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VCU STAT 210 - Lecture22(2) (1)

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Slide 1Practice ProblemsAdditional Reading and ExamplesTest 4Slide 5Z-Score TransformationZ-Score TransformationZ-Score TransformationZ-Score TransformationSteps for SolvingExample 51Example 51Example 51Example 51Example 51Example 51Example 52Example 52Example 52Example 52Example 52Example 52Example 52Example 53Example 53Example 53Example 53Example 53Example 53Example 54Example 54Example 54Example 54t-Distributionst-Distributionst-DistributionsShapeCenterSpreadUnusual FeaturesSlide 41t-DistributionsExample 55Example 55Example 55Sampling DistributionsSampling DistributionSampling Distribution ExampleSampling Distribution ExampleSampling Distribution ExampleSampling Distribution ExampleSampling DistributionSlide 53Extra Practice ProblemsExtra Practice Problems - AnswersSlide 56STAT 210Lecture 22Normal Distributions,t-Distributions, and Sampling DistributionsOctober 17, 2016Practice ProblemsPages 162 through 165Relevant problems: VI.16 and VI.17Recommended problems: VI.16 and VI.17Additional Reading and ExamplesRead pages 158 through 160Test 4Wednesday, October 19Questions for the first 10 minutes, then test – papers due promptly at the end of class!Covers chapter 6 (pages 139 – 168)Combination of multiple choice questions and written/short answer problems.Formulas and Z-table provided; Bring a calculator!Practice Tests and Formula Sheet on Blackboard.ClickerZ-Score TransformationSuppose 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 TransformationFor a problem that asks to find a probabi l i ty, we convert to a standard normal variable Z ~ N(0, 1).Z = X - m = value - mean s standard deviationThen find the probability for Z using the normal table.Z-Score TransformationX ~ N(m, s) Z = (X - m) / sSuppose we want to find the value x such the probability of being in some interval is as specified.Z-Score Transformation1. Find the value z that satisfies the probability 2. Then calculate x by substituting into the following formula: X = m + Z sSteps for Solving1. Identify the variable X.2. Write out the distribution: X ~ N(m, s).3. Determine the type of problem: probability or value.4. Solve the problem.5. Make sure the answer makes sense.ClickerExample 51X = time people stir sugar into their iced tea X ~ N(12.3, 3.1)Find P(19.833 < X < 22.53)X = time people stir sugar into their iced tea X ~ N(12.3, 3.1)Find P(19.833 < X < 22.53) = P(19.833 - 12.3 < Z < 22.53 - 12.3) 3.1 3.1 = P(2.43 < Z < 3.30)Example 51X = time people stir sugar into their iced tea X ~ N(12.3, 3.1)Find P(19.833 < X < 22.53) = P(19.833 - 12.3 < Z < 22.53 - 12.3) 3.1 3.1 = P(2.43 < Z < 3.30) = P(Z < 3.30) - P(Z < 2.43)Example 51X = time people stir sugar into their iced tea X ~ N(12.3, 3.1)Find P(19.833 < X < 22.53) = P(19.833 - 12.3 < Z < 22.53 - 12.3) 3.1 3.1 = P(2.43 < Z < 3.30) = P(Z < 3.30) - P(Z < 2.43) = .9995 - .9925Example 51X = time people stir sugar into their iced tea X ~ N(12.3, 3.1)Find P(19.833 < X < 22.53) = P(19.833 - 12.3 < Z < 22.53 - 12.3) 3.1 3.1 = P(2.43 < Z < 3.30) = P(Z < 3.30) - P(Z < 2.43) = .9995 - .9925 = .007Example 51X = time people stir sugar into their iced tea X ~ N(12.3, 3.1)Find P(19.833 < X < 22.53) = P(19.833 - 12.3 < Z < 22.53 - 12.3) 3.1 3.1 = P(2.43 < Z < 3.30) = P(Z < 3.30) - P(Z < 2.43) = .9995 - .9925 = .007Calculator: normalcdf (19.833, 22.53, 12.3, 3.1) Example 51Example 52X = weight of leatherback turtleX ~ N(760, 98)Find the weight of a turtle such that the turtle falls in the bottom 20% of the distribution.Example 52X = weight of leatherback turtleX ~ N(760, 98)Find the weight of a turtle such that the turtle falls in the bottom 20% of the distribution.Find the value of X such that the probability of being less than this value is .20.Example 52X = weight of leatherback turtleX ~ N(760, 98)Find the value of X such that the probability of being less than this value is .20.(1) z: P(Z < z) = .20Example 52X = weight of leatherback turtleX ~ N(760, 98)Find the value of X such that the probability of being less than this value is .20.(1) z: P(Z < z) = .20(2) --------------------------(3) Find p=.20 in the body of the table, read across and up to find z = -0.84.Example 52X = weight of leatherback turtle X ~ N(760, 98)Find the value of X such that the probability of being less than this value is .20.(1) z: P(Z < z) = .20(2) --------------------------(3) Find p=.20 in the body of the table, read across and up to find z = -0.84.x = m + zs =760 + (- 0.84)(98)= 760 – 82.32 = 677.68Example 52X = weight of leatherback turtle X ~ N(760, 98)Find the value of X such that the probability of being less than this value is .20.(1) z: P(Z < z) = .20(2) --------------------------(3) Find p=.20 in the body of the table, read across and up to find z = -0.84.x = m + zs =760 + (- 0.84)(98) = 760 – 82.32 = 677.68The turtle would have to weight 677.68 pounds or less.Example 52X = weight of leatherback turtle X ~ N(760, 98)Find the value of X such that the probability of being less than this value is .20.(1) z: P(Z < z) = .20(2) Find p=.20 in the body of the table, read across and up to find z = -0.84.x = m + zs = 760 - 0.84(98) = 760 – 82.32 = 677.68The turtle would have to weight 677.68 pounds or less.Calculator: invNorm (.20, 760, 98)Example 53X = miles per gallon for new Pontiac G6’s that Oprah gave away on her September 13, 2004 showX ~ N(26.3, 3.6)Example 53X = miles per gallon for new Pontiac G6’s that Oprah gave away on her September 13, 2004 showX ~ N(26.3, 3.6)Find the miles per gallon to be in the top 20.19% of the distribution.ClickerExample 53X = miles per gallon for new Pontiac G6’s that Oprah gave away on her September 13, 2004 showX ~ N(26.3, 3.6)Find the miles per gallon to be in the top 20.19% of the distribution.Find the value x such that the miles per gallon is in the top20.19% of the distribution.Example 53X = miles per gallon for new Pontiac G6’s that Oprah gave away on her September 13, 2004 showX ~ N(26.3, 3.6)Find the value x such that the miles per gallon is in the top20.19% of the distribution.1. z: P(Z > z) = .20192. z: P(Z < z) = 1 - .2019 = .79813. Find p=.7981 in the body of the table, read across and up to find z = 0.83 (or z = 0.84).Example 53X = miles per gallon for new Pontiac G6’s that Oprah


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VCU STAT 210 - Lecture22(2) (1)

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