A student scores 78 on an exam with a class mean of 70 and a standard deviation of 8. What fraction of the class scored lower? You don’t need to survey every student. With one formula and a single table, the normal distribution gives you the answer in under a minute — and by the end of this lesson, so will you.
That formula works because of a pattern that 18th-century astronomers stumbled upon: every time they recorded many independent measurements of the same star and plotted the errors, the result was a perfect bell-shaped curve. Abraham de Moivre described this curve mathematically in 1733; Carl Friedrich Gauss later showed it was the inevitable shape of measurement error — which is why it is also called the Gaussian distribution. When many small, independent sources of variation combine, their total always tends toward this same bell shape.
Today, the normal distribution describes heights, exam scores, manufacturing tolerances, blood pressure readings, and the distribution of sample means (which you’ll see in the Inference module). Learning to work with the normal distribution is not optional in statistics — it is the core skill.
After this lesson, you will be able to:
By the end of this lesson, you will be able to:
Describe the properties of the normal distribution and apply the Empirical Rule (68–95–99.7).
Use the z-table to find , , and for the standard normal.
Standardize a normal variable using and find probabilities.
Solve inverse normal problems: find given a cumulative probability.
Recognize when the normal model is and is not appropriate.
Section 2: Prerequisites
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What you need coming in — and why it matters today:
Random variable notation (PR-4): The notation reads “X is distributed as.” Today we write . You also need the CDF interpretation — as cumulative area — which is how the z-table is organized.
Mean and standard deviation (PR-4, DS-5): is the center; measures spread as average distance from the mean. In PR-5 you computed and for a binomial — the normal uses the same parameters, just with a continuous bell curve instead of discrete bars.
Empirical Rule (DS-5): About 68%, 95%, and 99.7% of values fall within 1, 2, and 3 standard deviations of the mean for approximately normal data. Today we derive this from the z-table directly.
Z-score (DS-5): tells you how many standard deviations lies above or below the mean. In DS-5 you used z-scores for comparison. Today they become the key to computing probabilities.
Quick check — can you recall these?
If and the distribution is roughly symmetric about , about 68% of values fall within which range of ?
Success Factor:
What changes in this lesson: In DS-5 you used z-scores to compare values across different distributions. Here the z-score becomes the passport to a single probability table — the z-table — that works for every normal distribution ever. One table, infinite distributions. The key is the standardization formula; everything else follows mechanically from it. PR-7 will use everything you learn today to approximate binomial probabilities, and INF-1 will use the same z-table to reason about sample means.
Retrieval Warm-up — from earlier lessons
Heights of adult women in a city are approximately normally distributed with mean 163 cm and standard deviation 7 cm. A woman is 177 cm tall. Compute her z-score.
A binomial random variable has trials and success probability . What are the mean and standard deviation?
Section 3: Core Concepts
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How this section is organized: Five concepts build in order. Read C1–C2 for the shape and table structure, then C3 for the conversion formula that links any normal distribution to the table. C4 reverses the process (finding values from probabilities), and C5 teaches when not to use the normal model.
C1: Shape and Empirical Rule — what the normal curve looks like and the 68–95–99.7 pattern
C2: The standard normal Z and the z-table — how to read three types of probabilities
C3: Standardizing any normal variable — the bridge formula
C4: Inverse normal — finding a value from a probability
C5: Recognizing when the normal model fits
C0 — Areas as Proportions of Populations
The normal distribution is often introduced as a mathematical curve. But it is more intuitive to think of it as a description of a population.
Suppose the heights of adult women in a large city follow a normal distribution with mean cm and standard deviation cm. Imagine measuring every one of the city’s 100,000 adult women.
Range
Approx. count
Proportion
Between 156 cm and 170 cm (within 1σ)
68,270 women
68.3%
Between 149 cm and 177 cm (within 2σ)
95,450 women
95.5%
Between 142 cm and 184 cm (within 3σ)
99,730 women
99.7%
Shorter than 149 cm (more than 2σ below)
2,275 women
2.3%
Taller than 184 cm (more than 3σ above)
135 women
0.13%
This is the key insight: probabilities are areas, and areas are proportions of populations. When we say , we mean that if we drew a random woman from this population, she has a 2.3% chance of being shorter than 149 cm — equivalently, 2.3% of the population is below 149 cm.
