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The Normal Distribution Outline 1 Introduction: Continuous Random Variables (3.6) 2 Introduction: Normal Distribution (4.1-4.2) 3 Areas Under a Normal Curve (4.3) Standard Normal Table: Computing Probabilities Standard Normal Table: Finding Quantiles General Normal Distribution Using the Computer A Case Study 4 Assessing Normality (4.4) Outline 1 Introduction: Continuous Random Variables (3.6) 2 Introduction: Normal Distribution (4.1-4.2) 3 Areas Under a Normal Curve (4.3) Standard Normal Table: Computing Probabilities Standard Normal Table: Finding Quantiles General Normal Distribution Using the Computer A Case Study 4 Assessing Normality (4.4) Recall: Discrete Random Variables We can describe a discrete random variable with a table. For example: k 0 1 5 10 Pr {X = k } 0.1 0.5 0.1 0.3 But a continuous random variable can take any value in a interval; we could never list all of these values in a table Continuous Distributions A continuous random variable has possible values over a continuum. The total probability of one is not in discrete chunks at specific locations, but rather is ground up like a very fine dust and sprinkled on the number line. We cannot represent the distribution with a table of possible values and the probability of each. Continuous Distributions (continued) Instead, we represent the distribution with a probability density function which measures the thickness of the probability dust. Probability is measured over intervals as the area under the curve. A legal probability density f : 1 2 is never negative (f (x) ≥ 0 for −∞ < x <R ∞). ∞ has a total area under the curve of one ( −∞ f (x)dx = 1). For You Read 3.6 in your book Outline 1 Introduction: Continuous Random Variables (3.6) 2 Introduction: Normal Distribution (4.1-4.2) 3 Areas Under a Normal Curve (4.3) Standard Normal Table: Computing Probabilities Standard Normal Table: Finding Quantiles General Normal Distribution Using the Computer A Case Study 4 Assessing Normality (4.4) The Normal Distribution The Normal Distribution is the most important distribution of continuous random variables. The normal density curve is the famous symmetric, bell-shaped curve. The central limit theorem is the reason that the normal curve is so important. Essentially, many statistics that we calculate from large random samples will have approximate normal distributions (or distributions derived from normal distributions), even if the distributions of the underlying variables are not normally distributed. This fact is the basis of most of the methods of statistical inference we will study in the last half of the course. Chapter 4 introduces the normal distribution as a probability distribution. Chapter 5 culminates in the central limit theorem, the primary theoretical justification for most of the methods of statistical inference in the remainder of the textbook. The famous bell curve Used everywhere: very useful description for lots of biological (and other) random variables. Body weight, Crop yield, Protein content in soybean, Density of blood components σ σ Y ∼ N (µ, σ) values can, in principle, go