正态分布基础_英文

2
∫
−∞
σ 2π
e2
σ
dx = 1
Empirical Rule (Golden Rule)
The following diagram illustrates relevant areas and associated probabilities of the Normal Distribution. Approximate 68.3% of the area lies within µ±σ, 95.5% of the area lies within µ±2σ, and 99.7% of the area lies within µ±3σ.
P ( X ≥ 12) 12 − 10 = P Z ≥ 2 = 0.5 − 0.3413 = 0.1587
Solution: (b) For the distribution of X with µ=10, σ=2
P(9.5 ≤ X ≤ 11) = P(- 0.25≤ Z ≤ 0.5) = 0.0987 + 0.1915 = 0.2902
Z=
X −µ
σ
Standardizing Process
This can be done by means of the transformation.
z=
x−µ
σ
X −µ Var ( Z ) = Var σ 1 = 2 Var ( X − µ )
The mean of Z is zero and the variance is respectively,
The normal table can be used to find values like P(Z > a), P(Z < b) and P(a ≤ Z ≤ b). We illustrate with the following examples.
Example 1: Solution:
P(-1.28 < Z < 0) = ?
Example 8: If P(|Z| > c) = 0.1, fin the values of c?
Solution: P(|Z| > c) = 0.1 Q P(Z > c) = 0.05 ∴ P( c > Z > 0) = 0.5 – 0.05 = 0.45 From table, A(1.645) ≈ 0.45 ∴ c ≈ 1.645
For normal curves with the same σ, they are identical in shapes but the means µ are centered at different positions along the horizontal axis.
For normal curves with the same mean µ, the curves are centered at exactly the same position on the horizontal axis, but with different standard deviations σ, the curves are in different shapes, i.e. the curve with the larger standard deviation is lower and spreads out farther, and the curve with lower standard deviation and the dispersion is smaller.
P(-1.28 < Z < 0) = P(0 < Z < 1.28) = 0.3997
Example 2: P(Z < -1.28) = ?
Solution:
பைடு நூலகம்
P(Z < -1.28)
= P(Z > 1.28) = 0.5 – 0.3997 =0.1003
Example 3: P(Z > -1.28) = ?
Properties of Normal Distribution
1. The curve extends indefinitely to the left and to the right, approaching the x-axis as x increases in magnitude, i.e. as x → ±∝, f(x) → 0. 2. The mode occurs at x=µ. 3. The curve is symmetric about a vertical axis through the mean µ 4. The total area under the curve and above the horizontal axis is equal to 1. −1 x − µ i.e. ∞ 1
From the Normal Table A(0.84) ≈ 0.3 ∴ a ≈ - 0.84
Example 7: If P(Z < b) = 0.32, find the value of b? Solution:
P(Z < b) = 0.32 P(b < Z < 0) = 0.5 – 0.32 = 0.18 From table, A(0.47) ≈ 0.18 ∴ b ≈ -0.47
Transformation
Example 9: If X ~ N(10, 4), find a) P(X ≥ 12); b) P(9.5 ≤ X ≤ 11); c) P(8.5 ≤ X ≤ 9) ?
Solution: (a) For the distribution of X with µ=10, σ=2
Solution: (c) For the distribution of X with µ=10, σ=2
P(8.5 ≤ X ≤ 9) = P(- 0.75≤ Z ≤ - 0.5) = 0.2734 – 0.1915 = 0.0819
Solution: P(Z > -1.28) = P(Z < 1.28) = 0.5 + 0.3997 = 0.8997
Example 4: P(-2.28 < Z < -1.28) = ?
Solution: P(-2.28 < Z < -1.28)
= P(1.28 < Z < 2.28) = 0.4887 – 0.3997 = 0.0890
Normal Distribution
A random variable X having a probability density function given by the formula
1 f ( x) = e σ 2π
−1 x − µ 2 σ
2
,−∞ < x < ∞
is said to have a Normal Distribution with parameters µ and σ2. Symbolically, X ~ N(µ, σ2).
Example 5: P(-1.28 < Z < 2.28) = ? Solution:
P(-1.28 < Z < 2.28) = 0.3997 + 0.4887 = 0.8884
Example 6: If P(Z > a) = 0.8, find the value of a?
Solution:
X −µ E(Z ) = E σ 1 = E(X − µ )
σ
σ
=
1
=0
σ
[E( X ) − µ ]
= =
1
σ σ
2
Var ( X ) ⋅σ 2
1
2
=1
Diagrammatic of the Standardizing Process
Transforms X ~ N(µ, σ2) to Z ~ N(0, 1). Whenever X is between the values x=x1 and x=x2, Z will fall between the corresponding values z=z1 and z=z2, we have P(x1 < X < x2) = P(z1 < Z < z2). It illustrates by the following diagram:
Normal Table
If the random variable X ~ N(µ, σ2), then we can transform all the values of X to the standardized values Z with the mean 0 and variance 1, i.e. Z ~ N(0, 1), on letting