L-DS-5: Position and Distribution Shape

Module 1 · Descriptive Statistics

Section 1: Introduction

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A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so you know where to look back.

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.

This extends L-DS-4: there you used standard deviation and quartiles to describe a distribution’s spread; here you use those same quantities to locate individual values and judge the distribution’s shape.

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 from its tail direction, using the mean–median relationship as supporting evidence
Apply the Empirical Rule (68–95–99.7 rule) to approximately normal distributions and identify when it cannot be applied

Section 2: Prerequisites

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A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so you know where to look back.

The tools introduced in this lesson are built directly on top of the spread measures you learned in L-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 L-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 L-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 L-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 L-DS-3: Mean and median. Extreme values affect the mean more than the median. (This becomes central in S3 C4 — you will compare the summaries with a graph’s tail direction when describing shape.)
From L-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 practice this in S4 Ex4 and the Distribution Shape Explorer.)

Prerequisite self-check

Retrieval Checkpoint: Before moving forward, test your L-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 option correctly gives the interquartile range (IQR) and interprets it?

Check your comfort level with the remaining foundations:

I can compute

for a small dataset I can find Q1, Q2, and Q3 for a sorted list with both odd and even

I know the difference between

(population) and

(sample) and when to use each

Success Factor:

Where L-DS-5 diverges from L-DS-4: In L-DS-4, you measured how spread out data are as a whole — the IQR and standard deviation describe the distribution collectively. In L-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. L-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

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A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so you know where to look back.

Navigation guide — five 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.” Under the percentile-rank convention used here, that means approximately 83% of test-takers scored strictly below you. Percentiles turn raw values into relative positions.

Percentile

This lesson uses the nearest-rank convention for a -th percentile, written : sort observations, compute , and take the value at rank . The selected order statistic guarantees that at least of observations are at or below that value. It does not guarantee that exactly are strictly below it: finite sample size and ties matter. When mapping a particular value back to a percentile rank, this lesson reports the actual percentage strictly below it. Stating the convention matters because software and tie-handling rules can differ.

For Q1, Q2, and Q3—especially Q2, the median—continue using the L-DS-4 median-of-halves convention. On a small dataset, a software package using a different quantile algorithm may return a different boundary value, so always name the calculation convention.

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 : 6 of 8 values (75%) are at or below 27, while 5 of 8 are strictly below it.

Figure 1: Switch directions to see how a percentile rank maps to a score—and how a score maps back to the proportion below it.

A percentile rank tells you the proportion of data strictly below a value under this lesson’s convention — it is not the value itself. Saying “I scored at the 90th percentile” means approximately 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 . Whose score is farther above its class mean in standard-deviation units? 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’s score is farther above its class mean in standard-deviation units: 3 SDs compared with Maria’s 2 SDs. Without information about both distribution shapes, this comparison does not prove that James has the higher percentile rank.

Figure 2: Standard deviation becomes the ruler. The same raw-score gap is measured in standard-deviation units without assuming a bell curve or converting the result to a percentile.

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:

: the value equals the mean exactly
: one standard deviation above the mean
: one standard deviation below the mean
: two standard deviations above the mean
For an approximately normal distribution, lies in the outer approximately 0.3% of values combined. Without the normal-shape condition, a z-score still measures standardized distance but does not determine a probability.

Interpreting a Z-Score

A z-score records two pieces of positional information: its sign gives direction from the mean, and its absolute value gives distance from the mean in standard-deviation units. Comparing z-scores compares standardized distance, even when the original datasets use different units.

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.

Figure 3: Each raw score is measured with its own dataset’s standard-deviation ruler, then both positions are compared on one shared z-score scale. A higher z is a higher standardized position; a larger |z| is farther from the mean.

C4 — Distribution Shape and Skewness

Intuitively: imagine a household-income 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 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. A common pattern is 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. A common pattern is mean < median.

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

Evidence rule: The tail in a graph determines the skew direction. The mean–median comparison is a useful clue, not a stand-alone mathematical definition: unusual shapes can have mean above the median without a clean right tail, or vice versa.