The z-table you will use in this lesson is simply a lookup table for these proportions. Every area under the standard normal curve corresponds to a proportion of a standard normal population — and, after standardization, to a proportion of any normal population.
C1 — Properties of the Normal Distribution
Connect to PR-4: a continuous random variable like height or test score cannot use a PMF — probabilities correspond to areas under a curve, not point masses. The normal distribution provides the specific curve.
Normal Distribution
A random variable follows a normal distribution, written , if its probability density function has a symmetric bell shape centered at with spread controlled by .
Key properties:
Symmetric about — the left and right halves are mirror images.
Mean = Median = Mode = (all three coincide at the center).
Tails are asymptotic — they approach but never touch the horizontal axis.
Total area under the curve = 1 (as required for any probability distribution).
Fully described by two parameters: (location) and (spread).
Empirical Rule (68–95–99.7): For any :
— about 68% of values within 1σ
— about 95% within 2σ
— about 99.7% within 3σ
Example: Heights of adult women follow approximately (mean 162 cm, σ = 6 cm). About 95% of women are between 150 cm and 174 cm (162 ± 12 = 162 ± 2×6).
The second parameter in is the variance, not the standard deviation . Writing when you mean (with ) is a common error. Always check: is the number after the comma or ? Convention throughout this course is .
C2 — Standard Normal Distribution and the Z-Table
Rather than computing a different area formula for every possible and , we use a single reference distribution.
We write to mean “Z follows the standard normal distribution” — the special case where and .
Standard Normal Distribution
has mean 0 and standard deviation 1.
The z-table gives the left-tail cumulative area: for any value .
From this single entry, we derive three probability types:
Left-tail: — read directly from the table.
Right-tail: — complement rule.
Between two values: — subtract the smaller left-tail from the larger.
The z-table for this course is reproduced below. Rows give the ones and tenths digits of ; columns give the hundredths digit.
The z-table gives — a left-tail area. If you look up and get 0.9332, that means 93.32% of the distribution is below 1.50. The right-tail probability is . Using 0.9332 directly as a right-tail probability is the single most common z-table error.
When computing , you must subtract the two left-tail areas, not add them. . Adding them () would give a probability of 1 — clearly wrong. The intuition: you want the area in the middle strip, which is the big left-area minus the small left-area that you don’t want.
Try it interactively. Use the visualization below to shade normal curve areas. Enter any z-score to see the corresponding probability, or switch to a general normal distribution to practice standardization.
Figure 1: Interactive normal curve. Switch between Standard N(0,1) and General N(μ, σ²) using the tabs. In General mode, enter μ, σ, and x — the z-score and probability update live, showing how standardization works. Use the Shade buttons to explore left-tail, right-tail, and between-values probabilities.
C3 — Standardizing a Normal Variable
Any can be converted to the standard normal scale using the standardization formula.
Standardization (Z-Score Transformation)
For , compute:
Then , and you can use the z-table exactly as in C2.
Interpretation: tells you how many standard deviations is above (positive) or below (negative) the mean.
Mini-example: IQ scores follow (mean 100, ). Find .
Standardize: .
Look up .
About 88.5% of people score below 118.
You must standardize before using the z-table. The z-table works only for . Looking up directly in a table designed for produces a nonsense answer. Every non-standard normal problem requires the conversion step first.
Figure 2: The standardization transformation. Set μ and σ, then drag x. The orange marker on the X curve and the blue marker on the Z curve sit at the same horizontal position — the same fraction of the bell — confirming that z = (x − μ) / σ is an axis relabelling that leaves the shaded probability unchanged.
C4 — Finding Values Given a Probability (Inverse Normal)
Sometimes the question is reversed: given a probability, find the corresponding value. This is the percentile or quantile problem.
Inverse Normal Lookup
Given a cumulative probability , find such that :
Look up in the body of the z-table to find (the z-score corresponding to the th quantile).
Unstandardize:
We use (read “z-star”) to distinguish this critical value from an ordinary observed z-score.
Mini-example: Exam scores follow (). Find the 90th percentile.
Step 1: Find such that . Scanning the z-table body, the closest entry is 0.8997 at . So .
Step 2: Unstandardize: .
The 90th percentile score is approximately 80.24.
