Key Idea: The normal distribution is a symmetric, bell-shaped curve that models many naturally occurring quantities — heights, test scores, measurement errors. It is defined by two parameters: the mean μ (centre of the bell) and standard deviation σ (width of the bell). Almost all normal distribution work in IB Math AI SL is done on the GDC.
✅ Key properties of the normal distribution
- X ~ N(μ, σ²)
- X follows a Normal distribution with mean μ and variance σ². Note: the second parameter is variance (σ²), not standard deviation. Example: X ~ N(50, 16) means μ = 50, σ = 4.
- Symmetry
- The curve is symmetric about μ. So P(X < μ) = P(X > μ) = 0.5.
- 68-95-99.7 rule
- 68% of data lies within 1σ of μ. 95% within 2σ. 99.7% within 3σ. Useful for quick estimates without GDC.
- Standardisation: z-score
- z = (x − μ) / σ. Converts any normal variable to standard normal N(0, 1). Positive z = above mean; negative z = below mean.
📊 GDC functions for the normal distribution
| GDC function | What it gives | Note |
|---|---|---|
| normalcdf(lower, upper, μ, σ) | P(lower < X < upper) | For 'less than a', use −∞ as lower (type −1E99 on GDC) |
| invNorm(p, μ, σ) | Find x such that P(X < x) = p | p is the LEFT-TAIL probability. For upper tail P(X > x) = p, use invNorm(1−p, μ, σ) |
Example: X ~ N(60, 25), so μ = 60, σ = 5 P(X < 55) = normalcdf(−1E99, 55, 60, 5) = 0.159 P(55 < X < 70) = normalcdf(55, 70, 60, 5) = 0.819 P(X > 68) = normalcdf(68, 1E99, 60, 5) = 0.0548 Find x where P(X < x) = 0.90: x = invNorm(0.90, 60, 5) = 66.4
The second parameter in X ~ N(μ, σ²) is variance, not standard deviation. If σ² = 25, then σ = 5. Always enter σ (not σ²) into GDC functions. For 'at least x' problems: P(X > x) = 1 − P(X ≤ x), or use normalcdf with 1E99 as the upper bound.
Paper 2 (GDC allowed): Write X ~ N(μ, σ²) and the probability statement before calculating. Show the GDC function used and its output. Inverse normal context: 'Find the value of x such that 25% of scores exceed x.' This means P(X > x) = 0.25, so P(X < x) = 0.75. Use invNorm(0.75, μ, σ).
IB-style question [7 marks]
The lifetime of a brand of light bulb is normally distributed with a mean of 1200 hours and a standard deviation of 100 hours. Let X be the lifetime of a randomly chosen bulb. (a) Find the probability that a bulb lasts less than 1000 hours. (b) Find the probability that a bulb lasts between 1100 and 1300 hours. (c) The manufacturer offers a free replacement for the 5% of bulbs with the shortest lifetimes. Find the lifetime below which a bulb qualifies for a free replacement.
Step by step:
Write down the distribution. The mean is μ = 1200 and the standard deviation is σ = 100.
(a) 'Less than 1000' is a left-tail area; use a very small lower bound on the GDC.
Read off the GDC.
(b) Use both bounds directly for the 'between' area.
Read off the GDC.
(c) The shortest 5% lie below the cut-off, so the cumulative (left) area is 0.05. Use the inverse normal.
Read off the cut-off lifetime.
(a) 0.0228. (b) 0.683. (c) 1040 hours (3 s.f.) — bulbs lasting under about 1040 hours qualify.