The Standard Normal Distribution

The Standard Normal Distribution The Standard Normal Distribution By: OpenStaxCollege The standard normal distribution is a normal distribution of standardized values called z-scores A z-score is measured in units of the standard deviation For example, if the mean of a normal distribution is five and the standard deviation is two, the value 11 is three standard deviations above (or to the right of) the mean The calculation is as follows: x = μ + (z)(σ) = + (3)(2) = 11 The z-score is three The mean for the standard normal distribution is zero, and the standard deviation is one x−μ The transformation z = σ produces the distribution Z ~ N(0, 1) The value x comes from a normal distribution with mean μ and standard deviation σ Z-Scores If X is a normally distributed random variable and X ~ N(μ, σ), then the z-score is: z= x – μ σ The z-score tells you how many standard deviations the value x is above (to the right of) or below (to the left of) the mean, μ Values of x that are larger than the mean have positive z-scores, and values of x that are smaller than the mean have negative zscores If x equals the mean, then x has a z-score of zero Suppose X ~ N(5, 6) This says that x is a normally distributed random variable with mean μ = and standard deviation σ = Suppose x = 17 Then: z= x–μ σ = 17 – =2 1/14 The Standard Normal Distribution This means that x = 17 is two standard deviations (2σ) above or to the right of the mean μ = The standard deviation is σ = Notice that: + (2)(6) = 17 (The pattern is μ + zσ = x) Now suppose x = Then: z = x–μ σ = 1–5 = –0.67 (rounded to two decimal places) This means that x = is 0.67 standard deviations (–0.67σ) below or to the left of the mean μ = Notice that: + (–0.67)(6) is approximately equal to one (This has the pattern μ + (–0.67)σ = 1) Summarizing, when z is positive, x is above or to the right of μ and when z is negative, x is to the left of or below μ Or, when z is positive, x is greater than μ, and when z is negative x is less than μ Try It What is the z-score of x, when x = and X ~ N(12,3)? z= – 12 ≈ – 3.67 Some doctors believe that a person can lose five pounds, on the average, in a month by reducing his or her fat intake and by exercising consistently Suppose weight loss has a normal distribution Let X = the amount of weight lost(in pounds) by a person in a month Use a standard deviation of two pounds X ~ N(5, 2) Fill in the blanks a Suppose a person lost ten pounds in a month The z-score when x = 10 pounds is z = 2.5 (verify) This z-score tells you that x = 10 is standard deviations to the (right or left) of the mean _ (What is the mean?) a This z-score tells you that x = 10 is 2.5 standard deviations to the right of the mean five b Suppose a person gained three pounds (a negative weight loss) Then z = This z-score tells you that x = –3 is standard deviations to the (right or left) of the mean b z = –4 This z-score tells you that x = –3 is four standard deviations to the left of the mean Suppose the random variables X and Y have the following normal distributions: X ~ N(5, 6) and Y ~ N(2, 1) If x = 17, then z = (This was previously shown.) If y = 4, what is z? 2/14 The Standard Normal Distribution z= y−μ σ = 4−2 = where µ = and σ = The z-score for y = is z = This means that four is z = standard deviations to the right of the mean Therefore, x = 17 and y = are both two (of their own) standard deviations to the right of their respective means The z-score allows us to compare data that are scaled differently To understand the concept, suppose X ~ N(5, 6) represents weight gains for one group of people who are trying to gain weight in a six week period and Y ~ N(2, 1) measures the same weight gain for a second group of people A negative weight gain would be a weight loss Since x = 17 and y = are each two standard deviations to the right of their means, they represent the same, standardized weight gain relative to their means Try It Fill in the blanks Jerome averages 16 points a game with a standard deviation of four points X ~ N(16,4) Suppose Jerome scores ten points in a game The z–score when x = 10 is –1.5 This score tells you that x = 10 is _ standard deviations to the (right or left) of the mean (What is the mean?) 1.5, left, 16 The Empirical Rule If X is a random variable and has a normal distribution with