In the inverse problem, search the table body (the probability values) to find , then use the row/column labels to read . Students sometimes look up the probability in the margins (the z-values) instead of the body — this produces a random number unrelated to the problem. Remember: body → → unstandardize.
C5 — Recognizing Non-Normal Shapes
The normal model is powerful, but it is not universal. Applying it blindly to non-normal data produces incorrect probabilities.
When to Use (and Not Use) the Normal Model
Normal model is appropriate when:
Data are roughly symmetric and unimodal
Data are continuous (or nearly so)
A histogram shows a bell shape
The Empirical Rule holds approximately
Normal model is NOT appropriate when:
Data are strongly right-skewed or left-skewed (use skewed distributions)
Data are discrete counts (use binomial, Poisson, etc.)
Data have heavy tails or multiple modes
Data are bounded between 0 and 1 (proportions — use binomial or beta)
Looking ahead: PR-7 covers the case where discrete binomial data are approximately normal when and . INF-1 covers how the Central Limit Theorem makes sample means approximately normal regardless of the original population’s shape — which is why the normal distribution appears throughout all of inferential statistics.
Income distributions are a classic example where the normal model fails badly: they are strongly right-skewed with a long upper tail. Using the normal model for income would dramatically underestimate the probability of very high incomes. Similarly, the number of car accidents per day is a count — binomial or Poisson, not normal.
Section 4: Worked Examples
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Example 1 — Fully Worked: Three Probability Types for Z
Find (a) , (b) , and (c) .
(a) — left-tail, read directly.
I notice this is a left-tail query, so the z-table gives the answer immediately. Row 1.4, column 0.05: .
(b) — right-tail, use complement.
The table gives left-tail areas. I choose to use the complement: .
Look up : row , column 0.02 → .
.
Sanity check: Since , the right-tail area should be greater than 0.5. ✓ (0.7324 > 0.5)
(c) — between two values, subtract.
I need the area in the strip between and . The strategy is: big left-area minus small left-area.
(row 0.8, column 0.05)
(row , column 0.00)
.
Example 2 — Prediction Checkpoint: Standardize and Find Probability
IQ scores are modeled as (). Find .
Setup:
The question asks for a right-tail probability. I must standardize first: .
Pause here. Before reading the solution:
Is the probability you expect greater or less than 0.50? Why?
Will you use the left-tail or complement approach?
Write down your prediction, then continue.
Show Solution
Since , the value 118 is above the mean. So the right-tail probability is less than 0.50.
(from the z-table in Section 3).
.
Therefore — about 11.5% of people score above 118.
Sanity check: Since 118 is 1.2 standard deviations above the mean, the Empirical Rule says about 84% of scores are below . Being below 1.2σ from the right puts us between 84% and 97.5%, so a right-tail of about 11.5% is consistent. ✓
Example 3 — Minimally Scaffolded: Inverse Normal (90th Percentile)
Exam scores follow (). Find the 90th percentile.
Hint: For the inverse problem, search the z-table body for the target probability, read off , then unstandardize.
Show Solution
Step 1: The 90th percentile means .
Step 2: Search the z-table body for 0.9000 (or the closest entry). The entry 0.8997 appears at ; the entry 0.9015 appears at . The closer value is .
Step 3: Unstandardize: .
The 90th percentile is approximately 80.24. A score of about 80 separates the top 10% from the rest of the class.
Example 4 — Error Analysis: Empirical Rule Application
Read the following analysis carefully. It contains an error.
Analysis to examine:
Heights of adult males follow approximately ( cm). A student is asked: “What fraction of men are between 158 cm and 182 cm?”
The student writes: “158 and 182 are each standard deviations from the mean, so by the Empirical Rule, 95% of men fall in this range.”
A second student says: “The interval is 158 to 182, which is , so 95.45% of men are in this range.”
Show Full Analysis
Both students identified the correct interval: and , so the interval is indeed .
The first student says “95%” — this is the rounded Empirical Rule value and is acceptable for quick reasoning.
The second student says “95.45%” — this is the more precise value from the z-table: .
For exam purposes, either “95%” or “95.45%” (equivalently, ) will be accepted. When you have access to the z-table (as in all exercises in this lesson), use the precise value.
Lesson: The Empirical Rule gives quick estimates; the z-table gives exact probabilities. Know when to use each.
Section 5: Guided Practice
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Note: The z-table in Section 3 is scrollable and reusable. Keep it open in a separate browser tab or scroll back to Section 3 as needed while working through these problems.
