Probability & Statistics · Varsity Maths

The Normal Distribution

The bell curve, z-scores, the 68-95-99.7 rule, reading the z-table, inverse normal and the Central Limit Theorem, everything in one place.

12Worked Examples
10Quiz Questions
4Interactive Diagrams
Share this page

§ 01What Is the Normal Distribution?

The normal distribution is the most important probability distribution in all of statistics. It appears wherever a quantity is the sum of many small, independent effects: which describes almost everything in nature.

Heights of adults, IQ scores, measurement errors in physics experiments, the weight of manufactured parts, daily stock returns: all follow a distribution that is approximately bell-shaped and symmetric. This is not a coincidence. The Central Limit Theorem (covered in §08) guarantees that sums and averages of independent random variables converge to a normal distribution, regardless of the underlying distribution.

A normal distribution is fully specified by just two numbers: its mean μ (mu), which locates the centre of the bell, and its standard deviation σ (sigma), which controls the spread. We write X ~ N(μ, σ²). Note: the second parameter is the variance σ², not the standard deviation σ. This is a common source of confusion, always check which convention a text uses.

Why is it called "normal"? The name comes from the 19th-century idea that this was the "normal" pattern of natural variation. Carl Friedrich Gauss used it to model observational errors in astronomy — hence it is also called the Gaussian distribution. The name has stuck despite the fact that many real-world distributions are decidedly not normal.

§ 02The Probability Density Function (PDF)

The probability density function of the normal distribution looks complicated at first glance, but each piece has a clear geometric meaning.

Normal Distribution — PDF
f(x) = 1⁄σ√(2π) · exp (−(x − μ)²⁄2σ²)

where μ ∈ ℝ is the mean and σ > 0 is the standard deviation.

Breaking down the formula: the exponential term e−(x−μ)²/(2σ²) is what creates the bell shape, it is maximised at x = μ and decays symmetrically on both sides. The exponent is negative, so the further x is from μ, the smaller the value. The prefactor 1/(σ√(2π)) is the normalising constant, it ensures the total area under the curve equals exactly 1.

Key Properties of the Normal Distribution

The normal PDF has several important properties worth memorising:

Properties at a glance Symmetric about x = μ  ·  Mean = Median = Mode = μ  ·  Inflection points at x = μ ± σ  ·  As x → ±∞, f(x) → 0  ·  Tails never reach zero (the distribution has infinite support)  ·  Total area = 1
Interactive — adjust μ and σ to reshape the bell curve

N(0, 1): mean = 0, std dev = 1. The dashed lines mark μ ± σ (inflection points).

§ 03The Standard Normal Distribution & Z-Scores

There are infinitely many normal distributions, one for each pair (μ, σ). The z-score transformation converts any of them into a single standard form so we can use one table for all calculations.

The standard normal distribution is the special case with μ = 0 and σ = 1, written Z ~ N(0, 1). Its PDF is often denoted φ(z) and its CDF is denoted Φ(z).

Standardisation — the Z-Score Formula
Z = (X − μ)⁄σ

Converts X ~ N(μ, σ²) to Z ~ N(0, 1). Read: "how many standard deviations is X from the mean?"

A z-score of +2.3 means the value is 2.3 standard deviations above the mean. A z-score of −1.0 means one standard deviation below. The sign and magnitude together tell you where in the distribution you are.

Why Standardise?

Because the normal CDF Φ(z) has no closed-form antiderivative, we cannot compute P(a ≤ X ≤ b) algebraically. We must look it up in a table (or use software). By standardising first, one table covers every normal distribution. The workflow is always: convert to z-scores → look up Φ → combine.

Computing Probabilities via Standardisation
P(a ≤ X ≤ b) = Φ(b−μ⁄σ) − Φ(a−μ⁄σ)
Example 1 — Computing a z-score

Exam scores follow N(65, 12²). Alice scored 83. What is her z-score, and what does it mean?

