesson.)
  • From DS-4: Quartiles Q1, Q2, Q3 and the IQR. Q1 is the 25th percentile, Q2 is the median (50th percentile), Q3 is the 75th percentile; IQR Q3 Q1. (This lesson extends that framework to all 99 percentiles.)
  • From DS-4: The five-number summary. Minimum, Q1, Q2, Q3, Maximum — a compact description of spread and position. (You will see how all five numbers connect to the broader percentile framework.)
  • From DS-3: Mean and median. The relationship between mean and median signals whether a distribution is skewed. (This becomes central in S3 C4 — you will use the mean vs. median comparison to classify shape.)
  • From DS-2: Reading histograms. A histogram’s visual shape — which tail is longer, where the peak sits — is the primary tool for identifying skewness. (You will practise this in S4 Ex4 and the Distribution Shape Explorer.)

Prerequisite self-check

Retrieval Checkpoint: Before moving forward, test your DS-4 foundations.

A sample of 9 daily temperatures (°C) gives a mean of 22°C and a standard deviation of °C. Q1 and Q3 . Which of the following correctly describes the interquartile range and its meaning?

Check your comfort level with the remaining foundations:

Retrieval Warm-up — from DS-3 and DS-4

A dataset of 9 daily high temperatures (°C) is: . Without computing the full variance, a student argues: “The mean will be pulled upward by the 35° value, and the standard deviation will also be larger than it would be without that outlier.”

Which of the following best evaluates the student’s claim?

A sample of 12 apartment rents has: Q1 = 1,150, Q3 = $1,380 (all in dollars). The IQR fence method flags one rent as a potential outlier.

Which of the following additional facts would tell you the outlier is on the high end (above the upper fence) rather than the low end?

Success Factor:

  • Where DS-5 diverges from DS-4: In DS-4, you measured how spread out data are as a whole — the IQR and standard deviation describe the distribution collectively. In DS-5, you will use that spread to locate individual values precisely. A single data point will go from being “above average” to having an exact position: at the 84th percentile, or 1.25 standard deviations above the mean. DS-5 also asks a new question: what is the overall shape of the distribution, and what does that shape constrain or enable?

Section 3: Core Concepts

Navigation guide — six concepts, one coherent framework:

  • C1 (Percentiles): The ruler — locates any value’s rank in the distribution.
  • C2 (Z-Score): The calibrated ruler — expresses position in standard deviation units.
  • C3 (Interpreting Z-Scores): Reading the calibrated ruler — what the number actually means.
  • C4 (Skewness): The shape of the distribution — which tail is longer, and what that says about mean vs. median.
  • C5 (Empirical Rule): The benchmark — what proportions to expect within 1, 2, and 3 standard deviations of the mean in a bell-shaped distribution.

C1 — Percentiles, Quartiles, and Deciles

Intuitively: when you receive a standardized test score report, it often says something like “you scored at the 83rd percentile.” That means approximately 83% of test-takers scored below you. Percentiles turn raw values into relative positions.

Percentile

The -th percentile, written , is the value below which approximately of the observations fall. For example, (Q1) is the value below which 25% of the data falls; (Q2, the median) is the value below which 50% falls; (Q3) is the value below which 75% falls.

Deciles divide the distribution into tenths: , , …, .

Mini-example — finding P75 by nearest rank: Sorted data (n = 8): 12, 15, 18, 21, 24, 27, 30, 33.

Rank for P75: . The 6th value in the sorted list is 27. So — approximately 75% of the values fall below 27.

A percentile rank tells you the proportion of data below a value — it is not the value itself. Saying “I scored at the 90th percentile” means 90% of scores are below yours, not that you scored 90 out of 100. These are completely different statements.


C2 — Z-Score (Standardization)

Intuitively: consider two students. Maria scored 88 on a chemistry exam where and . James scored 91 on a physics exam where and . Who performed better relative to their class? The raw scores don’t answer this — but z-scores do.

We write to mean “the number of standard deviations a value lies above or below the mean.” A positive means above the mean; a negative means below.

Z-Score

Population z-score:

Sample z-score:

The z-score is unitless (the units of and or cancel with the units of or ), which allows meaningful comparisons across datasets measured in different units.

