1、The Normal Distribution Chapter 6ObjectivesIn this chapter,you learn:nTo compute probabilities from the normal distributionnHow to use the normal distribution to solve business problemsnTo use the normal probability plot to determine whether a set of data is approximately normally distributedContinu
2、ous Probability DistributionsnA continuous variable is a variable that can assume any value on a continuum(can assume an uncountable number of values)nthickness of an itemntime required to complete a taskntemperature of a solutionnheight,in inchesnThese can potentially take on any value depending on
3、ly on the ability to precisely and accurately measureThe Normal Distributionn Bell Shapedn Symmetrical n Mean,Median and Mode are EqualLocation is determined by the mean,Spread is determined by the standard deviation,The random variable has an infinite theoretical range:+to Mean=Median=ModeXf(X)The
4、Normal DistributionDensity Function2)(X21e21f(X)nThe formula for the normal probability density function isWheree=the mathematical constant approximated by 2.71828=the mathematical constant approximated by 3.14159=the population mean=the population standard deviationX=any value of the continuous var
5、iableBy varying the parameters and,we obtain different normal distributionsABCA and B have the same mean but different standard deviations.B and C have different means and different standard deviations.The Normal Distribution ShapeXf(X)Changing shifts the distribution left or right.Changing increase
6、s or decreases the spread.The Standardized NormalnAny normal distribution(with any mean and standard deviation combination)can be transformed into the standardized normal distribution(Z)nTo compute normal probabilities need to transform X units into Z unitsnThe standardized normal distribution(Z)has
7、 a mean of 0 and a standard deviation of 1Translation to the Standardized Normal DistributionnTranslate from X to the standardized normal(the“Z”distribution)by subtracting the mean of X and dividing by its standard deviation:XZThe Z distribution always has mean=0 and standard deviation=1The Standard
8、ized Normal Probability Density FunctionnThe formula for the standardized normal probability density function isWheree=the mathematical constant approximated by 2.71828=the mathematical constant approximated by 3.14159Z=any value of the standardized normal distribution2(1/2)Ze21f(Z)The Standardized
9、Normal DistributionnAlso known as the“Z”distributionnMean is 0nStandard Deviation is 1Zf(Z)01Values above the mean have positive Z-values.Values below the mean have negative Z-values.ExamplenIf X is distributed normally with mean of$100 and standard deviation of$50,the Z value for X=$200 isnThis say
10、s that X=$200 is two standard deviations(2 increments of$50 units)above the mean of$100.2.0$50100$200XZComparing X and Z unitsNote that the shape of the distribution is the same,only the scale has changed.We can express the problem in the original units(X in dollars)or in standardized units(Z)Z$100
11、2.00$200$X(=$100,=$50)(=0,=1)Finding Normal Probabilities Probability is measured by the area under the curveabXf(X)P aXb()P aXb()=(Note that the probability of any individual value is zero)Probability as Area Under the CurveThe total area under the curve is 1.0,and the curve is symmetric,so half is
12、 above the mean,half is belowf(X)X0.50.51.0)XP(0.5)XP(0.5)XP(The Standardized Normal TablenThe Cumulative Standardized Normal table in the textbook(Appendix table E.2)gives the probability less than a desired value of Z(i.e.,from negative infinity to Z)Z02.000.9772Example:P(Z 2.00)=0.9772The Standar
13、dized Normal Table The value within the table gives the probability from Z=up to the desired Z value.97722.0P(Z 2.00)=0.9772 The row shows the value of Z to the first decimal point The column gives the value of Z to the second decimal point2.0.(continued)Z 0.00 0.01 0.02 0.00.1General Procedure for