  1. Apply the formula: Z = (X − μ)/σ = (83 − 65)/12 = 18/12 = 1.5.
  2. Alice's score is 1.5 standard deviations above the mean.
  3. From the z-table: Φ(1.5) ≈ 0.9332, so Alice scored higher than approximately 93.3% of students.
Example 2 — Probability above a value

Adult male heights follow N(178, 7²) cm. Find P(X > 185).

  1. Standardise: z = (185 − 178)/7 = 7/7 = 1.00.
  2. P(X > 185) = P(Z > 1.00) = 1 − Φ(1.00).
  3. From table: Φ(1.00) = 0.8413.
  4. P(X > 185) = 1 − 0.8413 = 0.1587
  5. About 15.9% of adult males are taller than 185 cm.
Example 3 — Probability between two values

Battery lifetimes follow N(200, 15²) hours. Find P(185 ≤ X ≤ 220).

  1. Standardise both bounds: z₁ = (185 − 200)/15 = −15/15 = −1.00 z₂ = (220 − 200)/15 = 20/15 = 1.33
  2. P(185 ≤ X ≤ 220) = Φ(1.33) − Φ(−1.00).
  3. From table: Φ(1.33) ≈ 0.9082 and Φ(−1.00) = 1 − Φ(1.00) = 1 − 0.8413 = 0.1587.
  4. = 0.9082 − 0.1587 = 0.7495
  5. About 74.9% of batteries last between 185 and 220 hours.

§ 04Reading the Z-Table — Φ(z) for z ≥ 0

The z-table gives Φ(z) = P(Z ≤ z) for the standard normal. The row gives the first decimal place, the column gives the second. The highlighted cell shows Φ(1.20) = 0.8849.

z .00.01.02.03.04 .05.06.07.08.09
0.0.5000.5040.5080.5120.5160.5199.5239.5279.5319.5359
0.5.6915.6950.6985.7019.7054.7088.7123.7157.7190.7224
1.0.8413.8438.8461.8485.8508.8531.8554.8577.8599.8621
1.2.8849.8869.8888.8907.8925.8944.8962.8980.8997.9015
1.5.9332.9345.9357.9370.9382.9394.9406.9418.9429.9441
1.6.9452.9463.9474.9484.9495.9505.9515.9525.9535.9545
1.8.9641.9649.9656.9664.9671.9678.9686.9693.9699.9706
1.9.9713.9719.9726.9732.9738.9744.9750.9756.9761.9767
2.0.9772.9778.9783.9788.9793.9798.9803.9808.9812.9817
2.5.9938.9940.9941.9943.9945.9946.9948.9949.9951.9952
3.0.9987.9987.9987.9988.9988.9989.9989.9989.9990.9990
Negative z-scores — use symmetry The z-table typically only lists z ≥ 0. For negative z, use: Φ(−z) = 1 − Φ(z). The standard normal is symmetric about zero, so the left tail beyond −z equals the right tail beyond +z.
Interactive — shade P(Z ≤ z) in real time

P(Z ≤ 1.00) = Φ(1.00) ≈ 0.8413, the gold shaded area.

§ 05The 68-95-99.7 Empirical Rule

The most memorable property of the normal distribution: fixed percentages of data fall within 1, 2 and 3 standard deviations of the mean.

Empirical Rule — exact values
P(μ − σ ≤ X ≤ μ + σ) = 0.6827   (≈ 68%) P(μ − 2σ ≤ X ≤ μ + 2σ) = 0.9545   (≈ 95%) P(μ − 3σ ≤ X ≤ μ + 3σ) = 0.9973   (≈ 99.7%)
μ ± 1σ
68.27%
68.27%
μ ± 2σ
95.45%
95.45%
μ ± 3σ
99.73%
99.73%
Example 4 — Applying the Empirical Rule

IQ scores follow N(100, 15²). Using the empirical rule, find: (a) the range containing the middle 95% of scores, (b) the probability a randomly chosen person has IQ above 130.

