- 2070 Asar · 6 marks
Standard normal with area property
Short answer
Standard normal distribution has mean 0 and variance 1, denoted Z ~ N(0,1), with pdf f(z) = (1/√(2π)) e^(−z²/2). Its total area under the curve is 1; the curve is symmetric about z=0. Standard normal table gives area P(0<Z<z), and P(Z<0)=0.5.
Define standard normal distribution with area property.
Standard normal distribution: the normal distribution with mean 0 and variance 1. Z = (X − μ)/σ ~ N(0, 1), with pdf
f(z) = (1/√(2π)) e^(−z²/2), −∞ < z < ∞
Area property
The total area under the curve is 1 and the curve is symmetric about z = 0. The standard normal table gives the area between 0 and z, P(0 < Z < z).
.-'''-.
.-' | '-.
.-' | '-.
_.--' | '--._
------+----+----+----+----+------
-2 -1 0 1 2 z
|<-0.3413->|<-0.3413->|
Key properties:
- P(Z < 0) = P(Z > 0) = 0.5.
- Symmetry: P(0 < Z < a) = P(−a < Z < 0), and P(Z > a) = P(Z < −a).
- P(Z < a) = 0.5 + P(0 < Z < a) for a > 0.
- P(a < Z < b) = P(0<Z<b) − P(0<Z<a) when both are positive; for a < 0 < b it is the sum of the two table areas.
- Empirical rule:
| Interval | Area |
|---|---|
| −1 < Z < 1 | 0.6827 (68.27%) |
| −2 < Z < 2 | 0.9545 (95.45%) |
| −3 < Z < 3 | 0.9973 (99.73%) |
| −1.96 < Z < 1.96 | 0.9500 (95%) |
Example: P(−1.2 < Z < 0.8) = P(0 < Z < 1.2) + P(0 < Z < 0.8) = 0.3849 + 0.2881 = 0.6730.
✓ Answer checked by Claude Sonnet 5.5
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