Problem 1 — Standard Normal Probabilities (C2)
Problem 2 — Standardize and Find Probability (C3)
Problem 3 — Inverse Normal (C4)
Problem 4 — Classify the Distribution (C5)
For each description below, decide whether a normal model is approximately appropriate or not. Give a one-sentence justification.
The ages at which people get their first driver’s licence in Quebec, recorded for all new licence holders in a given year.
The blood pressure readings (in mmHg) of a large sample of healthy adults, selected randomly from a national health survey.
The number of customer complaints received per day at a call centre, recorded over 200 working days.
The weights of apples harvested from a single large orchard, measured in grams for a random sample of 500 apples.
The annual household incomes of all residents in a city.
Section 6: Independent Practice
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Reminder: The z-table in Section 3 is scrollable and reusable. Open Section 3 in a separate tab or scroll back as needed.
Problem 1 — Standard Normal Probability (C2)
Problem 2 — Standardize and Find One-Tail Probability (C3)
Problem 3 — Between Two Values (C3)
Problem 4 — Inverse Normal (C4)
Problem 5 — Empirical Rule Application (C1)
The lengths of trout in a lake follow (mean 42 cm, cm).
(a) Using the Empirical Rule, what percentage of trout are between 32 cm and 52 cm?
(b) What percentage of trout are longer than 47 cm?
Show Solution
(a) and . By the Empirical Rule, approximately 95% of trout are between 32 cm and 52 cm (more precisely, 95.45%).
(b). By the Empirical Rule, 68% of trout are within 1σ of the mean. By symmetry, 16% are above cm. (More precisely: .)
The daily high temperatures in Montreal in July follow (mean 27°C, ).
(a) What percentage of July days have temperatures between 19°C and 35°C?
(b) What percentage of days exceed 31°C?
Show Solution
(a) and . By the Empirical Rule, approximately 95% of July days have highs between 19°C and 35°C.
(b). By symmetry of the Empirical Rule, about 16% of days exceed 31°C. (More precisely: .)
The weights of newborn babies in a hospital follow (mean 3400 g, g).
(a) Using the Empirical Rule, what percentage of babies weigh between 2800 g and 4000 g?
(b) A baby weighing less than 2500 g is classified as very low birth weight. Approximately what percentage fall in this category?
Show Solution
(a) and . By the Empirical Rule, approximately 95% of babies weigh between 2800 g and 4000 g.
(b). By the Empirical Rule, 99.7% of babies are within 3σ. By symmetry, 0.15% are below g. (More precisely: , so about 0.13%.)
The reaction times of trained athletes on a specific reflex test follow (mean 180 ms, ms).
(a) Using the Empirical Rule, what percentage of athletes have reaction times between 160 ms and 200 ms?
(b) A reaction time under 140 ms is considered elite. Using the Empirical Rule, what percentage of athletes qualify as elite?
Show Solution
(a) and . By the Empirical Rule, approximately 68% of athletes have reaction times in this range.
(b). By the Empirical Rule, 95% are within 2σ. By symmetry, only 2.5% fall below ms. (More precisely: , so about 2.28%.)
The scores on a standardized math placement test follow (mean 500, ).
(a) A university accepts students who score in the top 16% on this test. Using the Empirical Rule, what is the minimum score for admission?
(b) What percentage of students score between 200 and 800?
Show Solution
(a) The Empirical Rule says 68% of values fall within , leaving 32% outside; by symmetry, half of that — about 16% — falls above . The minimum admission score is 600.
Precision note: The Empirical Rule’s “16%” is an approximation. The z-table gives the exact figure: , so precisely the top 15.87% score above 600. These two values are consistent — 16% and 15.87% differ by less than one student in eight — and since the problem specifies using the Empirical Rule, 600 is the correct answer regardless of which precision you apply. There is no contradiction: the Empirical Rule rounds , and the z-table confirms it.
(b) and . By the Empirical Rule, approximately 99.7% of students score between 200 and 800.
Problem 6 — Find the Error
A student is asked to find .
Student’s work:
Look up in the z-table: .
What is the student’s error?
Full Correction
The z-table always gives — a left-tail area. For a right-tail probability, the complement rule must be applied:
Sanity check: Since , the right-tail probability must be less than 0.5. The value ✓. The student’s answer of would mean 91% of the distribution is above — clearly wrong for a positive z-score.