14、Finding Normal Probabilitiesn Draw the normal curve for the problem in terms of Xn Translate X-values to Z-valuesn Use the Standardized Normal TableTo find P(a X b)when X is distributed normally:Finding Normal ProbabilitiesnLet X represent the time it takes(in seconds)to download an image file from
15、the internet.nSuppose X is normal with a mean of18.0 seconds and a standard deviation of 5.0 seconds.Find P(X 18.6)18.6X18.0nLet X represent the time it takes,in seconds to download an image file from the internet.nSuppose X is normal with a mean of 18.0 seconds and a standard deviation of 5.0 secon
16、ds.Find P(X 18.6)Z0.12 0X18.6 18=18 =5=0=1(continued)Finding Normal Probabilities0.125.08.0118.6XZP(X 18.6)P(Z 0.12)Z0.12Solution:Finding P(Z 0.12)0.5478Standardized Normal Probability Table(Portion)0.00=P(Z 0.12)P(X 18.6)X18.618.0nNow Find P(X 18.6)(continued)Z0.12 0Z0.120.5478 01.0001.0-0.5478=0.4
17、522 P(X 18.6)=P(Z 0.12)=1.0-P(Z 0.12)=1.0-0.5478=0.4522Finding NormalUpper Tail ProbabilitiesFinding a Normal Probability Between Two ValuesnSuppose X is normal with mean 18.0 and standard deviation 5.0.Find P(18 X 18.6)P(18 X 18.6)=P(0 Z 0.12)Z0.12 0X18.6 18058118XZ0.1258118.6XZCalculate Z-values:Z
18、0.12Solution:Finding P(0 Z 0.12)0.04780.00=P(0 Z 0.12)P(18 X 18.6)=P(Z 0.12)P(Z 0)=0.5478-0.5000=0.04780.5000Z.00.010.0.5000.5040.5080.5398.54380.2.5793.5832.58710.3.6179.6217.6255.020.1.5478Standardized Normal Probability Table(Portion)nSuppose X is normal with mean 18.0 and standard deviation 5.0.
19、nNow Find P(17.4 X 18)X17.418.0Probabilities in the Lower Tail Probabilities in the Lower Tail Now Find P(17.4 X 18)X17.4 18.0 P(17.4 X 18)=P(-0.12 Z 0)=P(Z 0)P(Z -0.12)=0.5000-0.4522=0.0478(continued)0.04780.4522Z-0.12 0The Normal distribution is symmetric,so this probability is the same as P(0 Z 0
20、.12)Empirical Rule 1 encloses about 68.26%of Xsf(X)X+1-1What can we say about the distribution of values around the mean?For any normal distribution:68.26%The Empirical Rulen 2 covers about 95.44%of Xsn 3 covers about 99.73%of Xsx22x3395.44%99.73%(continued)nSteps to find the X value for a known pro
21、bability:1.Find the Z value for the known probability2.Convert to X units using the formula:Given a Normal ProbabilityFind the X ValueZXFinding the X value for a Known ProbabilityExample:nLet X represent the time it takes(in seconds)to download an image file from the internet.nSuppose X is normal wi
22、th mean 18.0 and standard deviation 5.0nFind X such that 20%of download times are less than X.X?18.00.2000Z?0(continued)Find the Z value for 20%in the Lower Tailn20%area in the lower tail is consistent with a Z value of-0.84Z.03-0.9.1762.1736.2033-0.7.2327.2296.04-0.8.2005Standardized Normal Probabi
23、lity Table(Portion).05.1711.1977.2266X?18.00.2000Z-0.84 01.Find the Z value for the known probability2.Convert to X units using the formula:Finding the X value8.130.5)84.0(0.18ZXSo 20%of the values from a distribution with mean 18.0 and standard deviation 5.0 are less than 13.80Both Minitab&Excel Ca
24、n Be Used To Find Normal ProbabilitiesFind P(X 9)where X is normal with a mean of 7and a standard deviation of 2 ExcelMinitabEvaluating NormalitynNot all continuous distributions are normalnIt is important to evaluate how well the data set is approximated by a normal distribution.nNormally distribut
25、ed data should approximate the theoretical normal distribution:nThe normal distribution is bell shaped(symmetrical)where the mean is equal to the median.nThe empirical rule applies to the normal distribution.nThe interquartile range of a normal distribution is 1.33 standard deviations.Evaluating Nor
