Slide 1Test 4Practice ProblemsAdditional Reading and ExamplesSlide 5NotationStandard Normal DistributionProbabilityProbabilityZ-Score TransformationZ-Score TransformationProbability Problems on CalculatorStandard Normal DistributionLess Than ProblemValue Problems on the CalculatorExample 42Example 43Greater Than ProblemExample 44Example 45Between ProblemBetween ProblemBetween ProblemBetween ProblemBetween ProblemBetween ProblemBetween ProblemBetween ProblemPractice ProblemsPractice Problem AnswersZ-Score TransformationZ-Score TransformationZ-Score TransformationZ-Score TransformationZ-Score TransformationExample 49Example 49Example 49Example 49Example 49Example 49Example 50Example 50Example 50Example 50Example 50Example 50Example 50Example 50Practice ProblemsPractice Problem AnswersSteps for Solving Application ProblemsExample 51Example 51Example 51Example 51Example 51Example 51Example 52Example 52Example 52Example 52Example 52Example 52Example 52Example 53Example 53Example 53Example 53Example 53Example 53Example 54Example 54Example 54Example 54Example 54STAT 210Lecture 21Normal DistributionsOctober 13, 2017Test 4Wednesday, October 18Covers chapter VI, pages 139 – 168Combination of short answer questions and problemsFormulas and normal table provided, please bring calculator and writing instrument.Practice ProblemsPages 162 through 165Relevant problems: VI.6 through VI.15Recommended problems: VI.14 and VI.15Additional Reading and ExamplesRead pages 158 through 160TOP HAT 2NotationX ~ N (m , s)“X is distributed normal with mean m and standard deviation s”Standard Normal Distribution•Denoted by Z•Has population mean m = 0 (center)•Has population standard deviation s = 1 (spread)•Shape is normal (symmetric bell curve)•No unusual features•Z ~ N(0, 1)•Probabilities are tabled on pages 338 - 339ProbabilityThe normal table gives the probability that the standard normal variable Z falls below some specified value z (less than problems).Read the value of z down the left-most column and across the top row, and read the probability from the body of the table.ProbabilitySuppose X is distributed normal with some mean m not equal to 0 and/or some standarddeviation s not equal to 1:X ~ N(m, s)Z-Score TransformationWe convert to a standard normal variable Z ~ N(0, 1).Z = X - m = value - mean s standard deviationZ-Score TransformationP(a < X < b) = P( a - m < X - m < b - m ) s s s = P( a - m < Z < b - m ) s sOnce converted from X to Z, the standard normal table on pages 338 and 339 is used to find the probability, just as in Examples 34 through 41.Probability Problems on Calculator1. Hit 2nd, then VARS – this gives a list of distributions.2. Choose option 2: normalcdf(3. Enter four numbers: the lower number of the interval, the upper number of the interval, the mean, and the standard deviation.Standard Normal DistributionAll of the problems so far asked us to find the probability given a value or values of Z.Now suppose the probability (or area or proportion or percentage) is given, and we want to find the corresponding value of Z (see page 145)There are three such problems.Less Than ProblemSuppose you want to find the value z such that the probability of being less than z (or less than and equal to z) is as specified.To solve: 1. Draw a normal curve and mark the information stated in the problem.2. In the normal table, find the specified less than probability in the body of the table and then read across and up to determine the appropriate z value.Value Problems on the CalculatorSee pages 160 and 161 for instructions for using thecalculator to determine normal probabilities and valuesof normal variables.1. Hit 2nd, then VARS – this gives a list of distributions2. Choose option 3: invNorm3. You must enter three numbers: the LESS THAN probability, then the mean (which is currently 0) and then the standard deviation (which is currently 1).Example 42Find the value of z such that the probability of being less than z is .8212.1. z: P(Z < z) = .8212 2. In the normal table, find .8212 in the body of the table. This corresponds to z = 0.92Calculator: invNorm(.8212, 0, 1)Example 43Find the value of z such that the probability of being less than z is .10.1. z: P(Z < z) = .10 2. In the normal table, find .10 in the body of the table. Closest is .1003, corresponding to z = -1.28Calculator: invNorm(.10, 0, 1)z= 10.10Greater Than ProblemSuppose you want to find the value z such that the probability of being greater than z (or greater than and equal to z) is as specified.To solve: 1. Draw a normal curve and mark the information stated in the problem.2. Convert the problem from a greater than problem to a less than problem by subtracting the given probability from 1.3. In the normal table, find the specified less than probability in the body of the table and then read across and up to determine the appropriate z value.Example 44Find the value of z such that the probability of being greater than z is .33.1. z: P(Z > z) = .332. z: P(Z < z) = 1 - .33 = .673. From the normal table, find .67 in the body of the table.p = .67 corresponds to z = 0.44Calculator: invNorm(.67, 0, 1)z=1.67.330Example 45Find the value of z such that the probability of being greater than z is .57.1. z: P(Z > z) = .572. z: P(Z < z) = 1 - .57 = .433. From the normal table, find .43 in the body of the table.p = .43 corresponds to z = -0.18Calculator: invNorm(.43, 0, 1)Between ProblemSuppose you want to find the values z1 and z2 such that the probability of being between the two values is as specified.For example, find the two values z1 and z2 such that the probability of being between the two values is .60.Between ProblemFind the two values z1 and z2 such that the probability of being between the two values is .60. Since the probability of being between the two values is .60, then the total probability outside (less than z1 and greater than z2) is 1 - .60 = .40.Between ProblemFind the two values z1 and z2 such that the probability of being between the two values is .60. Since the probability of being between the two values is .60, then the total probability outside (less than z1 and greater than z2) is 1 - .60 = .40.Since the distribution is symmetric, this probability is split evenly between being less than z1 and being greater than z2.Between ProblemSince the probability of being between the two values is .60, then the total
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