Mini-example — reading the tail and checking the summaries:

A histogram of final exam scores has a long left tail, mean , and median . The tail establishes that the distribution is left-skewed; mean < median is consistent with the low scores pulling the mean left.

“Right-skewed” does not mean the peak is on the right. It means the tail extends to the right. In the unimodal examples used here, most values and the peak sit toward the left while the long tail points rightward. Always identify skew from the tail, not from an assumed peak location.

Extreme values pull the mean toward the tail, not away from it. That is why right-skewed distributions commonly have mean > median and left-skewed distributions commonly have mean < median. Use the graph’s tail—not this comparison alone—to establish the skew direction.

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:

Figure 4: Compare the tail, bulk, mean, and median across left-skewed, symmetric, and right-skewed distributions. Select a distribution to focus it and see how the mean moves toward the tail.

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. Before you drag, predict what will happen as you push the skewness slider to the right: will the mean end up to the left or to the right of the median?

Pause and write. Record your answer and reasoning on paper or in your notes before continuing. Even if you are unsure, committing your thinking helps you see what you understood and what you need to revisit.

Figure 5: Adjust skewness and spread to see how a distribution’s shape separates or reunites its mean, median, and mode. The fixed 68–95–99.7 overlay is available only at the exactly symmetric slider setting.

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.

Figure 6: The nested regions show how a nearly normal distribution places about 68%, 95%, and 99.7% of its values within one, two, and three standard deviations of the mean. Arrows show that the two ≈0.15% tails continue beyond the chart edges.

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

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A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so you know where to look back.

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 (56 to 92). Sophie’s score is therefore relatively high, but the Empirical Rule alone does not give its exact percentile; that would require a normal table or software.

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

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. Committing first makes it easier to notice whether you understood the percentile direction and the nearest-rank step:

Show Solution

Using the nearest-rank method: .

The 3rd value in the sorted list is 55.

Therefore . Under nearest rank, 3 of 10 scores (30%) are at or below 55; 2 of 10 are strictly below it.

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:

Figure 7: Most house prices cluster at the lower end while a few expensive homes extend the right tail; use that 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 and write a corrected replacement conclusion.

Show Solution

Error 1 — Misidentifying where the peak is in this right-skewed distribution. In this unimodal histogram, the long tail extends to the right, while 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 this right-skewed housing distribution, the extreme high values in the right tail pull the mean upward. We would therefore expect mean > median, not the student’s reversed relationship.

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

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A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so you know where to look back.

Problem 1 — Relative Position and Z-Score Computation (C2 + C3)

Choose whether to work through the calculation step by step or solve the whole problem yourself.

Problem 2 — Interpreting Percentile Rank and Score (C1)

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

Which statement correctly interprets these percentile values?

Problem 3 — Application of the Empirical Rule (C5)

Choose whether to identify the standard-deviation region step by step or solve the whole problem yourself.

Problem 4 — Shape Classification and Measure Selection (C4)

A researcher collects data on annual salaries at a tech firm. The histogram has a long right tail; the mean salary is $112,000 and the median salary is $84,000.

Which description and reporting choice best match the evidence?

Problem 5 — Z-Score vs. Percentile Rank (C2 + C3 + C5)

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 option correctly evaluates the teacher's statement?

Section 6: Independent Practice

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A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so you know where to look back.

Problem 1 — Z-Score Generator (C2 + C3)

Problem 2 — Percentile Generator (C1)

Problem 3 — Empirical Rule (C3 + C4 + C5)

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 (C2 + C3)

Problem 5 — Multi-Step Synthesis (C2 + C3 + C5)

Mixed Review — Retrieval from Earlier Lessons

These problems draw on concepts from L-DS-3 and L-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 (L-DS-3)

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

Section 7: Mastery Check

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A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so you know where to look back.

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

Question 1 — Feynman Test

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—and what can it not tell you without knowing the distribution’s shape?

