Presentation on theme: "Normal Distribution Z-scores put to use!"— Presentation transcript:
1 Normal Distribution Z-scores put to use! Section 2.2Reference Text:The Practice of Statistics, Fourth Edition.Starnes, Yates, MooreLesson 2.2.1
2 Today’s Objectives The 68-95-99.7 Rule State mean an standard deviation for The Standard Normal DistributionGiven a raw score from a normal distribution, find the standardized “z-score”Use the Table of Standard Normal Probabilities to find the area under a given section of the Standard Normal curve.
3 The RuleHow many standard deviations do you think it would take for us to have the entire sample or population accounted for and just have a .03% uncertainty?In other words, how many standard deviations away from the mean encompasses almost all objects in the study?
4 The Rule3!The Rule describes the percent of observations fall within 1,2 or 3 standard deviations. Look at the visual:
5 The RuleSo,Approximately 68% of the observations fall within of the mean µApproximately 95% of the observations fall within 2 of the mean µApproximately 99.7% of the observations fall within 3 of the mean µ
6 The RuleIf I have data within 2 standard deviations, then I'm accounting for 95% of observationsQuestion: what percent is in the left tail?
7 You Try!The distribution of number of movies AP Statistic students watch in two weeks is close to normal. Suppose the distribution is exactly Normal with mean µ= 6.84 and standard deviation = 1.55 (this is non fiction data)A) Sketch a normal density curve for this distribution of movies watched. Label the points that are one, two, and three SD away from the mean.B) What percent of the movies is less that 3.74? Show your work!C) What percent of scores are between 5.29 and 9.94? Show work!Remember: Always put your answers back into context!
9 Standardizing Observations All normal distributions have fundamentally the same shape.If we measure the x axis in units of size σ about a center of 0, then they are all exactly the same curve.This is called the Standard Normal CurveWe abbreviate the normal dist. As N( µ, )To standardize observations, we change from x values (the raw observations) z values (the standardized observations) by the formula:
10 The Standard Normal Distribution Notice that the z-score formula always subtracts μ from each observation.So the mean is always shifted to zeroAlso notice that the shifted values are divided by σ, the standard deviation.So the units along the z-axis represent numbers of standard deviationsThus the Standard Normal Distribution is always N(0,1).
11 Example! The heights of young women are: N(64.5, 2.5) Use the formula to find the z-score of a woman 68 inches tall.A woman’s standardized height is the number of standard deviations by which her height differs from the mean height of all young women.
12 Normal Distribution Calculations What proportion of all young women are less than 68 inches tall?Notice that this does not fall conveniently on one of the σ bordersWe already found that 68 inches corresponds to a z-score of 1.4So what proportion of all standardized observations fall to the left of z = 1.4?Since the area under the Standard Normal Curve is always 1, we can ask instead, what is the area under the curve and to the left of z=1.4For that, we need a table!!
13 The Standard Normal Table Find Table A of the handoutIt is also in your textbook in the very backZ-scores (to the nearest tenth) are in the left columnThe other 10 columns round z to the nearest hundredthFind z = 1.4 in the table and read the areaYou should find area to the left = .9192So the proportion of observations less than z = 1.4 is about 92%Now put the answer in context: “About 92% of all young women are 68 inches tall or less.”
14 What about area above a value? Still using the N(64.5, 2.5) distribution, what proportion of young women have a height of 61.5 inches or taller?Z = (61.5 – 64.5)/2.5 = -1.2From Table A, area to the left of -1.2 =.1151So area to the right = = .8849So about 88.5% of young women are 61.5” tall or taller.
15 What about area between two values? What proportion of young women are between 61.5” and 68” tall?We already know 68” gives z = 1.4 and area to the left of .9192We also know 61.5” gives z = -1.2 and area to the left of .1151So just subtract: = .8041So about 80% of young women are between 61.5” and 68” tallRemember to write your answer IN CONTEXT!!!
16 Given a proportion, find the observation x SAT Verbal scores are N(505, 110). How high must you score to be in the top 10%?If you are in the top 10%, there must be 90% below you (to the left).Find .90 (or close to it) in the body of Table A. What is the z-score?You should have found z = 1.28Now solve the z definition equation for xSo you need a score of at least 646 to be in the top 10%.
17 How to Solve Problems Involving Normal Distribution State: Express the problem in terms of the observed variable xPlan: draw a picture of the distribution and shade the area of interest under the curve.Do: Preform the calculationsStandardize x to restate the problem in terms of standard normal variable zUse Table A and the fact that the total area under the curve is 1 to find the required area under the standard normal curveConclude: Write your conclusion in context of the problem.Lets look at TB pg 120 “Tiger on the Range”
18 Today’s Objectives The 68-95-99.7 Rule State mean an standard deviation for The Standard Normal DistributionGiven a raw score from a normal distribution, find the standardized “z-score”Use the Table of Standard Normal Probabilities to find the area under a given section of the Standard Normal curve.