# VCU STAT 210 - Lecture21 (76 pages)

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## Lecture21

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- Pages:
- 76
- School:
- Virginia Commonwealth University
- Course:
- Stat 210 - Basic Practice of Statistics

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STAT 210 Lecture 21 Normal Distributions October 13 2017 Test 4 Wednesday October 18 Covers chapter VI pages 139 168 Combination of short answer questions and problems Formulas and normal table provided please bring calculator and writing instrument Practice Problems Pages 162 through 165 Relevant problems VI 6 through VI 15 Recommended problems VI 14 and VI 15 Additional Reading and Examples Read pages 158 through 160 TOP HAT 2 Notation X 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 338 339 Probability The 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 Probability Suppose X is distributed normal with some mean m not equal to 0 and or some standard deviation s not equal to 1 X N m s Z Score Transformation We convert to a standard normal variable Z N 0 1 Z X m value mean s standard deviation Z Score Transformation P a X b P a m X m b m s P a m s s s Z b m s Once 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 Calculator 1 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 Distribution All 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 Problem Suppose 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 Calculator See pages 160 and 161 for instructions for using the calculator to determine normal probabilities and values of normal variables 1 Hit 2nd then VARS this gives a list of distributions 2 Choose option 3 invNorm 3 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 42 Find 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 92 Calculator invNorm 8212 0 1 Example 43 Find the value of z such that the probability of being less than z is 10 1 z P Z z 10 1 10 z 0 2 In the normal table find 10 in the body of the table Closest is 1003 corresponding to z 1 28 Calculator invNorm 10 0 1 Greater Than Problem Suppose 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 44 Find the value of z such that the probability of being greater than z is 33 1 z P Z z 33 1 67 2 z P Z z 1 33 67 33 0 z 3 From the normal table find 67 in the body of the table p 67 corresponds to z 0 44 Calculator invNorm 67 0 1 Example 45 Find the value of z such that the probability of being greater than z is 57 1 z P Z z 57 2 z P Z z 1 57 43 3 From the normal table find 43 in the body of the table p 43 corresponds to z 0 18 Calculator invNorm 43 0 1 Between Problem Suppose you want to find the values z 1 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 Problem Find 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 Problem Find 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 z 1 and being greater than z2 Between Problem Since the probability of being between the two values is 60 then the total probability outside less than z 1 and greater than z2 is 1 60 40 Since the distribution is symmetric this probability is split evenly between being less than z 1 and being greater than z2 40 2 20 so we have the probability of being less than z1 being 20 and the probability of being greater than z2 being 20 Between Problem 40 2 20 so we have the probability of being less than z1 being 20 and the probability of being greater than z2 being 20 20 60 20 Between Problem 40 2 20 so we have the probability of being less than z1 being 20 and the probability of being greater than z2 being 20 To find z1 look up 20 in the body of the table This gives z1 0 84 Between Problem 40 2 20 so we have the probability of being less than z1 being 20 and the probability of being greater than z2 being 20 To find z2 to use the table we must have a less than problem Since the probability of being greater than z2 is 20 then the probability of being less than z2 is …

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