Bài giảng Statistical Techniques in Business and Economics - Chapter 7 The Normal probability distribution

Chapter 7THENormalPROBABILITYDISTRIBUTION1When you have completed this chapter, you will be able to:1. Explain how probabilities are assigned to a continuous random variable.2. Explain the characteristics of a normal probability distribution.3.Define and calculate z value corresponding to any observation on a normal distribution.and...Chapter Goals2Use the normal probability distribution to approximate the binomial probability distribution.5. Determine the probability a random observation is in a given interval on a normal distribution using the standard normal distribution. 4.Chapter Goals3the set of all the values in any interval is uncountable or infinite!.we will now study the class of continuous probability distributions.RecallContinuous Random Variable4Continuous can assume any value within a specified range!e.g. - Pressure in a tire - Weight of a pork chop - Height of students in a classQuantitativeNumericalObservations can be classified as either Discrete orContinuousCharacteristicsVariablesAlso Recall that5Continuous Random Variablethe Total sum of probabilities should always be 1!When dealing with a Continuous Random Variable we assume that the probability that the variable will take on any particular value is 0!Instead, Probabilities are assigned to intervals of values!6Continuous Probability DistributionsP(a<X<b)f(x)xabRange of Values 7Data:Heights of adult Canadian males n = 50, class interval = 2 n = 500, class interval = 1 n = 5000, class interval = 0.4 Probability functionRelative Frequency Histogram  Probability FunctionSee Histograms80.08 -0.07 -0.06 -0.05 -0.04 -0.03 -0.02 -0.01 -0.00 -155 165 175 185 195 Chart 7-1(A)(A) n = 50 150 160 170 180 1900.08 -0.07 -0.06 -0.05 -0.04 -0.03 -0.02 -0.01 -0.00 -Chart 7-1(B)(B) n = 5000.08 -0.07 -0.06 -0.05 -0.04 -0.03 -0.02 -0.01 -0.00 - 145 155 165 175 185 195Chart 7-1(C)(C) n = 50000.07 -0.06 -0.05 -0.04 -0.03 -0.02 -0.01 -0.00 -140 150 160 170 180 190 200Chart 7-1(D)Probability FunctionHistograms9A probability function can have any shape, as long as it is non-negative (the curve is above the x-axis, and the total area under the curve is 1)that all probability functions are not normal!Reminder10The arithmetic mean, median, and mode of the distribution are equal and located at the peak. The normal curve is bell-shapedNormal Probability DistributionCharacteristicsand has a single peak at the exact centre of the distribution.thus half the area under the curve is above the mean and half is below it.11Normal Probability DistributionCharacteristicsThe normal probability distribution is symmetrical about its mean.The normal probability distribution is asymptotic, i.e. the curve gets closer and closer to the x-axis but never actually touches it.12Mean, median, and mode are equalNormal Probability DistributionCharacteristicsSummaryNormal curve is symmetricalTheoretically, curve extends to infinity+13part of a “family” of curvesHow does the Standard Deviation values affect the shape of f(x)?Question= 2, 3, 4ss = 2s =3s =4Normal Probability Distribution14QuestionHow does the following Expected values affect the location of f(x)?m = 10, 11, 12m = 10m = 11m = 12part of a “family” of curvesNormal Probability Distribution15The Standard Normal Probability Distribution is a normal distribution with a mean of 0 and a standard deviation of 1. Also called the Z DistributionA z-value is the distance between a selected value (designated X) and the population mean, divided by the population Standard Deviation, . s- µ)=(XzFormula A z-value is, therefore, a location on the standard normal curve.16Total Area under the curve is 100% or 1The Standard Normal Probability Distributionsince the mean is located at the peak of the curve, half the area under the curve is above the mean andArea = +0.5Area = + 0.5half is below the mean17The bi-monthly starting salaries of recent MBA graduates follows the normal distribution with a mean of $3,000 and a standard deviation of $300. What is the z-value for a salary of $3,300?