Mini-example — comparing two students:

Maria: . She is 2 standard deviations above her class mean.

James: . He is 3 standard deviations above his class mean.

James performed better relative to his class, even though Maria’s raw score is lower than James’s raw score. The z-score makes the comparison fair.

A z-score of +1 does not mean “in the top 68% of the data.” It means the value is exactly one standard deviation above the mean. The percentage of data above or below that point depends on the shape of the distribution — you cannot convert z-scores to percentile ranks without additional information about the distribution’s shape.


C3 — Interpreting Z-Scores

We read z-scores on a signed scale anchored at zero:

Because z-scores are unitless, they enable cross-dataset comparisons: a height of and a salary of both represent the same relative position in their respective distributions (halfway between the mean and one SD below it), even though heights are measured in centimetres and salaries in dollars.

Z-scores can be negative. A negative z-score simply means the raw value is below the mean — it does not indicate a “bad” measurement or an error. A temperature of is just 1.8 standard deviations cooler than average for that dataset.


C4 — Distribution Shape and Skewness

Intuitively: if you looked at household income in Canada, you would see a histogram that rises quickly on the left and then has a long, thin tail stretching far to the right. A small number of very high incomes pull the mean upward while the median — the “middle household” — sits much lower. This asymmetry is called skewness.

Skewness

Right-skewed (positive skew): The tail extends to the right. Most data cluster on the left, but a few unusually large values pull the mean to the right. Consequence: mean > median.

Left-skewed (negative skew): The tail extends to the left. Most data cluster on the right, but a few unusually small values pull the mean to the left. Consequence: mean < median.

Symmetric: The distribution has no dominant tail. Mean ≈ median. (A perfectly symmetric distribution has mean = median exactly.)

Mini-example — reading skewness from mean vs. median:

A dataset of final exam scores has mean and median . Because mean < median, the mean is being pulled to the left by a cluster of very low scores. The distribution is left-skewed: most students scored reasonably well, but a small group with very low scores drags the mean down.

“Right-skewed” does not mean the peak is on the right. It means the tail extends to the right. In a right-skewed distribution, the peak (mode) is on the left, and the long tail points rightward. Students who reverse this description will misread histograms and draw incorrect conclusions about mean vs. median relationships.

The mean is always pulled toward the tail, not away from it. In a right-skewed distribution (tail on the right), the mean is pulled right, so mean > median. In a left-skewed distribution (tail on the left), the mean is pulled left, so mean < median.

Now explore how shape, spread, and the positions of mean, median, and mode interact visually. First, three fixed shapes side by side as a reference — note where the tail points and how the mean and median order themselves in each:

Next, experiment freely: drag the sliders to morph a single distribution through every degree of skew and spread, and watch the mean, median, and mode separate and re-converge in real time.

Prediction Checkpoint: Before you drag — as you push the skewness slider to the right (making the distribution right-skewed), will the mean end up to the left or to the right of the median? Commit to an answer, then drag the slider to check it.


C5 — The Empirical Rule (68–95–99.7 Rule)

Intuitively: for data that follows an approximately bell-shaped (normal) distribution, there are three remarkably reliable benchmarks for how much data falls within 1, 2, or 3 standard deviations of the mean.

Empirical Rule

For an approximately normal (bell-shaped) distribution with mean and standard deviation :

  • Approximately 68% of the data falls within 1 standard deviation: between and
  • Approximately 95% of the data falls within 2 standard deviations: between and
  • Approximately 99.7% of the data falls within 3 standard deviations: between and

These are approximations — express them as “approximately 68%,” “approximately 95%,” and “approximately 99.7%,” never as exact values.

Mini-example — exam scores: Exam scores are approximately normally distributed with and .

  • Within 1 SD: 72 − 8 = 64 to 72 + 8 = 80 → approximately 68% of students scored between 64 and 80.
  • Within 2 SDs: 72 − 16 = 56 to 72 + 16 = 88 → approximately 95% scored between 56 and 88.
  • Within 3 SDs: 72 − 24 = 48 to 72 + 24 = 96 → approximately 99.7% scored between 48 and 96.