  1. (a) The middle 95% falls within μ ± 2σ = 100 ± 2(15) = 100 ± 30. Range: 70 to 130
  2. (b) IQ 130 = μ + 2σ. By the empirical rule, 95% of scores lie within 2σ, so 5% lie outside. By symmetry, 2.5% are above 130 and 2.5% below 70. P(X > 130) ≈ 0.025
  3. More precisely: z = (130−100)/15 = 2.00, Φ(2.00) = 0.9772, so P(X > 130) = 1 − 0.9772 = 0.0228.
Example 5 — Outside the normal range

A factory produces bolts with diameter X ~ N(10, 0.04) mm (σ = 0.2 mm). A bolt is rejected if its diameter is more than 2 standard deviations from the mean. What percentage of bolts are rejected?

  1. Acceptable range: μ ± 2σ = 10 ± 0.4, so 9.6 mm to 10.4 mm.
  2. By the empirical rule, 95.45% of bolts fall within this range.
  3. Rejected bolts: 100% − 95.45% = 4.55%.
  4. Of these, 2.275% are too wide and 2.275% are too narrow by symmetry.
Diagram — the 68-95-99.7 rule visualised

The three shaded bands show the probability captured within 1σ (darkest), 2σ and 3σ of the mean.

§ 06Inverse Normal — Finding the Value for a Given Probability

Sometimes the question runs in reverse: instead of "what is the probability of X being less than b?", we are asked "what value b gives a probability of 0.90?" This is the inverse normal (also called a quantile or percentile calculation).

Inverse Normal — Procedure
1.   Find z* such that Φ(z*) = p   (look up p in the z-table body) 2.   Convert back: x = μ + z* · σ
Example 6 — Finding a Percentile

A student needs to score in the top 10% of an exam to receive a distinction. Scores follow N(62, 11²). What is the minimum score needed?

  1. Top 10% means P(X > x) = 0.10, so P(X ≤ x) = 0.90.
  2. Look up 0.90 in the z-table body: Φ(1.28) ≈ 0.8997 ≈ 0.90, so z* = 1.28.
  3. Convert back: x = μ + z* · σ = 62 + 1.28 × 11 = 62 + 14.08 = 76.08.
  4. The student needs a score of at least 77 (rounding up to the nearest whole number) for a distinction.
Example 7 — Symmetric Interval

Find the symmetric interval around the mean of N(50, 9) that contains 90% of the distribution.

  1. Symmetric means P(μ − c ≤ X ≤ μ + c) = 0.90, leaving 5% in each tail.
  2. Need Φ(z*) = 0.95: from table, z* ≈ 1.645.
  3. Note: σ = √9 = 3. So c = z* × σ = 1.645 × 3 = 4.935.
  4. Interval: (50 − 4.935, 50 + 4.935) = (45.065, 54.935).
  5. This is the 90% confidence interval construction used heavily in statistics.
Example 8 — Two-Sided Inverse

X ~ N(0, 1). Find a such that P(−a ≤ Z ≤ a) = 0.95.

  1. By symmetry, P(−a ≤ Z ≤ a) = 2Φ(a) − 1 = 0.95.
  2. So Φ(a) = (1 + 0.95)/2 = 0.975.
  3. From z-table: Φ(1.96) = 0.9750, so a = 1.96.
  4. P(−1.96 ≤ Z ≤ 1.96) = 0.95
  5. The critical value z = 1.96 is the most important number in statistics, it underlies the 95% confidence interval for a mean.

§ 07Further Worked Examples

Example 9 — Probability Below a Negative Z

Petrol consumption of cars follows N(8.2, 0.6²) litres/100km. Find P(X < 7.5).

  1. Standardise: z = (7.5 − 8.2)/0.6 = −0.7/0.6 = −1.167 ≈ −1.17.
  2. P(X < 7.5) = Φ(−1.17) = 1 − Φ(1.17).
  3. Φ(1.17) ≈ 0.8790 (interpolate between 1.1 and 1.2 rows).
  4. P(X < 7.5) = 1 − 0.8790 = 0.1210
  5. About 12.1% of cars use less than 7.5 L/100km.
Example 10 — Sum of Independent Normals

X ~ N(30, 4) and Y ~ N(20, 9) are independent. Find the distribution of X + Y and compute P(X + Y > 55).

