正态分布基础_英文

1、1Normal DistributionA random variable X having a probability density function given by the formulaxexfx,21)(221is said to have a Normal Distribution with parameters and 2.Symbolically, X N(, 2).2Properties of Normal DistributionThe curve extends indefinitely to the left and to the right, approaching

2、 the x-axis as x increases in magnitude, i.e. as x , f(x) 0. The mode occurs at x=.The curve is symmetric about a vertical axis through the mean The total area under the curve and above the horizontal axis is equal to 1. i.e.121221dxex3Empirical Rule (Golden Rule) The following diagram illustrates r

3、elevant areas and associated probabilities of the Normal Distribution. Approximate 68.3% of the area lies within , 95.5% of the area lies within 2, and 99.7% of the area lies within 3.4For normal curves with the same , they are identical in shapes but the means are centered at different positions al

4、ong the horizontal axis.5For normal curves with the same mean , the curves are centered at exactly the same position on the horizontal axis, but with different standard deviations , the curves are in different shapes, i.e. the curve with the larger standard deviation is lower and spreads out farther

5、, and the curve with lower standard deviation and the dispersion is smaller.6Normal TableIf the random variable X N(, 2), 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 XZ7Standardizing ProcessxzThis can be done by me

6、ans of the transformation.The mean of Z is zero and the variance is respectively,0)(11)(XEXEXEZE11)(1)(1)(2222XVarXVarXVarZVar8Diagrammatic of the Standardizing ProcessTransforms X N(, 2) 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

7、and z=z2, we have P(x1 X x2) = P(z1 Z a), P(Z b) and P(a Z b). We illustrate with the following examples.Example 1:P(-1.28 Z 0) = ?Solution:P(-1.28 Z 0) = P(0 Z 1.28)= 0.399710Example 2: P(Z -1.28) = ?Solution: P(Z 1.28)= 0.5 0.3997=0.100311Example 3: P(Z -1.28) = ?Solution:P(Z -1.28) = P(Z 1.28)= 0

8、.5 + 0.3997= 0.899712Example 4: P(-2.28 Z -1.28) = ?Solution: P(-2.28 Z -1.28) = P(1.28 Z 2.28)= 0.4887 0.3997= 0.089013Example 5: P(-1.28 Z 2.28) = ?Solution:P(-1.28 Z a) = 0.8, find the value of a?Solution: From the Normal TableA(0.84) 0.3 a - 0.8415Example 7: If P(Z b) = 0.32, find the value of b

9、?Solution:P(Z b) = 0.32P(b Z c) = 0.1, fin the values of c?Solution: P(|Z c) = 0.1P(Z c) = 0.05 P( c Z 0) = 0.5 0.05 = 0.45From table, A(1.645) 0.45 c 1.645 17TransformationExample 9: If X N(10, 4), findP(X 12);P(9.5 X 11);P(8.5 X 9) ?18Solution: (a) For the distribution of X with =10, =21587. 03413. 05 . 021012)12(ZPXP19Solution: (b) For the distribution of X with =10, =2P(9.5 X 11)=