商業(yè)應(yīng)用統(tǒng)計(英文版)ch6

1、Copyright 2011John Wiley & Sons, Inc. Understand concepts of the continuous distribution, especially the normal distribution.Recognize normal distribution problems, and know how to solve them.Decide when to use the normal distribution to approximate binomial distribution problems, and know how to wo

2、rk them.Decide when to use the exponential distribution to solve problems in business, and know how to work them.Copyright 2011John Wiley & Sons, Inc. 2Learning ObjectivesContinuous DistributionsContinuous distributionsContinuous distributions are constructed from continuous random variables which c

3、an be any values over a given intervalWith continuous distributions, probabilities of outcomes occurring between particular points are determined by calculating the area under the curve between these pointsUnlike discrete probability distributions, the probability of being exactly at a given point i

4、s 0 (since you can measure it more precisely)Copyright 2011 John Wiley & Sons, Inc. 3Properties of the Normal DistributionCharacteristics of the normal distribution:Continuous distribution - Line does not breakBell-shaped, symmetrical distributionRanges from - to Mean = median = modeArea under the c

5、urve = total probability = 168% of data are within one std dev of mean, 95% within two std devs, and 99.7% within three std devsCopyright 2011 John Wiley & Sons, Inc. 4Probability Density Function ofthe Normal DistributionThere are a number of different normal distributions, they are characterized b

6、y the mean and the std devCopyright 2011 John Wiley & Sons, Inc. 5Probability Density Function ofthe Normal Distribution. . . 2.71828 . . . 3.14159 = Xof dev std Xof mean eWherexxfe:21)(221XCopyright 2011 John Wiley & Sons, Inc. 6Rather than create a different table for every normal distribution (wi

7、th different mean and std devs), we can calculate a standardized normal distribution, called ZA z-score gives the number of standard deviations that a value x is above the mean.Z distribution is normal distribution with a mean of 0 and a std dev of 1 Normal Distribution Calculating ProbabilitiesCopy

8、right 2011 John Wiley & Sons, Inc. 7xzStandardized Normal Distribution - ContinuedZ distribution probability values are given in table A5 or can be calculated using softwareTable A5 gives the total area under the Z curve between 0 and any point on the positive Z axisSince the curve is symmetric, the

9、 area under the curve between Z and 0 is the same whether the Z curve is positive or negativeCopyright 2011 John Wiley & Sons, Inc. 8Z TableSecond Decimal Place in Z Z0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.000.00000.00400.00800.01200.01600.01990.02390.02790.03190.03590.100.03980.04380.0

10、4780.05170.05570.05960.06360.06750.07140.07530.200.07930.08320.08710.09100.09480.09870.10260.10640.11030.11410.300.11790.12170.12550.12930.13310.13680.14060.14430.14800.15170.900.31590.31860.32120.32380.32640.32890.33150.33400.33650.33891.000.34130.34380.34610.34850.35080.35310.35540.35770.35990.362

11、11.100.36430.36650.36860.37080.37290.37490.37700.37900.38100.38301.200.38490.38690.38880.39070.39250.39440.39620.39800.39970.40152.000.47720.47780.47830.47880.47930.47980.48030.48080.48120.48173.000.49870.49870.49870.49880.49880.49890.49890.49890.49900.49903.400.49970.49970.49970.49970.49970.49970.4

12、9970.49970.49970.49983.500.49980.49980.49980.49980.49980.49980.49980.49980.49980.4998Copyright 2011 John Wiley & Sons, Inc. 9Table Lookup of a StandardNormal Probability-3-2-10123PZ().010 3413 Z0.00 0.01 0.02 0.000.00000.00400.00800.100.03980.04380.04780.200.07930.08320.08711.000.34130.34380.34611.1

13、00.36430.36650.36861.200.38490.38690.3888Copyright 2011 John Wiley & Sons, Inc. 10Applying the Z FormulaX is normally distributed with = 485, and =105PXPZ()(.) .48560001103643For X = 485 ,Z=X-485 485105010. 1105485600-X=Z600, = XFor Z0.00 0.01 0.02 0.000.00000.00400.00800.100.03980.04380.04781.000.3

14、4130.34380.34611.100.36430.36650.36861.200.38490.38690.3888Copyright 2011 John Wiley & Sons, Inc. 11Applying the Z Formula7123.)56. 0()550(100= and 494,= with ddistributenormally is XZPXP56. 0100494550-X=Z550 = XFor Copyright 2011 John Wiley & Sons, Inc. 12Applying the Z Formula0197.)06. 2()700(100=

15、 and 494,= with ddistributenormally is XZPXP06. 2100494700-X=Z700 = XFor Copyright 2011 John Wiley & Sons, Inc. 13Applying the Z Formula94. 1100494300-X=Z300 = XFor 8292.)06. 194. 1()600300(100= and 494,= with ddistributenormally is XZPXP06. 1100494600-X=Z600 = XFor Copyright 2011 John Wiley & Sons,

16、 Inc. 14Demonstration Problem 6.9These types of problems can be solved quite easily with the appropriate technology. The output shows the MINITAB solution. Suppose we know that X is normally distributed with mean 3.58 and std dev 1.04, and we want P(X3.10172), we calculateCopyright 2011 John Wiley &

17、 Sons, Inc. 15Normal Approximation of theBinomial DistributionFor certain types of binomial distributions, thenormal distribution can be used to approximate the probabilitiesAt large sample sizes, binomial distributions approach the normal distribution in shape regardless of the value of pThe normal

18、 distribution is a good approximate for binomial distribution problems for large values of nCopyright 2011 John Wiley & Sons, Inc. 16Normal Approximation of Binomial:Parameter ConversionConversion equationsConversion example:n pn p q55. 3)70)(.30)(.60(18)30)(.60().30. and 60|25(find on,distributi bi

19、nomial a has X Given thatqpnpnpnXPCopyright 2011 John Wiley & Sons, Inc. 1765.28335. 7365.1018)55. 3(31830102030405060n70Normal Approximation of Binomial:Interval CheckCopyright 2011 John Wiley & Sons, Inc. 18Normal Approximation of Binomial:Correcting for Continuity Values Being DeterminedCorrectio

20、nX X X X X X +.50-.50-.50+.05-.50 and +.50+.50 and -.50The binomial probability , and is approximated by the normal probabilityP(X24.5| and P Xnp(|.).).256030183 55Copyright 2011 John Wiley & Sons, Inc. 19252627282930313233Total0.01670.00960.00520.00260.00120.00050.00020.00010.00000.0361XP(X)The normal approximation,P(X24.5| and 18355245 1