The Empirical Rule applies only to approximately normal (bell-shaped) distributions. Applying it to a heavily skewed distribution, a bimodal distribution, or any distribution with extreme outliers will produce badly wrong estimates. Always check that the distribution is approximately symmetric and bell-shaped before invoking this rule.

Section 4: Worked Examples

Example 1 — Compute and Interpret a Z-Score (Fully Worked)

Problem: In a statistics class of 30 students, the exam scores are approximately normally distributed with mean and standard deviation . Sophie scored 91. Compute Sophie’s z-score and interpret it fully.

Step 1 — Identify what is given and what is asked.

I notice this is a population (the whole class), so I use and , not and . I am asked for the z-score and its interpretation.

Given: , , . Formula: .

Step 2 — Compute.

Step 3 — Interpret.

I notice the z-score is positive and close to 2. I choose the following interpretation because it communicates both direction and magnitude in context:

Sophie’s score of 91 is approximately 1.89 standard deviations above the class mean. Since the distribution is approximately normal, the Empirical Rule tells us approximately 95% of scores fall within 2 SDs of the mean (55 to 93). Sophie’s score places her in the top few percent of the class.

The z-score is unitless — it would be meaningful to compare Sophie’s performance to another exam with different scoring if we had that exam’s z-score.


Example 2 — Locating a Percentile (Prediction Checkpoint)

Problem: A dataset of 10 sorted quiz scores is: 42, 48, 55, 61, 67, 72, 78, 83, 89, 95. What value is at the 30th percentile ()?

Before computing, form your own prediction:

Prediction Checkpoint: Look at the sorted list above. If means “the value below which 30% of the data falls,” roughly which value do you expect? Commit to a guess before revealing the solution.

Show Solution

Using the nearest-rank method: .

The 3rd value in the sorted list is 55.

Therefore — approximately 30% of quiz scores fall below 55.

Sanity check: (Q1) should be around the 2nd–3rd values; our answer of 55 (3rd value) is consistent with that. (median) would average the 5th and 6th values: . Our 55 is below the median, which makes sense for the 30th percentile.


Example 3 — Applying the Empirical Rule

Problem: Adult male heights in a region are approximately normally distributed with cm and cm. (a) What percentage of men are between 164 cm and 192 cm tall? (b) A man is 199 cm tall — approximately what percentage of men are taller than him?

Show Solution

(a) First, express the bounds in terms of and :

  • 164 = 178 − 14 =
  • 192 = 178 + 14 =

By the Empirical Rule, approximately 95% of men are between 164 cm and 192 cm.

(b) 199 = 178 + 21 = .

By the Empirical Rule, approximately 99.7% of men fall within 3 SDs of the mean, so approximately fall outside 3 SDs on both sides combined. By symmetry, approximately of men are taller than 199 cm.

This man’s height is exceptionally rare — roughly 1 in 667 adult males.


Example 4 — Find the Error

A student is given a histogram of house prices in a suburban area. The histogram has a long right tail with a few very expensive properties. The student writes the following analysis:

Histogram of house prices. Most homes cluster at the lower end on the left, forming a tall peak, while a few very expensive homes form a long thin tail stretching to the right — a right-skewed shape. House price (low → high)
The suburb's house prices. Use the shape to evaluate the student's analysis below.

Student’s analysis:

“The histogram shows a right-skewed distribution. Because the tail is on the right, I know that the peak of the distribution (the mode) is on the right side of the histogram and the mean is pulled to the left of the median. Therefore, the mean is less than the median. I will use the mean as my measure of centre because it is the most accurate measure.”

Identify every error in this analysis.

Show Solution

Error 1 — Misidentifying where the peak is in a right-skewed distribution. In a right-skewed distribution, the long tail extends to the right, and the peak (mode) is on the left. The student wrote the opposite. The peak clusters where most values are — toward the lower end — and the tail stretches toward the few extreme high values.

Error 2 — Wrong direction of mean vs. median in right-skewed data. In a right-skewed distribution, the extreme high values (the right tail) pull the mean upward toward the right. Therefore mean > median in right-skewed data. The student wrote “mean < median,” which describes left-skewed data.

Error 3 — Inappropriate measure of centre for skewed data. For skewed distributions, the median is more representative than the mean because it is resistant to the extreme values in the tail. The mean is pulled toward the outliers and overstates the “typical” house price. The student chose the wrong measure and gave an incorrect justification.