mean µ and standard deviation σ, then the Empirical Rule says the following: • About 68% of the x values lie between –1σ and +1σ of the mean µ (within one standard deviation of the mean) • About 95% of the x values lie between –2σ and +2σ of the mean µ (within two standard deviations of the mean) • About 99.7% of the x values lie between –3σ and +3σ of the mean µ (within three standard deviations of the mean) Notice that almost all the x values lie within three standard deviations of the mean • The z-scores for +1σ and –1σ are +1 and –1, respectively • The z-scores for +2σ and –2σ are +2 and –2, respectively • The z-scores for +3σ ... CHAPTER 9 The Normal Distribution Introduction A branch of mathematics that uses probability is called statistics. Statistics is the branch of mathematics that uses observations and measurements called data to analyze, summarize, make inferences, and draw conclusions based on the data gathered. This chapter will explain some basic concepts of statistics such as measures of average and measures of variation. Finally, the relationship between probability and normal distribution will be explained in the last two sections. 147 Copyright © 2005 by The McGraw-Hill Companies, Inc. Click here for terms of use. Measures of Average There are three statistical measures that are commonly used for average. They are the mean, median, and mode. The mean is found by adding the data values and dividing by the number of values. EXAMPLE: Find the mean of 18, 24, 16, 15, and 12. SOLUTION: Add the values: 18 þ 24 þ 16 þ 15 þ 12 ¼ 85 Divide by the number of values, 5: 85 Ä 5 ¼ 17 Hence the mean is 17. EXAMPLE: The ages of 6 executives are 48, 56, 42, 52, 53 and 52. Find the mean. SOLUTION: Add: 48 þ 56 þ 42 þ 52 þ 53 þ 52 ¼ 303 Divide by 6: 303 Ä 6 ¼ 50.5 Hence the mean age is 50.5. The median is the middle data value if there is an odd number of data values or the number halfway between the two data values at the center, if there is an even number of data values, when the data values are arranged in order. EXAMPLE: Find the median of 18, 24, 16, 15, and 12. SOLUTION: Arrange the data in order: 12, 15, 16, 18, 24 Find the middle value: 12, 15, 16, 18, 24 The median is 16. EXAMPLE: Find the median of the number of minutes 10 people had to wait in a checkout line at a local supermarket: 3, 0, 8, 2, 5, 6, 1, 4, 1, and 0. SOLUTION: Arrange the data in order: 0, 0, 1, 1, 2, 3, 4, 5, 6, 8 The middle falls between 2 and 3; hence, the median is (2 þ 3) Ä 2 ¼ 2.5. CHAPTER 9 The Normal Distribution 148 The third measure of average is called the mode. The mode is the data value that occurs most frequently. EXAMPLE: Find the mode for 22, 27, 30, 42, 16, 30, and 18. SOLUTION: Since 30 occurs twice and more frequently than any other value, the mode is 30. EXAMPLE: Find the mode for 2, 3, 3, 3, 4, 4, 6, 6, 6, 8, 9, and 10. SOLUTION: In this example, 3 and 6 occur most often; hence, 3 and 6 are used as the mode. In this case, we say that the distribution is bimodal. EXAMPLE: Find the mode for 18, 24, 16, 15, and 12. SOLUTION: Since no value occurs more than any other value, there is no mode. A distribution can have one mode, more than one mode, or no mode. Also, the mean, median, and mode for a set of values most often differ somewhat. PRACTICE 1. Find the mean, median, and mode for the number of sick days nine employees used last year. The data are 3, 6, 8, 2, 0, 5, 7, 8, and 5. 2. Find the mean, median, and mode for the number of rooms seven hotels in a large city have. The data are 332, 256, 300, 275, 216, 314, and 192. 3. Find the mean, median, and mode for the number of tornadoes that occurred in a specific state over the last 5 years. The data are 18, 6, 3, 9, and 10. 4. Find the mean, median, and mode for the number of items 9 people purchased at the express checkout register. The data are 12, 8, 6, 1, 5, 4, 6, 2, and 6. 