Compare with a model answer

“Below average” gives only direction. A z-score of −1.5 gives direction and magnitude: the value is 1.5 standard deviations below the mean, so its distance is expressed relative to the distribution’s spread and can be compared with z-scores from other datasets. The z-score alone does not give an exact percentile or probability. Those claims require information about the distribution’s shape and, for an exact normal percentile, a normal table or software.

Question 2 — Novel Application

Two scholarship applicants took different standardized assessments. Decide which result is farther above its own comparison-group mean in standard-deviation units and which probability statement, if any, the supplied distribution evidence supports. Choose your method rather than relying on the raw scores.

Question 3 — Error Analysis

Self-Assessment

How confident are you with the material in this lesson?

Still unsureFully confident

Section 8: Boss Fight

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A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so you know where to look back.

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. Scores are approximately normally distributed with and . Three records have been flagged for a routine data-quality review: Student 1: 89, Student 2: 45, Student 3: 71.

Complete the full review and recommendation on paper or in your notes before entering the numerical checkpoint. Choose an appropriate comparison method, compute the needed positions, compare all three records, and state what the statistics can—and cannot—justify. The optional step review then lets you compare the complete reasoning.

Metacognitive reflection: What does this analysis reveal about relying on raw scores alone? Why must an unusual z-score trigger a check of the record or process rather than an automatic grade change?

Compare with the model reflection

Raw scores are not comparable until they are placed relative to their own distribution. Z-scores show that Student 2 is slightly farther from the mean than Student 1, while Student 3 is close to average. Statistical unusualness is only a flag to verify scoring, data entry, or circumstances under an established review policy; it is not evidence that a grade is wrong and must not determine treatment automatically.

🏗️ 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.

Complete the full operations recommendation on paper or in your notes before entering the numerical checkpoint. Decide which method is justified for each line, compare the specification limits with the supplied summaries, estimate what the evidence permits, and identify what the summaries cannot establish. The optional step review then lets you compare the complete reasoning.

Metacognitive reflection: What could go wrong operationally if a quality manager applied a normal-distribution benchmark to Line B without checking its shape? What additional evidence would the manager need before making an operational decision?

Compare with the model reflection

A normal benchmark would distribute expected failures symmetrically and could seriously understate Line B’s long upper tail. That could hide excessive-thickness risk and produce misleading quality reports. The manager needs the raw measurements or interval counts, a well-labelled histogram, and direct counts outside each specification limit rather than a normal approximation.

Section 9: Challenge Problems

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A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so you know where to look back.

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 increasing 93, while it remains 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 . Hold fixed and vary while preserving . Show when the mean must be greater than the median.

Show Solution

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

Because , every order statistic that defines lies among the fixed values . Therefore remains fixed while varies.

Now solve the inequality:

Among values of that preserve the sorted order, the mean is therefore greater than the median exactly when . This proves the claim for this constructed family; it does not make mean > median a definition of every right-skewed distribution.

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 these upper extremes pull it toward the long tail. In this example the mean therefore lies above the median; more generally, the graph’s tail is the primary evidence of skew direction.

D. Generalize with a new dataset

The generated problem gives the first four values of a five-value sorted dataset. Determine exactly how large the fifth value must be for the mean to exceed the unchanged median, then explain what the threshold does—and does not—prove.

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

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A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so you know where to look back.

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 uses the same method. Example 2 covers percentiles (C1), while Example 1 covers z-score computation and interpretation (C2 + C3). The solutions page shows the reasoning behind every step, not just the final answer.

Quick-Reference Formulas

Z-Score (Standardization):

Z-Score Range	Empirical Rule (Normal Distributions)
Between -1 and 1	Contains of data
Between -2 and 2	Contains of data
Between -3 and 3	Contains of data

Term	Meaning
Percentile ()	Nearest rank ; take the value at rank
Quartile positions	marks 25%, is the median, marks 75%; compute them with the L-DS-4 convention
Deciles	, , …,