Scores on a placement test follow (). A student is asked to find .
Student’s work:
Look up in the z-table: .
“A score above 118 is virtually impossible.”
What is the student’s error?
Full Correction
The z-table is built for the standard normal only. A raw value cannot be looked up directly — standardize first.
Step 1 — Standardize:
Step 2 — Look up:
Step 3 — Complement:
About 11.5% of test-takers score above 118 — not “virtually impossible.” The student’s answer of came from treating the raw score as if it were a z-score, which placed it impossibly far in the tail.
A student is asked to find .
Student’s work:
“The entire distribution lies between and .”
What is the student’s error?
Full Correction
The area between two z-values equals the larger left-tail area minus the smaller — never the sum:
Adding the two left-tail areas double-counts everything to the left of , inflating the result to — an impossible probability for a proper subinterval of the distribution.
Sanity check: The Empirical Rule says about 95% of values fall within . An interval of should capture less than 95.45% but more than 68.27%. The value is consistent with this. ✓
Problem 7 — Multi-Step Synthesis: Quality Control at a Bottling Plant
A bottling machine fills bottles with (mean 500 mL, mL). Bottles are rejected if they contain less than 490 mL or more than 514 mL.
(a) Find the probability that a randomly selected bottle is rejected.
Show Solution (a)
A bottle is rejected if or .
Standardize the lower bound: .
Standardize the upper bound: .
About 5.74% of bottles are rejected.
(b) The plant manager wants to reduce the rejection rate. She proposes re-centering the machine to mL (keeping mL) so bottles are less likely to be underfilled. Will this reduce the total rejection rate? Show your work.
Show Solution (b)
With , :
Lower bound: . .
Upper bound: . .
The new rejection rate is about 4.56%, down from 5.74%. Yes, re-centering reduces the total rejection rate — and here is the precise reason: the midpoint of the acceptance interval is . Moving the machine mean to the midpoint of the acceptance interval places it equidistant from both bounds ( in each direction), which minimises the total rejection rate for any fixed-width acceptance interval. With the original , the bounds were asymmetric: the lower bound was only away while the upper bound was away — the machine was losing probability mass to the nearer lower tail. Re-centering at 502 balances both tails at , giving the lowest possible rejection rate for this machine and these bounds.
(c) The plant manager wants to find the fill level below which only 1% of bottles fall (the 1st percentile). Find this value with , .
Show Solution (c)
Find such that .
Scanning the z-table body: the closest entry is 0.0099 at .
Unstandardize: mL.
The 1st percentile fill level is approximately 486.02 mL — only 1% of bottles are filled below this amount.
Mixed Review — Retrieval from Earlier Lessons
These problems draw on concepts from earlier in the course. Attempting them without re-reading prior lessons is the point — retrieval practice strengthens long-term memory more than re-reading.
Review Problem 1 — Empirical Rule applied to data
Test scores for a standardized exam are approximately normally distributed with mean 500 and standard deviation 100.
(a) According to the Empirical Rule, between what two scores do the middle 99.7% of test-takers fall?
(b) A score of 350 corresponds to what z-score? Is this score unusually low, or within the typical range?
(c) Approximately what percentage of students scored above 600?
Show Solution
(a) The Empirical Rule: 99.7% of values fall within ±3σ.
The middle 99.7% fall between 200 and 800.
(b).
A z-score of −1.5 lies within ±2σ (the middle 95%), so this score is somewhat below average but not unusually low. The Empirical Rule places roughly 95% of scores between 300 and 700, and 350 falls inside this range.
(c) 600 = 500 + 1(100), so 600 is exactly 1σ above the mean. The Empirical Rule says about 68% of values fall within ±1σ (400–600), leaving 32% outside. Since the distribution is symmetric, about 16% of students scored above 600.
Review Problem 2 — Binomial cumulative probability
A multiple-choice quiz has 10 questions, each with 4 options. A student who did not study guesses randomly on every question. Let = number of correct answers.
(a) Verify that follows a binomial distribution by checking the BINS conditions.
(b) What are the mean and standard deviation of ?
(c) Compute — the probability of getting every question wrong.