$300$3,000$3,300-=A z-value of 1 indicates that the value of $3,300 is one standard deviation above the mean of $3,000.QuestionThe Standard Normal Probability Distributionz = 1.00Show curvezFormula s- µ)=(Xz18 z = 0 z = 1.0$3000$3300A z-value of 1 indicates that the value of $3,300 is one standard deviation above the mean of $3,000.19The Standard Normal Probability DistributionThe bi-monthly starting salaries of recent MBA graduates follows the normal distribution with a mean of $3,000 and a standard deviation of $300. $300$3,000$2,550-=A z-value of –1.50 indicates that the value of $2,550 is one and a half (1.5) standard deviations below the mean of $3,000.QuestionZ = -1.50What is the z-value for a salary of $2,550?Show curvezFormula s- µ)=(Xz20 z = 0 z = -1.50A z-value of –1.50 indicates that the value of $2,550 is one and a half standard deviations below the mean of $3,000.$3000$2550The Standard Normal Probability Distribution21The Standard Normal Probability Distribution there is an infinite # of normal distributions, but only one table each distribution is determined by  &  work with  &  to calculate normal probabilities, by figuring out how many standard deviations is x away from the mean? Z becomes the metre stick for measuring22Understanding Standard Deviations230 20 40 50 60 80 100QuestionHow many  is 40 away from  ?If = 50  = 8+40 is 10 units(-10)away from 50If 1 standard deviation () is 8 units, then 10 units must be 1.25 SDUsing z40 is –1.25  away from 50Understanding Standard Deviations24The Standard Normal Probability Distributionthe Z DistributionA z-value is the distance between a selected value (designated X) and the population mean µ, divided by the population Standard Deviation, . Recall distance from the  divide by s to get zsm-=XzFormula 40 – 508z ==– 108= -1.25Note25The Standard Normal Probability Distributionthe Z DistributionNote+ z-valueRight of mean- z-valueLeft of mean0 20 40 50 60 80 100RecallThe mean µ, of a standard normal distribution is 0, and Standard Deviation  is 1.Question0Z-scale26The Standard Normal Probability DistributionQuestionCalculate P(50 < X < 60)0 20 40 50 60 80 100If = 50  = 8Transferring each value into z-scores,01.25Z1= (60-50)/8 = 1.25P(50 < X < 60)= P(0 < z < 1.25)z-tables- µ)=(Xz27Normal Distribution TableTo find the area between a z score of 0 and a score of 1.25Provides the z-score to 1 decimal placeProvides the second decimal place1.200.050.3944thus28The Standard Normal Probability Distributionthe Z Distribution01.250 20 40 50 60 80 100 0.3944 is the area between the mean and a positive z score 1.250.3944Therefore, the Probability of 50<X<60 is 39.44%New29andThe Standard Normal Probability DistributionQuestionCalculate P(40 < X < 60)If = 50  = 801.25-1.25Z1= (40-50)/8 = -1.25Z2= (60-50)/8 = 1.25Transferring each value into z-scores,0 20 40 50 60 80 100s- µ)=(Xz30The Standard Normal Probability Distribution= for a Total area of: 0.3944 + 0.3944 or 0.7888 Z2= 1.25 (determined earlier)0 20 40 50 60 80 10001.25-1.250.3944 because both sides are symmetrical, the left side (the area between the mean and a negative z score -1.25) must have the same area.. 0.3944.31The Standard Normal Probability Distributionthe Z DistributionQuestionCalculate P(40 < X < 60)0 20 40 50 60 80 100If = 50  = 8P(40 < X < 60) =NoteP(40  X  60)since P(X= 40 or 60) is zero32Using the Normal Distribution TablePractise Finding Areas33Practise using the Normal Distribution Table Look up 1.00 in TableLocate Area on the normal curve01.001Be