Slide 1Practice ProblemsAdditional Reading and ExamplesSlide 4PropertiesNotationStandard Normal DistributionNormal TableNormal VariablesZ-Score TransformationZ-Score TransformationZ-Score TransformationZ-Score TransformationExample 47Example 47Example 47Example 47Example 47Example 47Example 48Example 48Example 48Example 48Example 48Practice ProblemsPractice Problem AnswersStandard Normal DistributionLess Than ProblemLess Than ProblemLess Than ProblemFinding Values of ZValue Problems on the CalculatorExample 42Example 42Example 42Example 43Example 43Example 43Example 43Greater Than ProblemGreater Than ProblemGreater Than ProblemGreater Than ProblemExample 44Example 44Example 44Example 44Example 45Example 45Example 45Example 45Between ProblemBetween ProblemBetween ProblemBetween ProblemBetween ProblemBetween ProblemBetween ProblemBetween ProblemExample 46Example 46Example 46Example 46Example 46Example 46Example 46Z-Score TransformationZ-Score TransformationZ-Score TransformationZ-Score TransformationZ-Score TransformationSTAT 210Lecture 20Normal DistributionsOctober 11, 2017Practice ProblemsPages 162 through 165Relevant problems: VI.4Recommended problems: VI.4Additional Reading and ExamplesRead pages 158 through 160TOP HAT 2Properties•The normal curve is bell-shaped•The peak of the curve is the population mean m•The normal curve is symmetric about m•The center and spread are completely specified by specifying the values of the population mean m and the population standard deviation s•The total area under the normal curve is 1 (or 100%)NotationX ~ 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 - 339Normal TableGives 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.Normal VariablesA standard normal (Z) distribution requires that the mean is 0 and that the standard deviation is 1.How many real variables (weight, height, grades on a test, etc) do you think have a mean of 0 and a standard deviation of 1?Answer: I cannot think of any!Z-Score TransformationSkip to page 147Suppose 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 TransformationSubtraction of m converts the mean to 0.Division by s converts the standard deviation to 1.Z = X - m s msm = 0s = 1Z-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.Example 47X ~ N (10, 5)P(12 < X < 20) = ????1220?s = 5Example 47X ~ N (10, 5)P(12 < X < 20) = P( 12 - 10 < Z < 20 - 10 ) 5 512 20?s = 5Example 47X ~ N (10, 5)P(12 < X < 20) = P( 12 - 10 < Z < 20 - 10 ) 5 5 = P(0.40 < Z < 2.00) 12 20?s = 50.4 2?s = 1Example 47X ~ N (10, 5)P(12 < X < 20) = P( 12 - 10 < Z < 20 - 10 ) 5 5 = P(0.40 < Z < 2.00) = P(Z < 2.00) - P(Z < 0.40)12 20?s = 50.4 2.00?s = 1Example 47X ~ N (10, 5)P(12 < X < 20) = P( 12 - 10 < Z < 20 - 10 ) 5 5 = P(0.40 < Z < 2.00) = P(Z < 2.00) - P(Z < 0.40) = .9772 - .6554 = .321812 20?s = 50.4 2.00.3218s = 1Example 47X ~ N (10, 5)P(12 < X < 20) = P( 12 - 10 < Z < 20 - 10 ) 5 5 = P(0.40 < Z < 2.00) = P(Z < 2.00) - P(Z < 0.40) = .9772 - .6554 = .3218Calculator: normalcdf(12, 20, 10, 5)Example 48Suppose X ~ N( 78, 12)P(X > 84) = ????78s =1284?Suppose X ~ N( 78, 12)P(X > 84) = P(Z > 84 - 78 ) = P(Z > 0.50) = ???? 12 =Example 4878s =1284?0s =1.50?Suppose X ~ N( 78, 12)P(X > 84) = P(Z > 84 - 78 ) = P(Z > 0.50) 12 = 1 - P(Z < 0.50) = 1 -Example 480s =1.50?0s =1.50?Suppose X ~ N( 78, 12)P(X > 84) = P(Z > 84 - 78 ) = P(Z > 0.50) 12= 1 - P(Z < 0.50)= 1 - .6915= .3085Example 48Suppose X ~ N( 78, 12)P(X > 84) = P(Z > 84 - 78 ) = P(Z > 0.50) 12= 1 - P(Z < 0.50)= 1 - .6915= .3085Calculator: normalcdf(84, 1E99, 78, 12)Example 48Practice ProblemsSuppose X ~ N(70, 12)1. Find the probability that X is greater than 88. P(X > 88) = ???2. Find the probability that X is less than 94. P(X < 94) = ???3. Find the probability that X is between 40 and 64. P(40 < X < 64) = ???Practice Problem AnswersSuppose X ~ N(70, 12)1. Find the probability that X is greater than 88. P(X > 88) = P(Z > (88-70)/12) = P(Z > 1.50) = 1 – P(Z < 1.50) = 1-.9332 = .0668Calculator: normalcdf(88, 1E99, 70, 12)2. Find the probability that X is less than 94. P(X < 94) = P(Z < (94-70)/12) = P(Z < 2.00) = .9772Calculator: normalcdf(-1E99, 94, 70, 12)3. Find the probability that X is between 40 and 64. P(40 < X < 64) = P((40-70)/12 < Z < (64-70)/12) = P(-2.50 < Z < -0.50)= P(Z < -0.50) – P(Z < -2.50) = .3085 - .0062 = .3023Calculator: normalcdf(40, 64, 70, 12)Standard Normal DistributionAll of the problems from the last lecture 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.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.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.Finding Values of ZBody of tableValues of ZValue Problems on the
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