Show Solution
(a) BINS check:
Binary: each question is correct or incorrect. ✓
Independent: guessing on one question does not affect another. ✓
N fixed: questions. ✓
Same : each question has probability of being correct by guessing. ✓
All conditions hold, so .
(b)
(c)
There is about a 5.6% chance of getting every question wrong when guessing randomly.
Section 7: Mastery Check
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Question 1 — Feynman Test
In your own words, explain why using only the symmetry of the normal distribution. Do not use the z-table or specific numbers — just the concept of symmetry. Aim for 200–500 characters.
0 / 500
Model Answer
The standard normal curve is perfectly symmetric about zero. The left tail below is the mirror image of the right tail above . Because the total area under the curve is 1 and the two halves are identical mirror images, the area in the left tail at any value must equal the area in the right tail at the corresponding value . Therefore .
More formally: by symmetry, for any . This is one of the most important symmetry properties of the standard normal.
Question 2 — Apply: Commute Times
A city planner models daily commute times as (mean 35 min, min). A new transit policy is evaluated for commuters who spend between 27 and 51 minutes commuting.
Part A: What fraction of commuters have times in this range?
Part B: Find the 95th percentile commute time.
Show Full Solution
Part A: Standardize both bounds:
About 81.85% of commuters fall in this range. (The interval is asymmetric — the upper bound is 2σ above the mean, the lower bound is only 1σ below.)
Part B: For the 95th percentile, find such that .
From the z-table: the body entry 0.9500 falls between (entry 0.9495) and (entry 0.9505). Using (midpoint):
minutes.
The 95th percentile commute time is approximately 48.16 minutes.
Question 3 — Error Analysis
Flawed solution:
A student is asked to find for ().
The student writes: “Look up in the z-table. , so . There is essentially no probability above 112.”
Identify the error and compute the correct answer.
Show Full Analysis
The error: The student used the raw value (112) directly as a z-score. The z-table works only for . For any other normal distribution, you must standardize first.
Correct solution:
(from the z-table)
About 21.19% of values exceed 112 — far from zero. The student’s answer of approximately 0 was wrong because a z-score of 0.80 is only 0.8 standard deviations above the mean, a very ordinary result.
Self-Assessment
How confident do you feel about the normal distribution and z-table lookups?
Still confusedReady for the Boss Fight
Section 8: Boss Fight
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Choose your path. Both require standardization, table lookup, and inverse normal reasoning.
⚙️ Path A: The Engineer
A manufacturing process produces bolts whose diameters are normally distributed. You must find the rejection rate and set a threshold that rejects exactly the most extreme bolts.
📊 Path B: The Analyst
A professor wants to set the grade cutoff for an A. You must find the minimum score, then determine what spread would be needed if the cutoff must be at least 80.
⚙️ Path A: The Engineer
A manufacturing process produces bolts whose diameters follow (mean 10.0 mm, mm). Bolts are rejected if their diameter falls outside the interval mm.
Task 1. Find the probability that a randomly selected bolt is accepted (diameter falls within ). Show all standardization steps.
Show Guidance for Task 1
Standardize both bounds:
About 86.64% of bolts are accepted.
Task 2. What is the rejection rate? How many bolts out of 5,000 produced per day would you expect to be rejected?
Show Guidance for Task 2
Rejection rate = , so about 13.36% of bolts are rejected.
Expected daily rejections: bolts.
Task 3. The engineering team wants to adjust the upper rejection threshold (keeping the lower threshold at 9.7 mm) so that only the top 5% of bolt diameters are rejected on the upper end. Find the new upper threshold diameter.
Show Guidance for Task 3
We want , i.e., .
Find such that . From the z-table: .
Unstandardize: mm.
The new upper threshold is approximately 10.33 mm.
Task 4. With the original bounds and , the process engineer proposes reducing to 0.1 mm (by upgrading equipment). How does this change the acceptance rate? Is the investment worthwhile if each 0.1 mm reduction in costs $50,000?
Show Guidance for Task 4
With :
Acceptance rate improves from 86.64% to 99.74%. Rejection rate drops from 13.36% to 0.26%.
Out of 5,000 bolts/day: rejections drop from 668 to only 13 bolts/day — a reduction of 655 bolts/day.
The cost-benefit analysis depends on the value of each rejected bolt and the production cost. If each rejected bolt costs more than 76 in waste, the investment pays off daily. For typical manufacturing, this precision upgrade is often worthwhile.