sure to always sketch the curve, insert the given values and shade the required area.Notez-tableFindthe area between:Z1 = 0 and Z2 = 1.00 234Normal Distribution TableTo find the area between a z score of 0 and a score of 1.00Provides the z-score to 1 decimal placeProvides the second decimal place1.000.000.3413The Area between 0<Z<1.00 is 0.3413Therefore, the Probability of 0<Z<1.00 is 34.13%35Practise using the Normal Distribution Table 1.64The Area between 0<Z<1.64 is 0.4495Therefore, the Probability of 0<Z<1.64 is 44.95%Look up 1.64 in TableLocate Area on the normal curve12Findthe area between:Z1 = 0 and Z2 = 1.64036Practise using the Normal Distribution Table 0.47The Area between 0<Z<0.47 is 0.1808Therefore, the Probability of 0<Z<0.47 is 18.08%Look up 0.47 in TableLocate Area on the normal curve12Findthe area between:Z1 = 0 and Z2 = 0.47037Practise using the Normal Distribution Table -1.64Look up 1.64 in TableLocate Area on the normal curve12Findthe area between:Z1 = 0 and Z2 = -1.640The Area between -1.64<Z<0 is 0.4495Therefore, the Probability of -1.64<Z<0 is 44.95% Because of symmetry, this area is the same as between z of 0 and positive 1.64Note38Practise using the Normal Distribution Table -2.22Locate Area on the normal curve1Look up 2.22 in Table2Findthe area between:Z1 = 0 and Z2 = -2.220The Area between –2.22<Z<0 is 0.4868Therefore, the Probability of –2.22<Z<0 is 48.68% 2.2239 A z-value is a location on the standard normal curve! Therefore, a z-value(also called z-score) can have a positive or negative value! However, areas and probabilities are always positive values!Practise using the Normal Distribution Table 40Practise using the Normal Distribution Table -2.96Look up 2.96 in TableLocate Area on the normal curve12Findthe area between:Z1 = 0 and Z2 = -2.960The Area between –2.96<Z<0 is 0.4985Therefore, the Probability of –2.96<Z<0 is 49.85% 2.9641Practise using the Normal Distribution Table -1.65Look up 1.65 in TableLocate Area on the normal curve12Findthe area between:Z1 = -1.65 and Z2 = 1.650The Area (a1) between –1.65<Z<0 is 0.4505Therefore, the required Total Area is 0.9010 1.65The Area (a2) between 0<Z<1.65 is 0.4505Add together42Practise using the Normal Distribution Table Locate Area on the normal curve1Look up – 2.00 then 1.00 in Table2Findthe area between:Z1 = -2.00 and Z2 = 1.00The Area (a1) between –2.00<Z<0 is 0.4772Therefore, the required Total Area is 0.8185The Area (a2) between 0<Z<1.00 is 0.3413Add together-2.000 1.0043Locate Area on the normal curve1Look up – 0.44 then 1.96 in Table2Findthe area between:Z1 = -0.44 and Z2 = 1.96The Area (a1) between –0.44<Z<0 is 0.1700Therefore, the required Total Area is 0.6450The Area (a2) between 0<Z<1.96 is 0.4750Add together-0.440 1.96Practise using the Normal Distribution Table 44The Standard Normal Probability DistributionTotal Area under the curve is 100% or 1since the mean is located at the peak of the curve, half the area under the curve is above the mean andArea = 0.5Area = 0.5half is below the mean45a1Locate Area on the normal curve1Look up 2.00 in Table2Findthe area to the Left of Z1 = 2.00 The Area to the left of Z1 is a1+.5 = 0.4772 + 0.5Therefore, the required Total Area is 0.97720 2Area = 0.5The Area (a1) between 0<Z<2.0 is 0.47723Adjust as neededPractise using the Normal Distribution Table 46a1Locate Area on the normal curve1Look up 1.96 in Table2Findthe area to the Left of Z1 = 1.96 The Area to the left of Z1 is a1+.5= 0.4750 + 0.5Therefore, the required Total Area is 0.9750 1.96The Area (a1) between 0<Z<1.96 is 0.47503Adjust as neededPractise using the Normal Distribution Table 0Area = 0.547Area = 0.5Locate Area on the normal curve1Look up 1.64 in Table2Findthe area to the Left of Z1 = -1.64The Area to the left of Z1 is 0.5 - a1 = 0.5 - 0.4495Therefore, the required