Section 5: Guided Practice

Problem 1 — Relative Position and Z-Score Computation (C1, C2)

First, commit to each part of the calculation. Then try a fresh instance solo.


Problem 2 — Interpreting Percentile Rank and Score (C3)

The distribution of scores on a national mathematics test has , , , and .


Problem 3 — Application of the Empirical Rule (C5)

Identify the standard-deviation region before using the coverage rule. The first item guides the sequence; the second lets you attempt it solo.


Problem 4 — Shape Classification and Measure Selection (C5)

A researcher collects data on annual salaries at a tech firm. The mean salary is $112,000 and the median salary is $84,000. Which of the following best describes the distribution shape, and what does it imply for choosing a measure of centre?


Problem 5 — Z-Score vs. Percentile Rank (C2, C3)

A national exam is approximately normally distributed with and . A student scores 82. Her teacher reports: “Your z-score is , which means you are at the 68th percentile.”

Which of the following correctly evaluates the teacher’s statement?

Section 6: Independent Practice

Problem 1 — Z-Score Generator


Problem 2 — Percentile Generator


Problem 3 — Empirical Rule

A. Central coverage

B. One-tail coverage

C. Is the rule applicable?

D. Three-standard-deviation bounds

E. An unusual value


Problem 4 — Find the Error


Problem 5 — Multi-Step Synthesis


Mixed Review — Retrieval from Earlier Lessons

These problems draw on concepts from DS-3 and DS-4. Attempting them without re-reading prior lessons is the point — retrieval practice strengthens long-term memory more than re-reading.

Review Problem 1 — Mean, Median, and Outlier Sensitivity (DS-3)


Review Problem 2 — IQR, Outlier Detection, and Spread Comparison (DS-4)

Section 7: Mastery Check

No hints. No guidance. These three items measure whether the core ideas have actually landed.


Question 1 — Feynman Test (Relative Position)

Unscored self-check: write your response before comparing it with the model answer.

Explain, in your own words and without using formulas, why a z-score of −1.5 tells you more about a value’s position than simply saying “it’s below average.” What information does the z-score communicate that “below average” does not?

0 / 400

Question 2 — Apply (Empirical Rule Applicability)

A human resources analyst collects salary data from a small company. The median annual salary is $68,000 and the mean is $94,000. The analyst wants to use the Empirical Rule to estimate what percentage of employees earn within one standard deviation of the mean.

Which of the following best describes what the analyst should do?

Show Solution

The correct action is to refrain from applying the Empirical Rule. The mean ($94,000) is substantially higher than the median ($68,000) — this difference of $26,000 signals significant right skew, almost certainly driven by a small number of highly-paid executives. The Empirical Rule applies only to approximately normal (bell-shaped) distributions.


Question 3 — Analyze (Z-Score Computation Error)


Question 4 — Retrieve (Nearest-Rank Percentile)


Self-Assessment

How confident are you with the material in this lesson?

Still unsureFully confident

Section 8: Boss Fight

Boss Fight — operational cases. Both cases assess the same lesson outcomes in different settings. Attempt either case first, then use the other as an equivalent rematch.

🔬 Case A: Exam review

The final exam results are in. Decide what evidence would justify special attention to a student’s performance.

🏗️ Case B: Production review

You are a quality engineer auditing two production lines. Decide what can responsibly be concluded from the limited information.

🔬 Case A: Exam review

The final exam results are in. The professor tells you that exam scores are approximately normally distributed with and . Three students are asking whether their scores are “unusual enough” to warrant a grade review. The scores are: Student 1: 89, Student 2: 45, Student 3: 71.

Before writing a memo, make and defend a decision about one student’s result. A fresh checkpoint is available each attempt.


Write a brief memo.

Write 3–5 sentences addressed to the professor summarizing your findings. State which students, if any, are performing in a range that might warrant special attention, and explain why — using z-scores as evidence.


0 / 800

🏗️ Case B: Production review

You are a quality engineer auditing two production lines. You have verbal descriptions of their output distributions and limited data.

Line A produces ceramic tiles. A sample of 200 tiles has mean thickness mm and mm. The histogram of tile thicknesses looks approximately bell-shaped and symmetric.

