4 · IB Maths
The normal distribution
A bell curve set by its mean and standard deviation. Shade a region to read the probability as an area, convert bounds to z-scores, and work backwards with the inverse normal.
Predict first
Start from the defaults (μ = 50, σ = 8, region between a = 42 and b = 58). If you raise σ from 8 to 12, what happens to P(42 < X < 58)?
A larger σ spreads the curve out, so less area lies in the fixed interval. The z-values shrink from ±1 to ±0.667, and the probability falls from 0.683 to about 0.495.
Check it in the model: Move the Standard deviation, σ slider from 8 to 12 and compare the probability and z readouts.
density curve shaded probability inverse-normal value
P(a < X < b) = Φ(z_b) − Φ(z_a)
inverse: x = μ + σ z, where Φ(z) = p
Try this
- Use μ = 50, σ = 8 and bounds 42 and 58 (one σ either side). Tick the σ bands and compare the probability with 68%. Then widen the bounds to μ ± 2σ.
- Keep the same bounds and raise σ from 8 to 12. Does the probability between 42 and 58 rise or fall? Explain from the shape of the curve.
- Tick Inverse normal and set p = 95%. Read x, then choose Right tail with a equal to that x and check the shaded probability is about 5%.
Common error: forgetting that the curve is symmetric about the mean, so P(X < μ − kσ) = P(X > μ + kσ); sketch the shaded area first and check that your answer is sensible for its size.
More on 4: Venn diagrams and conditional probability · Discrete random variables · Continuous random variables · Bayes' theorem: medical test · Standard deviation and outliers · Binomial distribution · Markov chains · Poisson distribution · Scatter and regression · Tree diagrams and Bayes’ theorem · Box plots and histograms