to infinity, both ways. µ − 4σ µ − 2σ µ µ + 2σ µ + 4σ The Normal Density Normal curves have the following bell-shaped, symmetric density. f (x) = √ 1 2πσ − 21 e x−µ σ 2 , (−∞ < x < ∞) Parameters: The parameters of a normal curve are the mean µ and the standard deviation σ. Notation: a random variable is normal with mean µ and standard deviation σ, Y ∼ N (µ, σ) Standard Normal. We often consider µ = 0 and σ = 1. Z ∼ N (0, 1) Recall: Empirical Rule The 68–95–99.7 Rule. For every normal curve: 1 2 3 ≈ 68% of the area is within one SD of the mean ≈ 95% of the area is within two SDs of the mean ≈ 99.7% of the area is within three SDs of the mean Standard Normal Density Standard Normal Density Area within 1 = 0.68 Area within 2 = 0.95 Density Area within 3 = 0.997 −3 −2 −1 0 Possible Values 1 2 3 Outline 1 Introduction: Continuous Random Variables (3.6) 2 Introduction: Normal Distribution (4.1-4.2) 3 Areas Under a Normal Curve (4.3) Standard Normal Table: Computing Probabilities Standard Normal Table: Finding Quantiles General Normal Distribution Using the Computer A Case Study 4 Assessing Normality (4.4) Recall: the Normal Density f (x) = √ 1 2πσ − 12 e (can you integrate that?) x−µ σ 2 , (−∞ < x < ∞) Compute areas There is no formula to calculate general areas under the normal curve. (The integral of the density has no closed form solution.) We will learn to use normal tables for this. We will also learn how to use R. However, you will have to use normal tables on the exams. The Standard Normal Table We have tables for N (0, 1) The standard normal table lists the area to the left of z under the standard normal curve for each value from −3.49 to 3.49 by 0.01 increments. The normal table is located: 1 2 inside front cover of your textbook. Table 3 (p. 675-676) Numbers in the margins represent z. Numbers in the middle of the table are areas to the left of z. Warning! Learn the normal table today. Make it part of your being. Standard Normal Table: Computing Probabilities The standard normal Z ∼ N (0, 1) Whole area = 1 (total probability rule) Pr {Z ≤ 0} = by symmetry Pr {Z = 0} = 0 Pr {Z < 1} = .8413 (table, front cover) Pr {0 ≤ Z ≤ 1} = Pr {−1 ≤ Z ≤ 1} = Pr {Z > 1.5} = −4 −2 0 1 2 3 4 Draw a picture and make use of symmetry. Pr {−0.5 ≤ Z ≤ 0.3} = The standard normal Z ∼ N (0, 1) Whole area = 1 (total probability rule) Pr {Z ≤ 0} = by symmetry Pr {Z = 0} = 0 Pr {Z < 1} = .8413 (table, front cover) Pr {0 ≤ Z ≤ 1} = Pr {−1 ≤ Z ≤ 1} = Pr {Z > 1.5} = −4 −2 0 1 2 3 4 Draw a picture and make use of symmetry. Pr {−0.5 ≤ Z ≤ 0.3} = The standard normal Z ∼ N (0, 1) Whole area = 1 (total probability rule) Pr {Z ≤ 0} = by symmetry Pr {Z = 0} = 0 Pr {Z < 1} = .8413 (table, front cover) Pr {0 ≤ Z ≤ 1} = Pr {−1 ≤ Z ≤ 1} = Pr {Z > 1.5} = −4 −2 0 1 2 3 4 Draw a picture and make use of symmetry. Pr {−0.5 ≤ Z ≤ 0.3} = The standard normal Z ∼ N (0, 1) Whole area = 1 (total probability rule) Pr {Z ≤ 0} = .5 by symmetry Pr {Z = 0} = 0 Pr {Z < 1} = .8413 (table, front cover) Pr {0 ≤ Z ≤ 1} = Pr {−1 ≤ Z ≤ 1} = Pr {Z > 1.5} = −4 −2 0 1 2 3 4 Draw a picture and make use of symmetry. Pr {−0.5 ≤ Z ≤ 0.3} = The standard normal Z ∼ N (0, 