26、malityComparing data characteristics to theoretical propertiesnConstruct charts or graphsnFor small-or moderate-sized data sets,construct a stem-and-leaf display or a boxplot to check for symmetrynFor large data sets,does the histogram or polygon appear bell-shaped?nCompute descriptive summary measu
27、resnDo the mean,median and mode have similar values?nIs the interquartile range approximately 1.33?nIs the range approximately 6?(continued)Evaluating NormalityComparing data characteristics to theoretical propertiesn Observe the distribution of the data setnDo approximately 2/3 of the observations
28、lie within mean 1 standard deviation?nDo approximately 80%of the observations lie within mean 1.28 standard deviations?nDo approximately 95%of the observations lie within mean 2 standard deviations?n Evaluate normal probability plotnIs the normal probability plot approximately linear(i.e.a straight
29、line)with positive slope?(continued)ConstructingA Normal Probability PlotnNormal probability plotnArrange data into ordered arraynFind corresponding standardized normal quantile values(Z)nPlot the pairs of points with observed data values(X)on the vertical axis and the standardized normal quantile v
30、alues(Z)on the horizontal axisnEvaluate the plot for evidence of linearityA normal probability plot for data from a normal distribution will be approximately linear:306090-2-1012ZXThe Normal Probability PlotInterpretationNormal Probability PlotInterpretationLeft-SkewedRight-SkewedRectangular306090-2
31、-1 012ZX(continued)306090-2-1 012ZX306090-2-1 012ZXNonlinear plots indicate a deviation from normalityEvaluating NormalityAn Example:Mutual Fund ReturnsThe boxplot is skewed to the right.(The normal distribution is symmetric.)Evaluating NormalityAn Example:Mutual Fund ReturnsDescriptive Statistics(c
32、ontinued)nThe mean(21.84)is approximately the same as the median(21.65).(In a normal distribution the mean and median are equal.)nThe interquartile range of 6.98 is approximately 1.09 standard deviations.(In a normal distribution the interquartile range is 1.33 standard deviations.)nThe range of 59.
33、52 is equal to 9.26 standard deviations.(In a normal distribution the range is 6 standard deviations.)n77.04%of the observations are within 1 standard deviation of the mean.(In a normal distribution this percentage is 68.26%.)n86.79%of the observations are within 1.28 standard deviations of the mean
34、.(In a normal distribution this percentage is 80%.)n96.86%of the observations are within 2 standard deviations of the mean.(In a normal distribution this percentage is 95.44%.)nThe skewness statistic is 1.698 and the kurtosis statistic is 8.467.(In a normal distribution,each of these statistics equa
35、ls zero.)Evaluating Normality Via ExcelAn Example:Mutual Fund Returns(continued)Plot is not a straight line and shows the distribution is skewed to the right.(The normal distribution appears as a straight line.)Excel(quantile-quantile)normal probability plotEvaluating Normality Via MinitabAn Example
36、:Mutual Fund Returns(continued)Plot is not a straight line,rises quickly in the beginning,rises slowly at the end and shows the distribution is skewed to the right.Normal Probability Plot From Minitab70605040302010099.99995908070605040302010510.13YrReturn%PercentProbability Plot of 3YrReturn%Normal
37、Evaluating NormalityAn Example:Mutual Fund ReturnsnConclusionsnThe returns are right-skewednThe returns have more values concentrated around the mean than expectednThe range is larger than expectednNormal probability plot is not a straight linenOverall,this data set greatly differs from the theoreti
38、cal properties of the normal distribution(continued)Chapter SummaryIn this chapter we discussed:nComputing probabilities from the normal distributionnUsing the normal distribution to solve business problemsnUsing the normal probability plot to determine whether a set of data is approximately normally distributed