Line B produces custom-order glass panels. A sample of 200 panels has mean thickness mm and mm. The histogram has a sharp peak at 10 mm and a long right tail extending to panels thicker than 18 mm. The median is 10.5 mm.

Specification limits: Line A accepts tiles between 7.1 mm and 8.9 mm. Line B accepts panels between 10.0 mm and 14.0 mm.

Before writing a recommendation, make and defend a decision about a single tile. A fresh checkpoint is available each attempt.


Write an engineering recommendation.

In 4–6 sentences, recommend whether each line’s output can be described using the Empirical Rule, what the implications are for quality reporting, and what additional data you would collect to better assess Line B.


0 / 800

Section 9: Challenge Problems

Optional — stretch beyond the lesson objectives. These problems require connecting concepts or applying mathematical reasoning. They are not required to complete the lesson.


Challenge 1 — Why Can Mean Exceed Median?

A. Explain an unfamiliar dataset

Consider the dataset . What feature of the data explains why the two common measures of centre differ? Explain how changing the largest value would affect each measure.

Show Solution

The upper value is far from the tight cluster of the other four values. It changes the sum, so it can pull the mean far upward. The median depends only on the middle rank, so making the largest value still larger leaves the median unchanged.

B. Make the general argument

Let a dataset have sorted values , and suppose is much larger than all other values. Show when the mean must be greater than the median.

Show Solution

Let (the sum of all values except ). Then:

The median is fixed — it depends only on rank position, not on the value of . As :

while remains constant. Therefore, for sufficiently large , .

More precisely, whenever — that is, whenever the largest value exceeds a threshold determined by how far the median already sits above the sum of the other values. Since right-skewed distributions are defined by the presence of such extreme upper values, mean > median follows structurally.

C. Explain the location of the centre

In a distribution with a few values far to the right of the main cluster, explain why the two measures of centre need not agree.

Show Solution

Imagine a rigid number line acting as a see-saw, with each data value as a unit weight placed at its position. The mean is the balance point — the fulcrum location where the total torque (weight × distance) sums to zero on both sides.

In a right-skewed distribution, a few extreme values sit far to the right of the bulk of the data. Even though they are few in number, their large distance from the center generates a disproportionate rightward torque. To restore balance, the fulcrum (mean) must shift right — well past the point that simply splits the data in half by count (the median).

The median is insensitive to how far those extreme values lie — it only asks “which value is in the middle rank?” The mean must account for their actual distance, so it is always pulled toward the long tail. In a right-skewed distribution the tail extends rightward, so mean > median.


Challenge 2 — Z-Scores Preserve Relative Order (Proof)

Let two values from the same dataset satisfy . Their z-scores are and , where .

(a) Prove algebraically that . (This shows that standardizing data preserves the relative ordering of all values.)

(b) Explain in one sentence why this property is necessary for z-scores to be useful for comparison across datasets.

(c) Does the same property hold if ? Explain what means about the dataset and why z-scores cannot be defined in that case.

Show Solution

(a) Given and :

Since , we have . Since , the fraction . Therefore , which means .

(b) If standardizing reversed the order of some values, z-scores could not be used to rank or compare observations across datasets — a higher z-score would not reliably indicate a higher relative position.

(c) If , every value in the dataset is identical: for all . The formula would require dividing by zero, which is undefined. This case is degenerate — a dataset with zero spread carries no positional information, and the concept of “relative position” is meaningless.

Section 10: Solutions Reference

Complete, step-by-step solutions for all problems in Sections 5–9 are available on the solutions page. Solutions include worked arithmetic, common mistakes to watch for, and interpretation guidance.

View Full Solutions →

If you’re stuck: Re-read the relevant Core Concept in Section 3, then find the Worked Example that maps to that concept (e.g., Example 1 maps to Concept 1). The solutions page shows the reasoning behind every step, not just the final answer.

Quick-Reference Formulas

Z-Score (Standardization):

Z-Score RangeEmpirical Rule (Normal Distributions)
Between -1 and 1Contains of data
Between -2 and 2Contains of data
Between -3 and 3Contains of data
TermMeaning
Percentile ()Approximately of values fall below this point
Quartiles, (Median),
Deciles, , …,