Reflection: Write two sentences explaining how reducing σ affects the rejection rate, and why the acceptance interval [9.7, 10.3] matters for this calculation.
0 / 500
📊 Path B: The Analyst
A professor wants the top 15% of students to earn an A. Class scores follow (mean 72, ).
Task 1. Find the minimum score required to earn an A. Show the full inverse normal procedure.
Show Guidance for Task 1
We want , i.e., .
Find such that . From the z-table: (entry 0.8508, closest to 0.85).
Unstandardize: .
The minimum A score is approximately 82.4 (or 83 rounded up to a whole number).
Task 2. A student scores 85. What percentile does this correspond to? Interpret your answer in plain language.
Show Guidance for Task 2
Standardize: .
.
The student is at approximately the 90th percentile — scoring higher than about 90.3% of the class.
Plain language: “A score of 85 puts this student in the top 10% of the class, comfortably within A territory.”
Task 3. The professor teaches a second section where the mean stays at 72 but the spread is unknown. She wants the minimum A score to be at least 80. What is the maximum value of that achieves this?
Show Guidance for Task 3
We need the 85th percentile to be at least 80:
The standard deviation must be at most approximately 7.69 for the A cutoff to be 80 or higher. If scores are more spread out (), the top 15% threshold rises above 80.
Task 4. The professor considers a different grading policy: award an A to any student more than 1 standard deviation above the mean (i.e., ). Under , what percentage of students earn an A under this policy? Is this more or less than the “top 15%” policy?
Show Guidance for Task 4
.
About 15.87% of students would earn an A — slightly more than the 15% policy.
This makes sense: the “top 15%” policy requires , which is slightly above 1.00. Setting the cutoff at exactly corresponds to the 84.13th percentile, giving 15.87% of students an A versus exactly 15% under the stricter policy.
Reflection: Explain in two sentences why reducing σ (making the class more homogeneous) would raise the minimum A score when the top 15% rule is used.
0 / 500
Section 9: Challenge Problems
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Ready for more? These go beyond the lesson objectives.
Challenge 1 — Symmetry Proof
Prove, using only the complement rule and the symmetry of , that for any .
Then verify numerically for : look up both and in the z-table and confirm the relationship holds.
Show Solution
Proof:
The standard normal is symmetric about 0: for all .
By this symmetry, the area to the left of equals the area to the right of :
Now apply the complement rule — the total area is 1:
(For a continuous distribution, , so .)
Combining: .
Numerical verification for :
From the z-table: .
.
Also from the z-table: . ✓
The values match. This also confirms the familiar fact that the critical values each cut off 2.5% in their respective tails — the basis for the 95% confidence interval (covered in INF-2).
Challenge 2 — Two-Equation Inverse Normal
A symmetric distribution has and . Assuming , find and .
Hint: Translate each probability statement into a z-score equation using the inverse normal. You will get a system of two equations in two unknowns ( and ).
Show Solution
Setting up the equations:
From : This means .
From the z-table, , so .
The standardization gives: → → Equation 1:
From : From the z-table, , so .
→ → Equation 2:
Solving the system:
Add Equation 1 and Equation 2:
→
Substitute into Equation 1: → →
Answer:, . The distribution is .
Check: By symmetry, both tail probabilities should be equal (both 10%) when the distribution is symmetric about , and indeed . ✓
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Quick-Reference Formula Card
Formula
Purpose
When to use
Standardize to
Before any z-table lookup for a non-standard normal
Right-tail from left-tail
When the table gives left-tail but you need right-tail
Between two values
Subtract smaller left-area from larger
Unstandardize (inverse normal)
After finding from a given cumulative probability
Z-Table Reading Guide
The z-table in Section 3 gives — always a left-tail area.
Row = ones and tenths digit of (e.g., row 1.4 for ).
Column = hundredths digit (e.g., column 0.05 for ).
For inverse problems: search the table body for the target probability, then read the row and column labels to get .
Symmetry shortcut: — you can always convert negative z-values using the positive half of the table.
Common Errors to Avoid
P1 (Right-tail confusion): The table gives . For , use the complement: .
P2 (Forgetting to standardize): Always compute before looking up the table. Never look up raw values.
P3 (Adding instead of subtracting): For , subtract — never add — the two left-tail areas.