1) Whole area = 1 (total probability rule) Pr {Z ≤ 0} = .5 by symmetry Pr {Z = 0} = 0 Pr {Z < 1} = .8413 (table, front cover) Pr {0 ≤ Z ≤ 1} = Pr {−1 ≤ Z ≤ 1} = Pr {Z > 1.5} = −5 −4 −2 0 1 2 4 Draw a picture and make use of symmetry. Pr {−0.5 ≤ Z ≤ 0.3} = The standard normal Z ∼ N (0, 1) Whole area = 1 (total probability rule) Pr {Z ≤ 0} = .5 by symmetry Pr {Z = 0} = 0 Pr {Z < 1} = .8413 (table, front cover) Pr {0 ≤ Z ≤ 1} = Pr {−1 ≤ Z ≤ 1} = Pr {Z > 1.5} = −4 −2 0 1 2 4 Draw a picture and make use of symmetry. Pr {−0.5 ≤ Z ≤ 0.3} = The standard normal Z ∼ N (0, 1) Whole area = 1 (total probability rule) Pr {Z ≤ 0} = .5 by symmetry Pr {Z = 0} = 0 Pr {Z < 1} = .8413 (table, front cover) Pr {0 ≤ Z ≤ 1} = 0.8413 − 0.5 = .3413 Pr {−1 ≤ Z ≤ 1} = Pr {Z > 1.5} = −4 −2 0 1 2 4 Draw a picture and make use of symmetry. Pr {−0.5 ≤ Z ≤ 0.3} = The standard normal Z ∼ N (0, 1) Whole area = 1 (total probability rule) Pr {Z ≤ 0} = .5 by symmetry Pr {Z = 0} = 0 Pr {Z < 1} = .8413 (table, front cover) Pr {0 ≤ Z ≤ 1} = 0.8413 − 0.5 = .3413 Pr {−1 ≤ Z ≤ 1} = Pr {Z > 1.5} = −4 −2 −1 0 1 2 4 Draw a picture and make use of symmetry. Pr {−0.5 ≤ Z ≤ 0.3} = The standard normal Z ∼ N (0, 1) Whole area = 1 (total probability rule) Pr {Z ≤ 0} = .5 by symmetry Pr {Z = 0} = 0 Pr {Z < 1} = .8413 (table, front cover) Pr {0 ≤ Z ≤ 1} = 0.8413 − 0.5 = .3413 Pr {−1 ≤ Z ≤ 1} = 2 ∗ 0.3413 = .6826 Pr {Z > 1.5} = −4 −2 −1 0 1 2 4 Draw a picture and make use of symmetry. Pr {−0.5 ≤ Z ≤ 0.3} = The standard normal Z ∼ N (0, 1) Whole area = 1 (total probability rule) Pr {Z ≤ 0} = .5 by symmetry Pr {Z = 0} = 0 Pr {Z < 1} = .8413 (table, front cover) Pr {0 ≤ Z ≤ 1} = 0.8413 − 0.5 = .3413 Pr {−1 ≤ Z ≤ 1} = 2 ∗ 0.3413 = .6826 Pr {Z > 1.5} = −4 0 1.5 4 5 Draw a picture and make use of symmetry. Pr {−0.5 ≤ Z ≤ 0.3} = The standard normal Z ∼ N (0, 1) Whole area = 1 (total probability rule) Pr {Z ≤ 0} = .5 by symmetry Pr {Z = 0} = 0 Pr {Z < 1} = .8413 (table, front cover) Pr {0 ≤ Z ≤ 1} = 0.8413 − 0.5 = .3413 Pr {−1 ≤ Z ≤ 1} = 2 ∗ 0.3413 = .6826 Pr {Z > 1.5} = 1 − Pr {Z ≤ −1.5} = 1 − 0.9332 = .0668 −4 0 1.5 4 5 Draw a picture and make use of symmetry. Pr {−0.5 ≤ Z ≤ 0.3} = The standard normal Z ∼ N (0, 1) Whole area = 1 (total probability rule) Pr {Z ≤ 0} = .5 by symmetry Pr {Z = 0} = 0 Pr {Z < 1} = .8413 (table, front cover) Pr {0 ≤ Z ≤ 1} = 0.8413 − 0.5 = .3413 Pr {−1 ≤ Z ≤ 1} = 2 ∗ 0.3413 = .6826 Pr {Z > 1.5} = 1 − Pr {Z ≤ −1.5} = 1 − 0.9332 = .0668 −4 −1.5 0 1.5 4 Draw a picture and make use of symmetry. Pr {−0.5 ≤ Z ≤ 0.3} = The standard normal Z ∼ N (0, 1) Whole area = 1 (total probability rule) Pr {Z ≤ 0} = .5 by symmetry Pr {Z = 0} = 0 Pr {Z < 1} = .8413 (table, front cover) Pr {0 ≤ Z ≤ 1} = 0.8413 − 0.5 = .3413 Pr {−1 ≤ Z ≤ 1} = 2 ∗ 0.3413 = .6826 Pr {Z > 1.5} = 1 − Pr {Z ≤ −1.5} = 1 − 0.9332 = .0668 −4 −0.50.3 4 Draw a picture and make use of symmetry. Pr {−0.5 ≤ Z ≤ 0.3} = Pr {Z ≤ 0.3} − Pr {Z ≤ −0.5} The standard normal Z ∼ N (0, 1) Whole area = 1 (total probability rule) Pr {Z ≤ 0} = .5 by symmetry Pr {Z = 0} = 0 Pr {Z < 1} = .8413 (table, front cover) Pr {0 ≤ Z ≤ 1} = 0.8413 − 0.5 = .3413 Pr {−1 ≤ Z ≤ 1} = 2 ∗ 0.3413 = .6826 Pr {Z > 1.5} = 1 − Pr {Z ≤ −1.5} = 1 − 0.9332 = .0668 −4 −0.50.3 4 Draw a picture and make use of symmetry. Pr {−0.5 ≤ Z ≤ 0.3} = Pr {Z ≤ 0.3} − Pr {Z ≤ −0.5} = .6179 − .3085 = .3094 Standard Normal Table: Finding Quantiles Basic Idea This is the inverse problem. Before. Question: here is the interval; Answer: Probability of the interval Now. Question: Here is the probability. Answer: What is the interval that matches with the probability. Problem: many intervals have the same area. We ask for a certain type of interval. The Reverse Question: Finding quantiles Pr {Z ≤?} = .975 the value ? is a quantile. Use Table 3 (or front cover) backward −3 0 ? 3 Pr {Z ≤?} = 0.20 Pr {Z ≥?} = 0.80 −3 ? 0 3 The Reverse Question: Finding quantiles Pr {Z ≤?} = .975 ? = 1.96 the value ? is a quantile. Use Table 3 (or front cover) backward −3 0 ? 3 Pr {Z ≤?} = 0.20 Pr {Z ≥?} = 0.80 −3 ? 0 3 The Reverse Question: Finding quantiles Pr {Z ≤?} = .975 ? = 1.96 the value ? is a quantile. Use Table 3 (or front cover) backward Pr {Z ≤?} = 0.20 Pr {Z ≥?} = 0.80 −3 0 ? 3 ? = −0.84 −3 ? 0 3 The Reverse Question: Finding quantiles Pr {Z ≤?} = .975 ? = 1.96 the value ? is a quantile. Use Table 3 (or front cover) backward Pr {Z ≤?} = 0.20 Pr {Z ≥?} = 0.80 −3 0 ? 3 ? = −0.84 ? = −0.84 −3 ? 0 3 zα Notation Instead of ? we introduce a fancy notation. Let zα be the number such that Pr {Z < zα } = 1 − α Draw Figure 4.19 (α to the right; 1 − α to the left) Example. z.025 = 1.96 Draw a picture! (Note: your book uses a capital, Zα ) General Normal Distribution General Normal? We are good if Y ∼ N (0, 1) What if Y ∼ N (10, 5) or N (11, 3), or ... (Directions from “Home”) Standardization All normal curves have the same shape, and are simply rescaled versions of the standard normal density. Consequently, every area under a general normal curve corresponds to an area under the standard normal curve. The key standardization formula is z= x −µ σ Solving for x yields x = µ + zσ which says algebraically that x is z standard deviations above the mean. Probability Example If X ∼ N(100, 2), find Pr {X > 97.5}. Solution: X − 100 97.5 − 100 Pr {X > 97.5} = Pr > 2 2 = Pr {Z > −1.25} = 1 − Pr {Z < −1.25} = ? Quantile Example If X ∼ N(100, 2), find the cutoff values for the middle 70% of the distribution. Solution: The cutoff points will be the 0.15 and 0.85 quantiles. From the table, 1.03 < z < 1.04 and z = 1.04 is closest. Thus, the cutoff points are the mean plus or minus 1.04 standard deviations. 100 − 1.04(2) = 97.92, 100 + 1.04(2) = 102.08 General form N (µ, σ) All normal distributions have the same shape. Transformation Y −µ ∼ N (0, 1) Z = σ Systolic blood pressure in healthy adults has a normal distribution with mean 112 mmHg and standard deviation 10 mmHg, i.e. Y ∼ N (112, 10). One day, I have 92 mmHg. 92 − 112 Y − 112 Pr {Y ≤ 92} = Pr ≤ 10 10 = Pr {Z ≤ −2} = 0.0227 General form N (µ, σ) µ = 112 mmHg, σ = 10 mmHg. Y − 112 122 − 112 102 − 112 ≤ ≤ Pr {102 ≤ Y ≤ 122} = Pr 10 10 10 = Pr {−1 ≤ Z ≤ 1} = .6826 68.3% of healthy adults have systolic blood pressure between 102 and 122 mmHg. A patient’s systolic blood pressure is 137 mmHg. Y − 112 137 − 112 Pr {Y ≥ 137} = Pr ≥ 10 10 = Pr {Z ≥ 2.5} = 1 − .9938 = 0.0062 This patient’s blood pressure is very high... General form N (µ, σ) µ = 112 mmHg, σ = 10 mmHg. What is “High blood pressure”? For instance, it could be the value BP∗ such that Pr {Y ≤ BP∗ } = .95 We need z∗ such that Pr {Z ≤ z∗ } = .95. Table in back cover (last line): z∗ = 1.645. Thus BP∗ lies 1.645 standard deviations above the mean: BP∗ = 112 + 1.645 ∗ 10 = 112 + 16.45 = 128.45 mmHg Formal approach: .95 = Pr {Y ≤ BP∗ } = Pr with BP∗ − 112 Y − 112 ≤ 10 10 BP∗ − 112 z∗ = i.e. BP∗ = 112 + 10 z∗ 10 = Pr {Z ≤ z∗ } Using the Computer Other Options Remember you need the tables for the exam. Statistical software (e.g. R) Online Calculators. See course website, handouts. Calculators for Various Statistical Problems Doing calculations with R norm binom p q d normal distribution binomial probability: Pr {Y ≤ . . .} quantile density, or probability mass function: Pr {Y = . . .} > pnorm(1) [1] 0.8413447 > pnorm(2)-pnorm(-2) [1] 0.9544997 > pnorm(3)-pnorm(-3) [1] 0.9973002 > 1- pnorm(137, mean=112, sd=10) [1] 0.006209665 > qnorm(.95, mean=112, sd=10) [1] 128.4485 > pbinom(1, size=6, prob=1/6) [1] 0.7367755 > dbinom(0:6, size=6, prob=1/6) [1] 0.335 0.402 0.201 0.054 0.008 0.001 0.000 A Case Study Exam Hint This how exam questions will often look. I give you some question in the context of a scientific area; you figure out the probabilty calculations to use. (word problems) Case Study Example Body temperature varies within individuals over time (it can be higher when one is ill with a fever, or during or after physical exertion). However, if we measure the body temperature of a single healthy person when at rest, these measurements vary little from day to day, and we can associate with each person an individual resting body temperture. There is, however, variation among individuals of resting body temperture. The Question Example In the population, suppose that: the mean resting body temperature is 98.25 degrees Fahrenheit; the standard deviation is 0.73 degrees Fahrenheit; resting body temperatures are normally distributed. Let X be the resting body temperature of a randomly chosen individual. Find: 1 Pr {X < 98}, the proportion of individuals with temperature less than 98. 2 Pr {98 < X < 100}, the proportion of individuals with temperature between 98 and 100. 3 The 0.90 quantile of the distribution. 4 The cutoff values for the middle 50% of the distribution. Outline 1 Introduction: Continuous Random Variables (3.6) 2 Introduction: Normal Distribution (4.1-4.2) 3 Areas Under a Normal Curve (4.3) Standard Normal Table: Computing Probabilities Standard Normal Table: Finding Quantiles General Normal Distribution Using the Computer A Case Study 4 Assessing Normality (4.4) Normal Probability Plots A standard question begins, “Assuming that variable Y has a normal distribution,. . . ”. But how do we know the distribution is approximately normal? Sometimes we can rely on the central limit theorem. If given data, there are tests for normality, but there are reasons not to do these. It is generally better to make a plot that sheds light on the question, “Is the data so far from normality