MathematicsGrade 12–13 · University

Normal distribution

How many bottles are underfilled? Under the bell curve every probability is an area, with the sigma rules as a rule of thumb.

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What is it about?

For a normally distributed random variable X∼N(μ; σ2)X \sim N(\mu;\,\sigma^2), the probability P(a≤X≤b)P(a \leq X \leq b) is the area under the bell curve between aa and bb, and equals F(b)−F(a)F(b) - F(a) with the distribution function F(x)=P(X≤x)F(x) = P(X \leq x); for N(0; 1)N(0;\,1) it is called Φ\Phi, and F(x)=Φ(x−μσ)F(x) = \Phi\left(\frac{x - \mu}{\sigma}\right). A single value has probability P(X=a)=0P(X = a) = 0.

The bell: symmetric about μ\mu, maximum 1σ2π\frac{1}{\sigma\sqrt{2\pi}} at μ\mu, points of inflection at μ±σ\mu \pm \sigma (their tangents meet the xx axis at μ±2σ\mu \pm 2\sigma), total area 11. Sigma rules: about 68.3 %68.3\,\%, 95.4 %95.4\,\% and 99.7 %99.7\,\% within μ±σ\mu \pm \sigma, ±2σ\pm 2\sigma, ±3σ\pm 3\sigma.

Standardising with z=x−μσz = \frac{x - \mu}{\sigma} shifts the bell by μ\mu and scales it by the factor 1σ\frac{1}{\sigma} into N(0; 1)N(0;\,1) – the marked area stays the same. Backwards (quantiles): from P(X≥x)=5 %P(X \geq x) = 5\,\% we get Φ(z)=0.95\Phi(z) = 0.95, z≈1.645z \approx 1.645 and x=μ+zσx = \mu + z\sigma.

Approximating the binomial: P(X≤k)≈Φ(k+0.5−μσ)P(X \leq k) \approx \Phi\left(\frac{k + 0.5 - \mu}{\sigma}\right) with μ=np\mu = np, σ=np(1−p)\sigma = \sqrt{np(1-p)}; rule of thumb (Laplace condition): σ>3\sigma > 3.

Try it

  1. Standard normal: drag the bounds and press “μ ± σ”, “μ ± 2σ”, “μ ± 3σ”.
  2. Properties of the bell and very small σ.
  3. Standardise: press “Standardise” and watch the IQ axis turn into the zz axis.
  4. Which IQ is exceeded by only 5 %? Press “Find the bound”.
  5. Two filling machines and narrow, but shifted.
  6. Approximating the binomial and a sample.

Tasks

Task 1: Standardising

IQ: μ=100\mu = 100, σ=15\sigma = 15. Find P(X≥130)P(X \geq 130). Load task

Show solution

z=2z = 2, so 1−Φ(2)≈0.02281 - \Phi(2) \approx 0.0228. Show

Task 2: Finding the bound

Which IQ is exceeded by only 5 %5\,\%? Load task

Show solution

Φ(z)=0.95\Phi(z) = 0.95 lies between Φ(1.64)=0.9495\Phi(1.64) = 0.9495 and Φ(1.65)=0.9505\Phi(1.65) = 0.9505, so z≈1.645z \approx 1.645 and x≈100+1.645⋅15≈124.7x \approx 100 + 1.645 \cdot 15 \approx 124.7. Show

Task 3: Two machines

A: N(500; 42)N(500;\,4^2), B: N(502; 2.52)N(502;\,2.5^2), tolerance 495495 to 505 ml505\,\text{ml}. Which is better? Load task

Show solution

A: 2Φ(1.25)−1≈0.78872\Phi(1.25) - 1 \approx 0.7887; B: Φ(1.2)−Φ(−2.8)≈0.8824\Phi(1.2) - \Phi(-2.8) \approx 0.8824. B is better. Show

Task 4: Approximating the binomial

100 coin tosses: approximate P(X≤55)P(X \leq 55). Load task

Show solution

μ=50\mu = 50, σ=5\sigma = 5: Φ(1.1)≈0.8643\Phi(1.1) \approx 0.8643 (exact 0.86440.8644; without correction 0.84130.8413). Show

Ideas for the classroom

  • Let students estimate an area before dragging; the bar in the chart of the distribution function explains the subtraction F(b)−F(a)F(b) - F(a).
  • The standardising animation shows why one table for Φ\Phi is enough.
  • Common stumbling blocks: reading the height of the curve as a probability; forgetting 1−F(a)1 - F(a) for P(X≥a)P(X \geq a); looking for Φ(z)=0.05\Phi(z) = 0.05 instead of 0.950.95; forgetting the continuity correction.