  1. Sums of independent normals are normal. Add means and variances: E[X + Y] = 30 + 20 = 50 Var(X + Y) = 4 + 9 = 13   (independence allows this)
  2. So X + Y ~ N(50, 13), meaning σ = √13 ≈ 3.606.
  3. P(X + Y > 55): z = (55 − 50)/√13 = 5/3.606 ≈ 1.387 ≈ 1.39.
  4. P(Z > 1.39) = 1 − Φ(1.39) ≈ 1 − 0.9177 = 0.0823.
  5. About 8.2% chance the combined value exceeds 55.
Example 11 — Working Backwards: Finding μ

X ~ N(μ, 4²). Given P(X < 60) = 0.8413, find μ.

  1. P(X < 60) = 0.8413 means Φ(z₀) = 0.8413 for some z₀.
  2. From the z-table: Φ(1.00) = 0.8413, so z₀ = 1.00.
  3. Use z = (x − μ)/σ: 1.00 = (60 − μ)/4.
  4. 60 − μ = 4  ⟹  μ = 56
  5. The distribution is N(56, 16).
Example 12 — Finding σ

X ~ N(100, σ²). Given P(X > 112) = 0.0668, find σ.

  1. P(X > 112) = 0.0668 means P(X ≤ 112) = 1 − 0.0668 = 0.9332.
  2. Φ(z₀) = 0.9332: from table, z₀ = 1.50.
  3. Use z = (x − μ)/σ: 1.50 = (112 − 100)/σ = 12/σ.
  4. σ = 12/1.50 = 8
  5. The distribution is N(100, 64).

§ 08The Central Limit Theorem

The Central Limit Theorem (CLT) is one of the most profound results in mathematics. It explains why the normal distribution appears so widely, and it underpins virtually all of classical statistics.

Central Limit Theorem — Statement
If X₁, X₂, …, Xₙ are i.i.d. with mean μ and variance σ², then: X̄ = (X₁ + X₂ + ⋯ + Xₙ)/n → N(μ, σ²/n)   as n → ∞

The distribution of the sample mean X̄ approaches normal regardless of the shape of the underlying distribution.

The CLT says three things simultaneously. First, the mean of the sampling distribution of X̄ is μ, the sample mean is an unbiased estimator of the population mean. Second, the variance of X̄ is σ²/n, it shrinks with sample size, so larger samples give more precise estimates. Third, the shape of the distribution of X̄ becomes normal, even if the individual observations are not.

How large does n need to be? As a rule of thumb: n ≥ 30 is generally sufficient for the CLT approximation to be reliable, even for moderately skewed distributions. For symmetric distributions the CLT kicks in much faster (n = 5 may suffice). For very heavy-tailed or bimodal distributions, you may need n > 100.
Interactive — CLT: sampling distribution of X̄ from a skewed population

With n = 30, the distribution of X̄ (teal) closely approximates a normal curve (gold dashed), even though the underlying population is right-skewed.

§ 09Normal Approximation to the Binomial

When n is large and p is not too close to 0 or 1, the binomial distribution B(n, p) can be approximated by a normal distribution. This is a direct consequence of the CLT, since each Bernoulli trial is an i.i.d. random variable.

Normal Approximation — with Continuity Correction
X ~ B(n, p)  ≈  N(μ, σ²)   where μ = np, σ² = np(1−p) Continuity correction: replace X = k with k − 0.5 ≤ X ≤ k + 0.5

Rule of thumb: approximation is good when np ≥ 5 and n(1−p) ≥ 5.

The continuity correction — why it matters The binomial is discrete; the normal is continuous. When approximating P(X ≤ k) with the normal, use P(X ≤ k + 0.5) instead. This "half-unit correction" accounts for the width of each discrete bar and significantly improves accuracy, especially for smaller n.
Example 13 — Normal Approximation to Binomial

A fair coin is tossed 100 times. Using a normal approximation, find P(X ≥ 55) where X is the number of heads.