as to bias a method that assumes normality?” There is no easy answer to the question, but a normal probability plot is much more informative than the result of a test. What is a Normal Probability Plot? A normal probability plot is a plot of the sorted sample data versus something close to the expected z-score for the corresponding rank of a random normal sample of the same size. For example, the expected z-score of the minumum from a random sample of size 10 from a normal population is about −1.54. If the plotted points are close to a straight line, there is evidence that the distribution is close to normal. If the plotted points are far from a straight line, there is evidence of non-normality. The Basic Idea A normal probability plot plots the ordered observations on the y-axis (y(1) , . . . , y(n) ) against some standard “normal scores” x1 , . . . , xn on the x-axis. If the data were truly from a normal distribution it would lie on a “straight” line. Why a straight line? Y = µ + σZ Normal Example: All three plots are normal (n = 50) −1 0 1 ● −2 −1 0 1 2 ● 2 −2 −1 0 1 2 Theoretical Quantiles Theoretical Quantiles Histogram of x Histogram of x Histogram of x −3 −2 −1 0 x 1 2 3 8 0 4 Frequency 15 0 5 5 10 Frequency 12 Theoretical Quantiles 0 Frequency 1 ● ●● ●●●● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ●● ● ●● ●●● ● ●● ● ●● ●●●● ●●●●● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ●● ●●●●● ●● ●● −1 ● Sample Quantiles 2 ● 0 2 Normal Q−Q Plot 15 −2 ● −2 2 0 ● ●● ●●● ●●● ●● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ●●● ●● ●●●● ●● Sample Quantiles Normal Q−Q Plot ● −2 Sample Quantiles Normal Q−Q Plot −3 −2 −1 0 x 1 2 3 −2 −1 0 1 x 2 Normal Example: All three plots are normal (n = 500) 0 1 2 −3 −2 −1 0 1 2 2 0 3 ● ●●●● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ●●● −3 −2 −1 0 1 2 Histogram of x Histogram of x Histogram of x 2 3 0 0 x 1 40 80 80 −1 0 3 80 Theoretical Quantiles Frequency Theoretical Quantiles Frequency Theoretical Quantiles 40 −3 ● −3 ●● ●●● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ●●● ●● Sample Quantiles 3 1 −1 3 0 Frequency −3 −2 −1 −3 ● ●●● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ●●●● Normal Q−Q Plot 40 ● Normal Q−Q Plot Sample Quantiles 3 1 −3 −1 Sample Quantiles Normal Q−Q Plot −3 −2 −1 0 x 1 2 3 −3 −1 0 x 1 2 3 Non-normal Example: One plot is not normal (n = 100) −2 −1 0 1 2 ● ● −2 −1 0 1 2 1 0 ●● ● ●●●● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ●●● ●● −2 2 0 ●●● ●●● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ●●●●●● ● −2 ● ● Normal Q−Q Plot Sample Quantiles Normal Q−Q Plot Sample Quantiles 0 2 4 6 8 Sample Quantiles Normal Q−Q Plot 2 ● ● ●● ● ● ●● ●● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ●●●● ●● −2 −1 0 1 2 Theoretical Quantiles Theoretical Quantiles Histogram of x Histogram of x Histogram of x 0 2 4 6 x 8 −2 −1 0 1 x 2 3 15 0 5 Frequency 15 0 5 Frequency 20 0 Frequency 40 Theoretical Quantiles −2 −1 0 x 1 2 For You Look at Figures 4.26 - 4.32 (p.136 - 139) See how different shaped histograms lead to different normal probability plots Warning! Learn the normal table today. Make it part of your being.