5. Find the mean, median, and mode for the ages of 10 children who participated in a field trip to the zoo. The ages are 7, 12, 11, 11, 5, 8, 11, 7, 8, and 6. CHAPTER 9 The Normal Distribution 149 ANSWERS 1. Mean ¼ 3 þ 6 þ 8 þ 2 þ 0 þ 5 þ A STUDY ON IMPROVING THE PERFORMANCE OF CONTROL CHARTS UNDER NON-NORMAL DISTRIBUTIONS SUN TINGTING NATIONAL UNIVERSITY OF SINGAPORE 2004 A STUDY ON IMPROVING THE PERFORMANCE OF CONTROL CHARTS UNDER NON-NORMAL DISTRIBUTIONS SUN TINGTING (B. Eng. & B. Sci., USTC) A THESIS SUBMITTED FOR THE DEGREE OF MASTER OF ENGINEERING DEPARTMENT OF INDUSTRIAL AND SYSTEMS ENGINEERING NATIONAL UNIVERSITY OF SINGAPORE 2004 Acknowledgement ACKNOWLEDGEMENT I would like to express full of my sincere gratitude to my main supervisor, Professor Xie Min, who is one of the most diligent, devoted and kind people I have ever met. I am very grateful to him not only for his invaluable guidance, patience and support throughout my study and research in NUS and the whole revision after I started working, but also for his great care and help during my staying in Singapore. I also would like to thank my second supervisor, Doctor Vellaisamy Kuralmani for his useful advice and consistent support for my research. Besides, I am indebted to all the faculty members of the ISE department for their kind assistance in my study and research. I am also indebted to Institute of High Performance Computing, which has sponsored me to accomplish my research work. I extend all of my gratitude to my friends Philippe, Chen Zhong, Tang Yong, Robin, Vivek, Igna, Jiying, Henri, Chaolan, Josephine, Reuben, Teena, Tao Zhen, Shujie, just name a few, who have made my two years’ stay in NUS an enjoyable memory. I treasure the precious friendships they have offered me, which will not fade away as time goes by. Finally, my wholehearted thankfulness goes to my dear parents for their continuous encouragement and support as well as their never-fading love for me. I TABLE OF CONTENTS TABLE OF CONTENTS ACKNOWLEDGEMENT ------------------------------------------------------------------------I TABLE OF CONTENTS ----------------------------------------------------------------------- II SUMMARY ----------------------------------------------------------------------------------------VI NOMENCLATURE ----------------------------------------------------------------------------VIII LIST OF FIGURES ------------------------------------------------------------------------------ IX LIST OF TABLES -------------------------------------------------------------------------------XII Chapter 1. Introduction 1.1. Research background and Motivations ------------------------------------------------------1 1.2. Objective of the Thesis ------------------------------------------------------------------------4 1.3. Organization of the Thesis --------------------------------------------------------------------5 Chapter 2. Literature Review 2.1. Introduction -------------------------------------------------------------------------------------6 2.2. Literature Review on Non-normality Issue on Control Charts ---------------------------7 2.2.1. Attribute Charts ------------------------------------------------------------------------7 2.2.2. Variable Charts -----------------------------------------------------------------------12 2.2.3. Individual and Multivariate Charts -------------------------------------------------15 II TABLE OF CONTENTS 2.3. Literature Review on Normalizing Transformations -------------------------------------18 2.3.1. Generic Distributions ------------------------------------------------------------------18 2.3.2. Specific Distributions and Statistic Families -------------------------------------- 20 2.3.2. Summary --------------------------------------------------------------------------------33 Chapter 3. Investigation of the Probability Limits in Traditional Shewhart R- Charts 3.1. Introduction ----------------------------------------------------------------------------------- 35 3.2. Positive LCL of R-chart for Improvement Detection ------------------------------------38 3.2.1. The need of a positive LCL for R-chart ---------------------------------------------38 3.2.2. The ... 3 The data cannot follow the uniform distribution The data cannot follow the exponential distribution The data cannot follow the normal distribution I only II only III only I, II, and III The. .. normal, Z ~ N(0, 1) The mean of the z-scores is zero and the standard deviation is one If z is the z-score for a value x from the normal distribution N(µ, σ) then z tells you how many standard deviations... does standardizing a normal distribution to the mean? The mean becomes zero Is X ~ N(0, 1) a standardized normal distribution? Why or why not? What is the z-score of x = 12, if it is two standard