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1、Normal DistributionA random variable X having a probability density function given by the formulais said to have a Normal Distribution with parameters and 2.Symbolically, X N(, 2).Properties of Normal DistributionThe curve extends indefinitely to the left and to the right, approaching the x-axis as
2、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.Empirical Rule (Golden Rule) The following diagram illustrates relevant areas and associa
3、ted 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.For normal curves with the same , they are identical in shapes but the means are centered at different positions along the horizontal axis.Fo
4、r 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, and the curve with lower
5、standard deviation and the dispersion is smaller.Normal 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 Standardizing ProcessThis can be done by means of the transformation.The mea
6、n of Z is zero and the variance is respectively,Diagrammatic 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 a), P(Z 1.28) = 0.5 0.3997 =0.1003Example 3: P(Z -1.28)
7、= ?Solution:P(Z -1.28) = P(Z a) = 0.8, find the value of a?Solution: From the Normal Table A(0.84) 0.3 a - 0.84Example 7: If P(Z c) = 0.1, fin the values of c?Solution: P(|Z c) = 0.1 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 TransformationExample 9: If X N(10, 4), fi
8、nd P(X 12); P(9.5 X 11); P(8.5 X 9) ?Solution: (a) For the distribution of X with =10, =2Solution: (b) For the distribution of X with =10, =2P(9.5 X 11) = P(- 0.25 Z 0.5) = 0.0987 + 0.1915 = 0.2902Solution: (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
