🔔 Normal Distribution
The bell curve underlying almost every continuous measurement in engineering — component dimensions, material strengths, sensor noise. Standardising to z-scores lets one single table answer any probability question.
2Topics
8Worked Examples
4Eng. Cases
12Practice Qs
📐z-scoresStandardising any normal variable
✅Probabilities & QCReading the standard normal table
⚙️ApplicationsReal engineering
🎮Quick-Fire GameTest your speed
📝PracticeGraded questions
🎯What you will learn
Convert any normally distributed value to a z-score; use the standard normal distribution to find probabilities; and apply the normal distribution to quality control problems involving tolerances and specification limits.
💡 Every normal distribution, whatever its mean and SD, becomes the exact same standard bell curve once converted to z-scores — that's why one table (or one formula) answers every question.
🧭How to use this resource
1. Read the rules banner at the top of each tab
2. Experiment with the interactive tool
3. Check the key vocabulary panel
4. Study the visual diagram, then the worked examples
📐
The Standard Normal & z-scores
Converting any value into "how many standard deviations from the mean" — the universal translator for normal distributions.
🔧z-score CalculatorEnter x, μ and σ
Enter x, μ and σ, then press Calculate to find the z-score…
📖 Key Vocabulary
z-score (standard score) — the number of standard deviations a value lies from the mean.
Standard normal distribution — the normal distribution with mean 0 and SD 1, used as the universal reference.
Φ(z) — standard notation for the cumulative probability up to z, i.e. P(Z≤z).
68-95-99.7 rule — roughly 68% of data lies within 1σ, 95% within 2σ, 99.7% within 3σ of the mean.
👁️ Visual aid: the bell curve's symmetry means the same 68-95-99.7 proportions apply to every normal distribution, whatever its mean and SD.
✏️Worked Example 1 — Finding a z-scoreType: Direct calculation
Question A shaft diameter is normally distributed with μ=25.0mm, σ=0.08mm. Find the z-score for x=25.15mm.
Substitute
z = (25.15 − 25.0) / 0.08Answer z = 0.15/0.08 = 1.875 — this shaft is 1.875 standard deviations above the mean.
➖Worked Example 2 — Negative z-scoreType: Below the mean
Question For the same distribution (μ=25.0, σ=0.08), find z for x=24.90mm.
Substitute
z = (24.90 − 25.0) / 0.08Answer z = −0.10/0.08 = −1.25 — a negative z-score simply means the value sits below the mean.
🔄Worked Example 3 — Working backwards from z to xType: Inverse
Question For μ=100, σ=15, find the x-value corresponding to z=2.
Rearrange
x = μ + zσ = 100 + 2(15)Answer x = 130
⚙️Worked Example 4 — Engineering context (comparing two batches)Type: Engineering
Question Batch A: μ=50N, σ=4N; Batch B: μ=200N, σ=15N. A test result of 58N from Batch A and 225N from Batch B — which is more unusual, using z-scores?
z for A
(58−50)/4 = 2.0z for B
(225−200)/15 = 1.667Answer The Batch A result (z=2.0) is more unusual than the Batch B result (z=1.667), even though 58N is numerically far closer to typical values — z-scores let you compare across totally different scales fairly.
🧠 Common slip: comparing raw values directly across two different distributions — always convert to z-scores first, since "unusual" only makes sense relative to each distribution's own mean and spread.
✅
Probabilities & Quality Control
Once you have z, the standard normal table (or an equivalent formula) gives the probability directly.
🔧Normal Probability CalculatorEnter x, μ, σ and choose the probability type
Enter x, μ, σ and choose P(X<x) or P(X>x), then press Calculate…
📖 Key Vocabulary
Standard normal table — a table of Φ(z) values, the cumulative probability up to z.
Tolerance limits — specification boundaries (e.g. ±0.05mm) whose "outside" probability can be found via z-scores.
Process capability — how well a manufacturing process's natural spread fits within tolerance limits.
Continuity correction — a small adjustment (±0.5) used when approximating a discrete distribution with the normal curve.
👁️ Visual aid: a normal probability is always the area under the curve — the shaded region here represents everything above the chosen value x.
✏️Worked Example 1 — Finding P(X < x)Type: Below a value
Question For μ=25.0, σ=0.08, find P(X < 25.15).
Find z
z = (25.15−25.0)/0.08 = 1.875Look up Φ(1.875)
Φ(1.875) ≈ 0.9696Answer P(X < 25.15) ≈ 0.9696 — about 97% of shafts are below this diameter.
➕Worked Example 2 — Finding P(X > x) using the complementType: Above a value
Question Using the same distribution, find P(X > 25.15).
Apply complement
P(X > 25.15) = 1 − Φ(1.875)Answer = 1 − 0.9696 = 0.0304 — about 3% of shafts exceed this diameter.
↔️Worked Example 3 — Probability between two valuesType: Between two limits
Question For μ=25.0, σ=0.08, find P(24.90 < X < 25.10).
Find both z
z₁=(24.90−25)/0.08=−1.25, z₂=(25.10−25)/0.08=1.25Apply symmetry
P = Φ(1.25) − Φ(−1.25) = Φ(1.25) − [1−Φ(1.25)] = 2Φ(1.25)−1Answer Φ(1.25)≈0.8944, so P ≈ 2(0.8944)−1 = 0.7888 — about 79% of shafts fall within this symmetric range.
⚙️Worked Example 4 — Engineering context (rejection rate)Type: Engineering
Question A bolt's tensile strength is normal with μ=450N, σ=20N. Bolts below 410N are rejected. What fraction is rejected?
