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1、第十二讲 正态分布序号第十二讲章节9 Normal distribution教学内容正态分布的性质，标准正态分布的计算方法，应用标准正态分 布计算正态分布，期望教学目的及 基本要求具备应用标准正态分布数表解决正态分布数据的概论问题，并 能应用期望的含义计算现实中的平均利润问题教学重点及教学难点标准正态分布数表的使用教学实现精讲标准正态分布数表的使用，现结合正态分布的实际问题进 行讲解，时间大约为50分钟教学时间90分钟Normal distribution、Getting started用概率来估计事件发生可能性的大小，可能针对不同类型的数据，有离散的， 如二项分布，Poisson分布；亦有连
2、续的，其中连续数据的概率分布中最常用的 是正态分布。而且在实际中可用正态分布来逼近二项分布和 Poisson 分布，所以 本章中我们重点来介绍正态分布的计算。Normal distribution：A random variable x follows a Normal distribution with mean 卩 and standarddeviation c，记为 x N(2).曲 线 形 状U XA对称的，以均值为中心，一般研究时，从均值向两端扩展，通常研 究卩土 kc情况。A = P(X x ) = 1 P(x x ) = P(x -x ) A A AP(x x x ) = P(x
3、 x ) P(x x )ABBB计算方法将所有正态分布转化成种特殊的正态分布标准正态分布，通过查表进行计算二.Standard normal distribution (section 9.3)表示用z表示标准正态分布的随机变量，zN(0,1)形状与正态分布基本一样，结合分布曲线图与数表可计算标准正态使用方法根据所要计算的随机变量的取值（区间的端点值）查数表，确定出 对应的概率。正 态 分 布 表z0.00.10.20.30.40.50.60.70.80.9A.500.540.580.618.655.692.726.758.788.816z1.01.11.21.31.41.51.61.71.8
4、1.9A.841.864.885.903.919.933.945.955.964.971z2.02.12.22.32.42.52.62.72.82.93.0A.977.982.986.989.992.994.995.996.997.998.999If a variable x is normally distributed with mean 0 and standard deviation 1,(A) find the probability that x is : less than 0.7 more than 0.7 less than -0.7 more than -0.7 lies
5、between 0.7 and 1.8 lies between -1.8 and -0.7 lies between -0.7 and 1.8 lies between -0.7 and 0.7(B) find the x value, such that the curve area(1) to the left of x is 0.580(2) to the right of x is 0.081(3) between -x and +x is 0.95(其中2, 4, 6, 8为练习)解题(A) 这些都是直接查数表的过程 直接查表，0.758 1-0.7580=0.242 0.242
6、1-第三问，0.758 查 1.8 得 0.964，查 0.7 得 0.758，结果 0.964-0.758=0.206 0.206 去掉比-0.7小和比1.8大的部分，结果是1-0.242-0.036=0.728 1-2*0.242=0.516(B) 反查数表，根据概率查变量的取值0.2；查概率 1-0.081=0.919,得 x 值为 1.4； 1-0.95=0.05,再除以 2 得 0.025， 0.95+0.025=0.975，查此概率得 z =2.0注意，在实用中，0.95查得变量值会取到更精确的结果，1.96； 0.99 对应的变量值为2.58三、Using the normal
7、dis tribu tiont ables(secti on 9.4 and 9.5)将正态分布转化成标准正态分布进行计算。If X N( PQ 2),公X HThe formula： z =, then z N(0,1).式Qltx Ux U且 P(X X X ) = P(t z )ABQQP205Worked example 9b (直接查数表)The monthly demand for a product is normally distributed with a mean of2000 and a standard deviation of 200.例Find the probab
8、ility that the demand in a given month will be :题(i) more than 23501(ii) less than 1600(iii) between 1600 and 2350解题 过 程(i) z二 + x HF x12n也服从正态分布，且均值为卩+卩 +卩，标准差为2 +Q 2 + Q 212n12n卩二2350 2000二1.8(因数表的设置，注意计算时保留一位小c200数)，from tables 0.964, the probability is 1-0.964=0.036.x _ p 1600 20001 i iii(ii) z
9、= 2, take numerical value +2. From tablesc2000.977, the probability is 1-0.977=0.023(iii) 0.941题例2P208Continuation of 9b.When an order is placed, a month is required for delivery. If the probability of running out of stock is to be less than 0.01, calculate the amount to order at the beginning of th
10、e month (assume that at the end of each month the stock will have fallen to zero.)解题Z value corresponding to 0.01 is 2.33.As x is the amount needed, use z = _,cx _ 2000 = 2.33 , thus x-2000=2.33*200, x=2000+466=2466 200练习 二 ： Invoice at a particular depot have amounts which follow a normal distribut
11、ion with a mean of 130.6 and a standard deviation of 8.75. Find the probability that the invoices will be:(a) over 120.5(b) less than 92.75(c) lies between 83.7 and 117.6(d) above what amount will 90% of invoices lie?四、 Sum of two or more variables all normally distributed设n个随机变量均服从正态分布，xNg q2),i二1,
12、2,n，贝廿ii i应用：P210，看书。五、Expected values概念Expected value = Probability X total解说这里，我们说的期望值通常指的是期望利润，即平均利润。如果是 总体的就是总体均值，如果是样本的，就是样本均值。这里只说明期望 利润的计算。Expected profit = probability of an event multiply by the profit if that event occurs.例题P212Worked example 9e工件长度服从正态分布，其均值为20cm，标准差为.025cm。工件合 格的标准是其长度必须
13、在19.96cm到20.05cm中间。每只工件的成本是 0.20。合格工件的购价是0.40。如果工件太短只能当费品卖得0.05/只, 如果工件太长必须截掉一段以致合格，成本会多用0.08。（a）计算最开始下面工件的百分比：（1）太长（2）太短（3）正好合格（b）如果做了 10000只工件，计算平均利润练习三： P214exercise 9D作业： The weight of a product is normally distributed with a mean of 500g and a standard deviation of 25g.1. If one product is selected at random, find the probability that the mean weight will be between 450g and 550g2. If the probability that the mean weight follow x and y is 0.9, calculate the weight x and y.
