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Chapter 6
The Normal Distribution and Other Continuous DistributionsStatistics for ManagersUsing Microsoft Excel 4th Edition
Chapter GoalsAfter completing this chapter, you should be able to: Describe the characteristics of the normal distributionTranslate normal distribution problems into standardized normal distribution problemsFind probabilities using a normal distribution tableEvaluate the normality assumptionRecognize when to apply the uniform and exponential distributions
Chapter GoalsAfter completing this chapter, you should be able to: Define the concept of a sampling distributionDetermine the mean and standard deviation for the sampling distribution of the sample mean, XDetermine the mean and standard deviation for the sampling distribution of the sample proportion, psDescribe the Central Limit Theorem and its importanceApply sampling distributions for both X and ps__(continued)
Probability DistributionsContinuous Probability DistributionsBinomialHypergeometricPoissonProbability DistributionsDiscrete Probability DistributionsNormalUniformExponentialCh. 5Ch. 6
Continuous Probability DistributionsA continuous random variable is a variable that can assume any value on a continuum (can assume an uncountable number of values)thickness of an itemtime required to complete a tasktemperature of a solutionheight, in inches
These can potentially take on any value, depending only on the ability to measure accurately.
The Normal DistributionProbability DistributionsNormalUniformExponentialContinuous Probability Distributions
The Normal Distribution Bell Shaped Symmetrical 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)
By varying the parameters and , we obtain different normal distributionsMany Normal Distributions
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The Normal Distribution ShapeXf(X)Changing shifts the distribution left or right.Changing increases or decreases the spread.
The Normal Probability Density FunctionThe 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 variable
The Standardized NormalAny normal distribution (with any mean and standard deviation combination) can be transformed into the standardized normal distribution (Z)
Need to transform X units into Z units
Translation to the Standardized Normal DistributionTranslate from X to the standardized normal (the Z distribution) by subtracting the mean of X and dividing by its standard deviation:Z always has mean = 0 and standard deviation = 1
The Standardized Normal Probability Density FunctionThe 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 distribution
The Standardized Normal DistributionAlso known as the Z distributionMean is 0Standard Deviation is 1Zf(Z)01Values above the mean have positive Z-values, values below the mean have negative Z-values
ExampleIf X is distributed normally with mean of 100 and standard deviation of 50, the Z value for X = 200 is
This says that X = 200 is two standard deviations (2 increments of 50 units) above the mean of 100.
Comparing X and Z unitsZ1002.00200XNote that the distribution is the same, only the scale has changed. We can express the problem in original units (X) or in standardized units (Z)( = 100, = 50)( = 0, = 1)
Probability as Area Under the Curvef(X)X0.50.5The total area under the curve is 1.0, and the curve is symmetric, so half is above the mean, half is below
Empirical Rules 1 encloses about 68% of Xsf(X)X+1-1What can we say about the distribution of values around the mean? There are some general rules:68.26%
The Empirical Rule 2 covers about 95% of Xs 3 covers about 99.7% of Xsx22x3395.44%99.72%(continued)
The Standardized Normal Table The Standardized Normal table in the textbook (Appendix table E.2) gives the probability less than a desired value for Z (i.e., from negative infinity to Z)
Z02.00.9772Example: P(Z < 2.00) = .9772
The Standardized Normal Table The value within the table gives the probability from Z = up to the desired Z value.97722.0P(Z < 2.00) = .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
0.00.1
General Procedure for Finding Probabilities Draw the normal curve for the problem in terms of X
Translate X-values to Z-values
Use the Standardized Normal TableTo find P(a < X < b) when X is distributed normally:
Finding Normal ProbabilitiesSuppose X is normal with mean 8.0 and standard deviation 5.0Find P(X < 8.6)X8.68.0
Suppose X is normal with mean 8.0 and standard deviation 5.0. Find P(X < 8.6)
Z0.12 0X8.6 8 = 8 = 10 = 0 = 1(continued)Finding Normal ProbabilitiesP(X < 8.6)P(Z < 0.12)
Solution: Finding P(Z < 0.12)Z0.12Z.00.010.0.5000.5040.5080.5398.54380.2.5793.5832.58710.3.6179.6217.6255.5478.020.1.5478Standardized Normal Probability Table (Portion)0.00= P(Z < 0.12)P(X < 8.6)
Upper Tail ProbabilitiesSuppose X is normal with mean 8.0 and standard deviation 5.0. Now Find P(X > 8.6)X8.68.0
Upper Tail ProbabilitiesNow Find P(X > 8.6)(continued)Z0.12 0Z0.12.5478 01.0001.0 - .5478 = .4522 P(X > 8.6) = P(Z > 0.12) = 1.0 - P(Z 0.12) = 1.0 - .5478 = .4522
Probability Between Two ValuesSuppose X is normal with mean 8.0 and standard deviation 5.0. Find P(8 < X < 8.6) P(8 < X < 8.6)= P(0 < Z < 0.12)Z0.12 0X8.6 8Calculate Z-values:
Solution: Finding P(0 < Z < 0.12)Z0.12.04780.00= P(0 < Z < 0.12)P(8 < X < 8.6)= P(Z < 0.12) P(Z 0)= .5478 - .5000 = .0478.5000Z.00.010.0.5000.5040.5080.5398.54380.2.5793.5832.58710.3.6179.6217.6255.020.1.5478Standardized Normal Probability Table (Portion)
Probabilities in the Lower Tail Suppose X is normal with mean 8.0 and standard deviation 5.0. Now Find P(7.4 < X < 8)X7.48.0
Probabilities in the Lower Tail Now Find P(7.4 < X < 8)X7.48.0 P(7.4 < X < 8) = P(-0.12 < Z < 0)= P(Z < 0) P(Z -0.12)= .5000 - .4522 = .0478(continued).0478.4522Z-0.12 0The Normal distribution is symmetric, so this probability is the same as P(0 < Z < 0.12)
Finding the X value for a Known ProbabilitySteps to find the X value for a known probability:1. Find the Z value for the known probability2. Convert to X units using the formula:
Finding the X value for a Known ProbabilityExample:Suppose X is normal with mean 8.0 and standard deviation 5.0. Now find the X value so that only 20% of all values are below this XX?8.0.2000Z? 0(continued)
Find the Z value for 20% in the Lower Tail20% 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 Probability Table (Portion).05.1711.1977.2266X?8.0.2000Z-0.84 01. Find the Z value for the known probability
Finding the X value2. Convert to X units using the formula:So 20% of the values from a distribution with mean 8.0 and standard deviation 5.0 are less than 3.80
Assessing NormalityNot all continuous random variables are normally distributedIt is important to evaluate how well the data set is approximated by a normal distribution
Assessing NormalityConstruct charts or graphsFor small- or moderate-sized data sets, do stem-and-leaf display and box-and-whisker plot look symmetric?For large data sets, does the histogram or polygon appear bell-shaped?Compute descriptive summary measuresDo the mean, median and mode have similar values?Is the interquartile range approximately 1.33 ?Is the range approximately 6 ?(continued)
Assessing NormalityObserve the distribution of the data setDo approximately 2/3 of the observations lie within mean 1 standard deviation?Do approximately 80% of the observations lie within mean 1.28 standard deviations?Do approximately 95% of the observations lie within mean 2 standard deviations?Evaluate normal probability plotIs the normal probability plot approximately linear with positive slope?(continued)
The Normal Probability PlotNormal probability plotArrange data into ordered arrayFind corresponding standardized normal quantile valuesPlot the pairs of points with observed data values on the vertical axis and the standardized normal quantile values on the horizontal axisEvaluate the plot for evidence of linearity
The Normal Probability PlotA normal probability plot for data from a normal distribution will be approximately linear:306090-2-1012ZX(continued)
Normal Probability PlotLeft-SkewedRight-SkewedRectangular306090-2-1012ZX(continued)306090-2-1012ZX306090-2-1012ZXNonlinear plots indicate a deviation from normality
The Uniform DistributionContinuous Probability DistributionsProbability DistributionsNormalUniformExponential
The Uniform DistributionThe uniform distribution is a probability distribution that has equal probabilities for all possible outcomes of the random variable
Also called a rectangular distribution
The Uniform DistributionThe Continuous Uniform Distribution:wheref(X) = value of the density function at any X valuea = minimum value of Xb = maximum value of X(continued)f(X) =
Properties of the Uniform DistributionThe mean of a uniform distribution is
The standard deviation is
Uniform Distribution ExampleExample: Uniform probability distribution over the range 2 X 6:26.25f(X) = = .25 for 2 X 66 - 21Xf(X)
The Exponential DistributionContinuous Probability DistributionsProbability DistributionsNormalUniformExponential
The Exponential DistributionUsed to model the length of time between two occurrences of an event (the time between arrivals)
Examples: Time between trucks arriving at an unloading dockTime between transactions at an ATM MachineTime between phone calls to the main operator
The Exponential DistributionDefined by a single parameter, its mean (lambda)The probability that an arrival time is less than some specified time X iswhere e = mathematical constant approximated by 2.71828 = the population mean number of arrivals per unitX = any value of the continuous variable where 0 < X <
Exponential Distribution ExampleExample: Customers arrive at the service counter at the rate of 15 per hour. What is the probability that the arrival time between consecutive customers is less than three minutes?The mean number of arrivals per hour is 15, so = 15Three minutes is .05 hoursP(arrival time < .05) = 1 e-X = 1 e-(15)(.05) = .5276So there is a 52.76% probability that the arrival time between successive customers is less than three minutes
Sampling DistributionsSampling DistributionsSampling Distributions of the MeanSampling Distributions of the Proportion
Sampling DistributionsA sampling distribution is a distribution of all of the possible values of a statistic for a given size sample selected from a population
Developing a Sampling DistributionAssume there is a population Population size N=4Random variable, X, is age of individualsValues of X: 18, 20, 22, 24 (years)ABCD
Developing a Sampling Distribution.3.2.1 0 18 20 22 24 A B C DUniform DistributionP(x)x(continued)Summary Measures for the Population Distribution:
Now consider all possible samples of size n=216 possible samples (sampling with replacement)(continued)Developing a Sampling Distribution16 Sample Means
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Sampling Distribution of All Sample Means18 19 20 21 22 23 240 .1 .2 .3 P(X) XSample Means Distribution16 Sample Means_Developing a Sampling Distribution(continued)(no longer uniform)_
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Summary Measures of this Sampling Distribution:Developing aSampling Distribution(continued)
Comparing the Population with its Sampling Distribution18 19 20 21 22 23 240 .1 .2 .3 P(X) X 18 20 22 24 A B C D0 .1 .2 .3 PopulationN = 4P(X) X_Sample Means Distributionn = 2_
Sampling Distributions of the MeanSampling DistributionsSampling Distributions of the MeanSampling Distributions of the Proportion
Standard Error of the MeanDifferent samples of the same size from the same population will yield different sample meansA measure of the variability in the mean from sample to sample is given by the Standard Error of the Mean:
Note that the standard error of the mean decreases as the sample size increases
If the Population is NormalIf a population is normal with mean and standard deviation , the sampling distribution of is also normally distributed with
and
(This assumes that sampling is with replacement or sampling is without replacement from an infinite population)
Z-value for Sampling Distributionof the MeanZ-value for the sampling distribution of :where:= sample mean= population mean= population standard deviation n = sample size
Finite Population CorrectionApply the Finite Population Correction if:the sample is large relative to the population (n is greater than 5% of N)andSampling is without replacement
Then
Sampling Distribution Properties
(i.e. is unbiased )
Normal Population DistributionNormal Sampling Distribution (has the same mean)
Sampling Distribution PropertiesFor sampling with replacement: As n increases, decreasesLarger sample sizeSmaller sample size(continued)
If the Population is not NormalWe can apply the Central Limit Theorem:Even if the population is not normal,sample means from the population will be approximately normal as long as the sample size is large enough.
Properties of the sampling distribution:
and
Central Limit TheoremnAs the sample size gets large enough the sampling distribution becomes almost normal regardless of shape of population
If the Population is not NormalPopulation DistributionSampling Distribution (becomes normal as n increases)Central TendencyVariation(Sampling with replacement)Larger sample sizeSmaller sample size(continued)Sampling distribution properties:
How Large is Large Enough?For most distributions, n > 30 will give a sampling distribution that is nearly normalFor fairly symmetric distributions, n > 15For normal population distributions, the sampling distribution of the mean is always normally distributed
ExampleSuppose a population has mean = 8 and standard deviation = 3. Suppose a random sample of size n = 36 is selected.
What is the probability that the sample mean is between 7.8 and 8.2?
ExampleSolution:Even if the population is not normally distributed, the central limit theorem can be used (n > 30) so the sampling distribution of is approximately normal with mean = 8 and standard deviation
(continued)
Example Solution (continued):
(continued)Z7.8 8.2-0.5 0.5Sampling DistributionStandard Normal Distribution.1915 +.1915Population Distribution????????????SampleStandardizeX
Sampling Distributions of the ProportionSampling DistributionsSampling Distributions of the MeanSampling Distributions of the Proportion
Population Proportions, p p = the proportion of the population having some characteristicSample proportion ( ps ) provides an estimate of p:
0 ps 1ps has a binomial distribution(assuming sampling with replacement from a finite population or without replacement from an infinite population)
Sampling Distribution of pApproximated by a normal distribution if:
where and(where p = population proportion)Sampling DistributionP( ps).3.2.1 0 0 . 2 .4 .6 8 1ps
Z-Value for ProportionsIf sampling is without replacement and n is greater than 5% of the population size, then must use the finite population correction factor:Standardize ps to a Z value with the formula:
ExampleIf the true proportion of voters who support Proposition A is p = .4, what is the probability that a sample of size 200 yields a sample proportion between .40 and .45?i.e.: if p = .4 and n = 200, what is P(.40 ps .45) ?
Example if p = .4 and n = 200, what is P(.40 ps .45) ?(continued)Find : Convert to standard normal:
Example if p = .4 and n = 200, what is P(.40 ps .45) ?Z.451.44.4251StandardizeSampling DistributionStandardized Normal Distribution(continued)Use standard normal table: P(0 Z 1.44) = .4251.400ps
Chapter SummaryPresented key continuous distributions normal, uniform, exponentialFound probabilities using formulas and tablesRecognized when to apply different distributions Applied distributions to decision problems
Chapter SummaryIntroduced sampling distributionsDescribed the sampling distribution of the meanFor normal populationsUsing the Central Limit TheoremDescribed the sampling distribution of a proportionCalculated probabilities using sampling distributions(continued)
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