Find z
z = (410−450)/20 = −2.0Find probability
P(X < 410) = Φ(−2.0) = 1 − Φ(2.0) = 1 − 0.9772Answer P ≈ 0.0228 — about 2.28% of bolts are rejected, information the manufacturer can use to plan scrap rates and, if needed, tighten the process.
🧠 Common slip: forgetting the symmetry relationship Φ(−z) = 1−Φ(z). Many tables only list positive z-values — for negative z, always convert using this identity.
⚙️
Engineering Applications
Four live scenarios — each with its own calculator. Enter your own numbers.
🔩 Tolerance Rejection Rate
Given a process mean and SD, find what fraction of production falls outside specification limits.
z = (limit − μ) / σ
📟 Try it — Rejection Rate Calculator
Enter mean, SD and both spec limits, then press Calculate…
📊 Setting a Specification Limit
Working backwards from a target rejection rate to find where a specification limit should be set.
Limit = μ + zσ (z from the target percentile)
📟 Try it — Limit Setter
Enter mean, SD and z-value, then press Calculate…
📡 Sensor Noise — Probability of Exceeding a Threshold
Random sensor noise (normally distributed around zero) can be checked against a trigger threshold using the same method.
P(noise > threshold) via z-score
📟 Try it — Threshold Probability
Enter the noise SD (mean=0 assumed) and threshold, then press Calculate…
🏗️ Load Rating — Probability of Exceeding Design Load
Structural loads vary randomly; the normal distribution estimates how often a design load might be exceeded.
z = (design load − mean load) / SD
📟 Try it — Load Exceedance Calculator
Enter mean load, SD and design load, then press Calculate…
🎮
Quick-Fire Normal Distribution Game
12 questions on z-scores and probabilities. How many can you get?
⚡ Normal Distribution Challenge
Score: 0Question: 1/12Streak: 0🔥
Press Start to begin!
📝
Practice Questions
Graded from warm-up to challenge. Use Φ(1.0)=0.8413, Φ(1.5)=0.9332, Φ(2.0)=0.9772, Φ(0.5)=0.6915 where needed.
Question 1 🥉2 marks
For μ=50, σ=4, find z for x=58.
(58−50)/4 = 2.0
Question 2 🥉2 marks
For μ=50, σ=4, find z for x=44.
(44−50)/4 = −1.5
Question 3 🥉2 marks
For μ=100, σ=10, find x when z=1.5.
x=100+1.5(10) = 115
Question 4 🥉2 marks
If Φ(1.0)=0.8413, find P(Z>1.0).
1−0.8413 = 0.1587
Question 5 🥈3 marks
For μ=60, σ=5, find P(X<65) using Φ(1.0)=0.8413.
z=(65−60)/5=1.0
P(X<65)=Φ(1.0)=0.8413
P(X<65)=Φ(1.0)=0.8413
Question 6 🥈4 marks
For μ=60, σ=5, find P(X>67.5) using Φ(1.5)=0.9332.
z=(67.5−60)/5=1.5
P(X>67.5)=1−0.9332
=0.0668
P(X>67.5)=1−0.9332
=0.0668
Question 7 🥈3 marks
For μ=200, σ=10, find P(190<X<210) using Φ(1.0)=0.8413.
z=±1.0
P=2Φ(1.0)−1=2(0.8413)−1
=0.6826
P=2Φ(1.0)−1=2(0.8413)−1
=0.6826
Question 8 🥈4 marks
A machine's output is normal with μ=500g, σ=8g. Find the probability a unit weighs less than 484g using Φ(2.0)=0.9772.
z=(484−500)/8=−2.0
P(X<484)=1−Φ(2.0)=1−0.9772
=0.0228
P(X<484)=1−Φ(2.0)=1−0.9772
=0.0228
Question 9 🥇4 marks
A shaft diameter is normal with μ=30.0mm, σ=0.1mm. Tolerance is 30.0±0.15mm. Using Φ(1.5)=0.9332, find the fraction rejected.
z=±1.5
Fraction within = 2Φ(1.5)−1 = 2(0.9332)−1 = 0.8664
Fraction rejected = 1−0.8664
=0.1336 (13.36%)
Fraction within = 2Φ(1.5)−1 = 2(0.9332)−1 = 0.8664
Fraction rejected = 1−0.8664
=0.1336 (13.36%)
Question 10 🥇5 marks
Bolt strength is normal, μ=450N, σ=20N. Using Φ(2.5)≈0.9938, find the strength value below which only 0.62% of bolts fall.
P(X<x)=0.0062 corresponds to z=−2.5 (since 1−0.9938=0.0062)
x = 450+(−2.5)(20) = 450−50
=400N
x = 450+(−2.5)(20) = 450−50
=400N
Question 11 🥇5 marks
Compare z-scores for x=90 from Distribution A (μ=80,σ=5) and x=300 from Distribution B (μ=250,σ=40). Which is more extreme?
z_A=(90−80)/5=2.0
z_B=(300−250)/40=1.25
Distribution A's value is more extreme (larger |z|) despite B's raw difference (50) being bigger than A's (10)
z_B=(300−250)/40=1.25
Distribution A's value is more extreme (larger |z|) despite B's raw difference (50) being bigger than A's (10)
Question 12 🥇5 marks
A structural load is normal, μ=100kN, σ=8kN. The structure is designed for 120kN. Using Φ(2.5)≈0.9938, find the probability the design load is exceeded, and comment on the safety margin.
z=(120−100)/8=2.5
P(X>120)=1−Φ(2.5)=1−0.9938
=0.0062 (0.62%)
A roughly 1-in-160 chance of exceeding the design load — engineers would compare this against required safety factors for the application.
P(X>120)=1−Φ(2.5)=1−0.9938
=0.0062 (0.62%)
A roughly 1-in-160 chance of exceeding the design load — engineers would compare this against required safety factors for the application.