  1. X ~ B(100, 0.5). Check: np = 50 ≥ 5 ✓, n(1−p) = 50 ≥ 5 ✓.
  2. Approximate: X ≈ N(μ, σ²) where μ = 100 × 0.5 = 50, σ² = 100 × 0.5 × 0.5 = 25, σ = 5.
  3. Apply continuity correction: P(X ≥ 55) ≈ P(X ≥ 54.5) in the normal. z = (54.5 − 50)/5 = 4.5/5 = 0.90
  4. P(Z ≥ 0.90) = 1 − Φ(0.90) = 1 − 0.8159 = 0.1841.
  5. About 18.4% chance of 55 or more heads in 100 fair coin flips.

§ 10Common Mistakes

Mistake 1 — Confusing σ and σ² When we write X ~ N(μ, σ²), the second parameter is the variance. So N(10, 25) has σ = 5, not σ = 25. Always check whether a formula uses σ or σ² before substituting.
Mistake 2 — Forgetting to standardise You cannot look up P(X ≤ 185) directly in the z-table if X ~ N(178, 7²). You must first compute z = (185 − 178)/7 = 1.00, then look up Φ(1.00).
Mistake 3 — Getting "above" vs "below" backwards The z-table gives P(Z ≤ z) = Φ(z). For P(Z > z), compute 1 − Φ(z). For P(a ≤ Z ≤ b), compute Φ(b) − Φ(a) — always bigger z minus smaller z.
Mistake 4 — Not applying the continuity correction for binomial approximation When approximating a discrete binomial with a continuous normal, omitting the continuity correction (±0.5) gives a noticeably worse answer, especially for probabilities involving exact values or small ranges.
Mistake 5 — Applying CLT to the sum without adjusting The CLT says X̄ ~ N(μ, σ²/n). The sum Sₙ = nX̄ ~ N(nμ, nσ²). These are different. Be careful whether the problem asks about the sample mean or the total sum.

§ 11Related Pages

The normal distribution sits at the intersection of probability theory, statistics and calculus. These pages build directly on or connect closely to this topic.

Probability
Continuous Distributions
PDFs, CDFs, Uniform and Exponential — the framework behind the Normal
Probability
Expected Value & Variance
E[X], Var(X), standard deviation and covariance from first principles
Probability
Binomial Distribution
When to use normal approximation and why it works
Probability
Basic Probability
Sample spaces, axioms and counting — the starting point
Probability
Bayes' Theorem
Updating beliefs — pairs naturally with normal likelihoods
Statistics
Hypothesis Testing
z-tests and t-tests rest on the normal distribution
Statistics
Confidence Intervals
The 1.96 critical value comes directly from §06
Statistics
Sampling & Estimation
The CLT and sampling distributions in full detail
Integration
Improper Integrals
How to prove ∫e^(−x²)dx = √π using polar coordinates
Statistics
Linear Regression
Residuals are assumed normally distributed in OLS

§ 12Quiz — Test Your Understanding

10 questions covering the PDF, z-scores, the z-table, the empirical rule, inverse normal and the CLT. Select an option for instant feedback.

Score: 0 / 0
Question 1 of 10

For X ~ N(μ, σ²), which TWO parameters fully determine the distribution?

Question 2 of 10

X ~ N(50, 16). What is the standard deviation σ?

Question 3 of 10

Z ~ N(0,1). What is Φ(−1.5)?

Question 4 of 10

Heights follow N(170, 6²) cm. Find the z-score for a height of 158 cm.

Question 5 of 10

According to the empirical rule, approximately what percentage of a normal distribution lies within 2 standard deviations of the mean?

Question 6 of 10

X ~ N(100, 25). Find P(X < 105). [Use Φ(1.00) = 0.8413]

Question 7 of 10

Which critical value z* satisfies P(−z* ≤ Z ≤ z*) = 0.95?

Question 8 of 10

The Central Limit Theorem says that if X̄ is the mean of n i.i.d. samples with mean μ and variance σ², then X̄ has approximately:

Question 9 of 10

X ~ N(200, 100). Find the 90th percentile of X. [Φ(1.28) ≈ 0.90]

Question 10 of 10

X ~ B(200, 0.4). Using a normal approximation with continuity correction, P(X ≤ 85) ≈ P(Z ≤ ?)  where Z is standard normal.

0/10

Cookie Settings