Total Area is 0.05050 -1.643Adjust as neededThe Area (a1) between –1.64<Z<0 is 0.4495Practise using the Normal Distribution Table a148Area = 0.5Locate Area on the normal curve1Look up 0.95 in Table20 -0.95The Area to the left of Z1 is 0.5 - a1 = 0.5 - 0.3289Therefore, the required Total Area is 0.17113Adjust as neededThe Area (a1) between –0.95<Z<0 is 0.3289Practise using the Normal Distribution Table Findthe area to the Left of Z1 = -0.95a149 Area = 0.5Locate Area on the normal curve1Look up 0.95 in Table20 -0.95The Area to the left of Z1 is 0.5 + a1 = 0.5 + 0.3289Therefore, the required Total Area is 0.82893Adjust as neededThe Area (a1) between –0.95<Z<0 is 0.3289 a1Practise using the Normal Distribution Table Findthe area to the Right of Z1 = -0.9550Locate Area on the normal curve1Look up 1.00 in Table20 1.00The Area to the left of Z1 is 0.5 - a1 = 0.5 - 0.3413Therefore, the required Total Area is 0.158703Adjust as neededThe Area (a1) between 0<Z<1.00 is 0.3413Practise using the Normal Distribution Table Findthe area to the Right of Z1 = 1.00Area = 0.5a1A51Locate Area on the normal curve1Findthe area between:2Look up 1.96 then 2.58 in TableThe Area (a1) between 0<Z<1.96 is 0.4750Therefore, the required Total Area is 0.0201The Area (a2) between 0<Z<2.58 is 0.4951Subtract3Adjust as neededPractise using the Normal Distribution Table 1.96 2.5800a2a1Z1 = 1.96 and Z2 = 2.58Find52Z1 = 0.55 and Z2 = 1.96Locate Area on the normal curve1Findthe area between:2Look up 0.55 then 1.96 in Table3Adjust as neededPractise using the Normal Distribution Table 1.96.550a2The Area (a1) between 0<Z<0.55 is 0.2088Therefore, the required Total Area is 0.2662The Area (a2) between 0<Z<1.96 is 0.4750Subtracta153 About 68 percent of the area under the normal curve is within one standard deviation of the mean.  +/-  About 95 percent is within two standard deviations of the mean.  +/- 2 Practically all are within three standard deviations of the mean.  +/- 3ExamplesAreas under the Normal CurveAreas under the Normal CurveBetween:95.44%99.74% - 1 s - 1 s - 2 s - 2 s - 3 s - 3 s68.26%55The daily water usage per person in Newmarket, Ontario is normally distributed with a mean of 80 litres and a standard deviation of 10 litres. Areas under the Normal Curve About 68% of the daily water usage will lie between 70 and 90 litresAbout 68 percent of those living in Newmarket will use how many litres of water?Probability thata person from Newmarket, selected at random, will use between 80 and 88 litres per day, is?10- 8080=0.00=sm)-=(Xz110- 8088=0.80=Locate Area on the normal curve12Calculate the Z scores8880Tablea1Z-scale 0.0 0.80sm)-=(Xz157Probability thata person from Newmarket, selected at random, will use between 80 and 88 litres per day, is?The area under a normal curve between a z-value of 80 and a z-value of 88 is 0.2881, therefore, we conclude that 28.81% of the residents will use between 80 to 88 litres per day! 3Look up 0.80 in TableA =.288158Probability thata person from Newmarket, selected at random, will use between 76 and 92 litres per day, is?10- 8076=0.40=10- 8092=1.20=Locate Area on the normal curve12Calculate the Z scores9276Tablea1a2Z-scale -.4 1.20sm)-=(Xz1sm)-=(Xz159Probability thata person from Newmarket, selected at random, will use between 76 and 92 litres per day, is?The area under a normal curve between a z-value of 76 and a z-value of 92 is 0.5403, therefore, we conclude that 54.03% of the residents will use between 76 to 92 litres per day! 3Look up 0.40 then 1.20 in Tablea2a1=.1554=.3849A =.5403Add togetherNew60Question Professor Mann has determined that the scores in his statistics course are approximately normally distributed with a mean of 72 and a standard deviation of 5. Let X be the score that separates an A from a B. SolutionHe announces to the class that the