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1、正态分布基础英文第1页，共20页，编辑于2022年，星期六Properties of Normal Distribution1.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=.3.The curve is symmetric about a vertical axis through the mean 4.The total area unde
2、r the curve and above the horizontal axis is equal to 1.i.e.第2页，共20页，编辑于2022年，星期六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,95.5%of the area lies within 2,and 99.7%of the
3、 area lies within 3.第3页，共20页，编辑于2022年，星期六For normal curves with the same,they are identical in shapes but the means are centered at different positions along the horizontal axis.第4页，共20页，编辑于2022年，星期六For normal curves with the same mean,the curves are centered at exactly the same position on the hori
4、zontal 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,and the curve with lower standard deviation and the dispersion is smaller.第5页，共20页，编辑于2022年，星期六Normal TableIf the random variable X
5、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 第6页，共20页，编辑于2022年，星期六Standardizing ProcessThis can be done by means of the transformation.The mean of Z is zero and the variance is respectively,第7页，共20页，编辑于2022年，星期六Di
6、agrammatic 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 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(
7、-1.28 Z 0)=P(0 Z 1.28)=0.3997第9页，共20页，编辑于2022年，星期六Example 2:P(Z -1.28)=?Solution:P(Z 1.28)=0.5 0.3997=0.1003第10页，共20页，编辑于2022年，星期六Example 3:P(Z -1.28)=?Solution:P(Z -1.28)=P(Z 1.28)=0.5+0.3997=0.8997第11页，共20页，编辑于2022年，星期六Example 4:P(-2.28 Z -1.28)=?Solution:P(-2.28 Z -1.28)=P(1.28 Z 2.28)=0.4887 0.3
8、997=0.0890第12页，共20页，编辑于2022年，星期六Example 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 -0.84第14页，共20页，编辑于2022年，星期六Example 7:If P(Z b)=0.32,find the value of b?Solution:P(Z b)=0.32P(b Z c)=0.1,fin the values of c?Solution:P(|Z c)=0.1P(Z c)=
9、0.05 P(c Z 0)=0.5 0.05 =0.45From table,A(1.645)0.45 c 1.645 第16页，共20页，编辑于2022年，星期六TransformationExample 9:If X N(10,4),finda)P(X 12);b)P(9.5 X 11);c)P(8.5 X 9)?第17页，共20页，编辑于2022年，星期六Solution:(a)For the distribution of X with=10,=2第18页，共20页，编辑于2022年，星期六Solution:(b)For the distribution of X with=10,=2P(9.5 X 11)=P(-0.25 Z 0.5)=0.0987+0.1915=0.2902第19页，共20页，编辑于2022年，星期六Solution:(c)For the distribution of X with=10,=2P(8.5 X 9)=P(-0.75 Z -0.5)=0.2734 0.1915=0.0819第20页，共20页，编辑于2022年，星期六