Normal Distribution Z-scores put to use! Section 2.2 Reference Text: The Practice of Statistics, Fourth Edition. Starnes, Yates, Moore Lesson 2.2.1

Today’s Objectives The 68-95-99.7 Rule State mean an standard deviation for The Standard Normal Distribution Given 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.

Standard Deviation In statistics the standard deviation (SD) (represented by the Greek letter sigma, σ) measures the amount of variation or dispersion from the average. A low standard deviation indicates that the data points tend to be very close to the mean A high standard deviation indicates that the data points are spread out over a large range of values.

The 68-95-99.7 Rule How 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? 3

Normal Distributions The 68-95-99.7 Rule Although there are many Normal curves, they all have properties in common. Normal Distributions Definition: The 68-95-99.7 Rule (“The Empirical Rule”) In the Normal distribution with mean µ and standard deviation σ: Approximately 68% of the observations fall within σ of µ. Approximately 95% of the observations fall within 2σ of µ. Approximately 99.7% of the observations fall within 3σ of µ.

The 68-95-99.7 Rule If I have data within 2 standard deviations, then I'm accounting for 95% of observations Question: what percent is in the left tail? 2.5%

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 σ=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! Complete sentence!

The distribution of Iowa Test of Basic Skills (ITBS) vocabulary scores for 7th grade students in Gary, Indiana, is close to Normal. Suppose the distribution is N(6.84, 1.55). Sketch the Normal density curve for this distribution. What percent of ITBS vocabulary scores are less than 3.74? What percent of the scores are between 5.29 and 9.94? Example, p. 113 Normal Distributions

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 Curve We abbreviate the normal dist. As 𝑁(𝜇,𝜎) To standardize observations, we change from x values (the raw observations)  z values (the standardized observations) by the formula:

The Standard Normal Distribution Notice that the z-score formula always subtracts μ from each observation. So the mean is always shifted to zero 𝑥−𝜇 Also notice that the shifted values are divided by σ, the standard deviation. So the units along the z-axis represent numbers of standard deviations 𝜎 Thus the Standard Normal Distribution is always N(0,1).

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.

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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 borders We already found that 68 inches corresponds to a z-score of 1.4 So 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=1.4 For that, we need a table!!

The Standard Normal Table Find Table A of the handout It is also in your textbook in the very back Z-scores (to the nearest tenth) are in the left column The other 10 columns round z to the nearest hundredth Find z = 1.4 in the table and read the area You should find area to the left = .9192 So the proportion of observations less than z = 1.4 is about 91.92% Now put the answer in context: “About 91.92% of all young women are 68 inches tall or less.”

The Standard Normal Table Normal Distributions Because all Normal distributions are the same when we standardize, we can find areas under any Normal curve from a single table. Definition: The Standard Normal Table Table A is a table of areas under the standard Normal curve. The table entry for each value z is the area under the curve to the left of z. Suppose we want to find the proportion of observations from the standard Normal distribution that are less than 0.81. We can use Table A: P(z < 0.81) = .7910 Z .00 .01 .02 0.7 .7580 .7611 .7642 0.8 .7881 .7910 .7939 0.9 .8159 .8186 .8212

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= 𝑥−𝜇 𝜎 = 61.5−64.5 2.5 =−1.2 From Table A, area to the left of -1.2 =.1151 So area to the right = 1 - .1151 = .8849 So about 88.49% of young women are 61.5” tall or taller.

What about area between two values? What proportion of young women are between 61.5” and 68” tall? The facts: We already know 68” gives z = 1.4 and area to the left of .9192 We also know 61.5” gives z = -1.2 and area to the left of .1151 So just subtract the two, and find the proportion between 61.5” and 68”: .9192 - .1151 = .8041 So about 80.41% of young women are between 61.5” and 68” tall Remember to write your answer IN CONTEXT!!!

Normal Distributions Finding Areas Under the Standard Normal Curve Example, p. 117 Normal Distributions Find the proportion of observations from the standard Normal distribution that are between -1.25 and 0.81. Can you find the same proportion using a different approach? 1 - (0.1056+0.2090) = 1 – 0.3146 = 0.6854

Given a proportion, find the observation x (Working Backwards!) 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 = 1.28 Now solve the z definition equation for x 𝑧= 𝑥−𝜇 𝜎 1.28= 𝑥−505 110 𝑥=1.28∗110+505 𝑥=645.8 So you need a score of at least 646 to be in the top 10%.

How to Solve Problems Involving Normal Distribution State: Express the problem in terms of the observed variable x (list what x represents, mean? SD?) (get your ducks in order to work with) Plan: Draw a picture of the distribution and shade the area of interest under the curve. Do: Preform the calculations Standardize x to restate the problem in terms of standard normal variable z Use Table A and the fact that the total area under the curve is 1 to find the required area under the standard normal curve Conclude: Write your conclusion in context of the problem. Lets look at TB pg 120 “Tiger on the Range”

Normal Distribution Calculations Normal Distributions When Tiger Woods hits his driver, the distance the ball travels can be described by N(304, 8). What percent of Tiger’s drives travel between 305 and 325 yards? Using Table A, we can find the area to the left of z=2.63 and the area to the left of z=0.13. 0.9957 – 0.5517 = 0.4440. About 44% of Tiger’s drives travel between 305 and 325 yards.

Video explaination https://www.youtube.com/watch?v=85G_PLBTX00 Test test Lesson 2.2.1

Today’s Objectives The 68-95-99.7 Rule State mean an standard deviation for The Standard Normal Distribution Given 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.

Test Results! Grade: Amount: Marginal % ……A......…….....5……….26% …….B…………...5……...26% 84% Passed …….C…………...6……...32% …….D…………...2……...11% …….F…..............1………..5% 16% Failed Mean: 81% Max: 100% Min: 50% No Outliers

Tracking AP Stats 2014-2015 (WHS) Ch. 1 Test Ch. 2 Test Ch. 3 Test A 5 B5 C6 D2 F1

14/15 VS 15/16 AP Stats 14/15 15/16 Chapter 1 Test A -5 B-5 C-6 D-2 F-1 14/15 15/16 Chapter 2 Test A - B- C- D- F- 14/15 15/16 Chapter 3 Test A - B- C- D- F- 14/15 15/16 Chapter 4 Test A - B- C- D- F- 14/15 15/16 Chapter 5 Test A - B- C- D- F-

Homework Worksheet 1-28 (Multiples of 3) EXTRA CREDIT: You MAY get ONLY 2 additional points added to your test grade for completing the entire assignment 1-28. However, regardless of your score, the maximum you may achieve is an 80%. Sooo, if you have a 79% and you do the extra credit…your test grade will max at 80%