top 15 percent of the scores will earn an A. What is the lowest score a student can earn and still receive an A?72XAF D C BSketch given information onto the normal curve161To begin, let X be the score that separates an A from a B.If 15 percent of the students score more than X, then 35 percent must score between the mean of 72 and X Solutiona1=.3572XAA=.15Now, use the table “backwards” to find the z-score corresponding to a1=.35Table2Determine the Z score from the given areas62Normal Distribution TableSearch in the centre of the table for the area of 0.351.000.35080.04Substitution63a1=.3572XAA =.15mean of 72 and a standard deviation of 5.He announces to the class that the top 15 percent of the scores will earn an A. What is the lowest score a student can earn and still receive an A?Z-scale 0 1.045- 72X1.04=X = 1.04 * 5 + 72 = 77.2 A student needs a score of 77.2 to receive an A Solution3Substitute values into equationsm)-=(Xz164Using the Normal Curve to Approximate the Binomial Distribution65The Normal Approximation to the BinomialThe normal distribution (a continuous distribution) yields a good approximation of the binomial distribution (a discrete distribution) for large values of n. Use when np and n(1- p ) are both greater than 5!CharacteristicsThe probability of success stays the same for each trial The trials are independent Interested in the number of successesThe experiment consists of n Bernoulli trialsThe outcomes are classified into one of two mutually exclusive categories, such as success or failureBinomial Experiment67Mean and Varianceof a Binomial Probability Distribution 2smnp=mFormula )1(pnp-=2sFormula From Chapter 6If .60 of students do their homework at night, find the probability that at least 50 students in a class of 80 did their homework last night.Qm = 80 * .60 = 48s 2 = 80 * .6 * .4 = 19.2Sketch given information onto the normal curve16849.5Locate 50 on the normal curve148If we want at least 50, we start at 49.5 on the normal curve!5050 students is represented by the narrow strip, not a single lineThis is adjustment of .5 is called the continuity correction factorfind the probability that at least 50 students in a class of 80 did their homework last night.Q69find the probability that at most 50 students in a class of 80 did their homework last night.QLocate 50 on the normal curve1If we want at most 50, we want to include all of 50, so we use 50.5 on the normal curve!50.5485049.54850Therefore if we want less than 50, we do not want to include all of 50, so we use 49.5 on the normal curveNEWMore continuity correction factor70The value .5 is subtracted or added, depending on the problem, to a selected value when a binomial probability distribution (a discrete probability distribution) is being approximated by a continuous probability distribution (the normal distribution)ExamplesContinuity Correction FactorThis is the mean of a binomial distribution A recent study by a marketing research firm showed that 15% of Canadian households owned a video camera. For a sample of 200 homes, how many of the homes would you expect to have video cameras? What is the variance?What is the standard deviation?5.25=s0498.5=Questions2 = np(1 – p) = (30)(1 - .15) = 25.5m= np =( .15)(200) = 30a1What is the probability that less than 40 homes in the sample have video cameras? “less than” does not include all of 40 – so use 39.5The likelihood that less than 40 of the 200 homes have a video camera is about 97%.3039.5Locate Area on the normal curve12Calculate the Z scores3Look up 1.88 in Table= .4699a1A =.96995.0498- 30 39.5=1.88=+ .5 =sm)-=(Xz173Test your learning www.mcgrawhill.ca/college/lindClick onOnline Learning Centrefor quizzesextra contentdata setssearchable glossaryaccess to Statistics Canada’s E-Stat dataand much more!74This completes Chapter 775