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M ARIO F . T RIOLA

S TATISTICS. E LEMENTARY. Chapter 5 Normal Probability Distributions. M ARIO F . T RIOLA. E IGHTH. E DITION. z Score (or standard score) the number of standard deviations that a given value x is above or below the mean. Measures of Position (Section 2.6). Sample.

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M ARIO F . T RIOLA

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  1. STATISTICS ELEMENTARY Chapter 5 Normal Probability Distributions MARIO F. TRIOLA EIGHTH EDITION

  2. z Score(or standard score) the number of standard deviations that a given value x is above or below the mean Measures of Position (Section 2.6)

  3. Sample Measures of Position z score Population x - µ x - x z = z =  s Round to 2 decimal places

  4. FIGURE 2-16 Interpreting Z Scores Unusual Values Ordinary Values Unusual Values - 3 - 2 - 1 0 1 2 3 Z

  5. 5-1 Overview 5-2 The Standard Normal Distribution 5-3 Normal Distributions: Finding Probabilities 5-4 Normal Distributions: Finding Values 5-5 The Central Limit Theorem 5-6 Normal Distribution as Approximation to Binomial Distribution (skipped) 5-7 Determining Normality Chapter 5Normal Probability Distributions

  6. Continuous random variable Graph is symmetric and bell shaped Area under the curve is equal to 1, therefore there is correspondence between area and probability. µ Normal Distribution

  7. Women: µ = 63.6  = 2.5 Heights of Adult Men and Women • Normal distributions (bell shaped) are a family of distributions that have the same general shape. • The mean, mu, controls the center • Standard deviation, sigma, controls the spread. Men: µ = 69.0  = 2.8 63.6 69.0 Figure 5-4 Height (inches)

  8. The Standard Normal Distribution 5-2

  9. Standard Normal Distribution a normal probability distribution that has a mean of 0 and a standard deviation of 1 Definition

  10. The Empirical Rule Standard Normal Distribution: µ = 0 and  = 1 0.1% 99.7% of data are within 3 standard deviations of the mean 95% within 2 standard deviations 68% within 1 standard deviation 34% 34% 2.4% 2.4% 0.1% 13.5% 13.5% z 0 -3 -2 -1 1 2 3

  11. Example: Using your knowledge of the Emperical rule, what is the probability that a value falls between the mean and 1 standard deviation above the mean?

  12. Example: Using your knowledge of the Emperical rule, what is the probability that a value falls between the mean and 1 standard deviation above the mean? 34% 0 1 z

  13. What if we need to find the probability of a value that does not fall on a standard deviation?

  14. Step 1: Draw a normal curve with the centerline labeled 0 on the x-axis. Step 2: Label given value(s) on the appropriate location on the x-axis. Draw vertical lines on the normal curve above the given values. Step 3: Shade the area of the region in question. Step 4: Use calculator function normalcdf() to find the area (probability) in question. normalcdf(left boundary, right boundary) Finding Probabilities from Standard(Z) scores

  15. Example: If thermometers have an average (mean) reading of 0 degrees and a standard deviation of 1 degree for freezing water and if one thermometer is randomly selected, find the probability that it reads freezing water between 0 degrees and 1.58 degrees. Since  = 0 and  =1, the values are standard or z-scores.

  16. Example: If thermometers have an average (mean) reading of 0 degrees and a standard deviation of 1 degree for freezing water and if one thermometer is randomly selected, find the probability that it reads freezing water between 0 degrees and 1.58 degrees. Since  = 0 and  =1, the values are standard or z-scores. 0 1.58

  17. Example: If thermometers have an average (mean) reading of 0 degrees and a standard deviation of 1 degree for freezing water and if one thermometer is randomly selected, find the probability that it reads freezing water between 0 degrees and 1.58 degrees. Since  = 0 and  =1, the values are standard or z-scores. P ( 0 < x < 1.58 ) = ? 0 1.58

  18. Table A-2 Standard Normal (z) Distribution z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3.0 .0000 .0398 .0793 .1179 .1554 .1915 .2257 .2580 .2881 .3159 .3413 .3643 .3849 .4032 .4192 .4332 .4452 .4554 .4641 .4713 .4772 .4821 .4861 .4893 .4918 .4938 .4953 .4965 .4974 .4981 .4987 .0040 .0438 .0832 .1217 .1591 .1950 .2291 .2611 .2910 .3186 .3438 .3665 .3869 .4049 .4207 .4345 .4463 .4564 .4649 .4719 .4778 .4826 .4864 .4896 .4920 .4940 .4955 .4966 .4975 .4982 .4987 .0080 .0478 .0871 .1255 .1628 .1985 .2324 .2642 .2939 .3212 .3461 .3686 .3888 .4066 .4222 .4357 .4474 .4573 .4656 .4726 .4783 .4830 .4868 .4898 .4922 .4941 .4956 .4967 .4976 .4982 .4987 .0120 .0517 .0910 .1293 .1664 .2019 .2357 .2673 .2967 .3238 .3485 .3708 .3907 .4082 .4236 .4370 .4484 .4582 .4664 .4732 .4788 .4834 .4871 .4901 .4925 .4943 .4957 .4968 .4977 .4983 .4988 .0160 .0557 .0948 .1331 .1700 .2054 .2389 .2704 .2995 .3264 .3508 .3729 .3925 .4099 .4251 .4382 .4495 .4591 .4671 .4738 .4793 .4838 .4875 .4904 .4927 .4945 .4959 .4969 .4977 .4984 .4988 .0199 .0596 .0987 .1368 .1736 .2088 .2422 .2734 .3023 .3289 .3531 .3749 .3944 .4115 .4265 .4394 .4505 .4599 .4678 .4744 .4798 .4842 .4878 .4906 .4929 .4946 .4960 .4970 .4978 .4984 .4989 .0239 .0636 .1026 .1406 .1772 .2123 .2454 .2764 .3051 .3315 .3554 .3770 .3962 .4131 .4279 .4406 .4515 .4608 .4686 .4750 .4803 .4846 .4881 .4909 .4931 .4948 .4961 .4971 .4979 .4985 .4989 .0279 .0675 .1064 .1443 .1808 .2157 .2486 .2794 .3078 .3340 .3577 .3790 .3980 .4147 .4292 .4418 .4525 .4616 .4693 .4756 .4808 .4850 .4884 .4911 .4932 .4949 .4962 .4972 .4979 .4985 .4989 .0319 .0714 .1103 .1480 .1844 .2190 .2517 .2823 .3106 .3365 .3599 .3810 .3997 .4162 .4306 .4429 .4535 .4625 .4699 .4761 .4812 .4854 .4887 .4913 .4934 .4951 .4963 .4973 .4980 .4986 .4990 .0359 .0753 .1141 .1517 .1879 .2224 .2549 .2852 .3133 .3389 .3621 .3830 .4015 .4177 .4319 .4441 .4545 .4633 .4706 .4767 .4817 .4857 .4890 .4916 .4936 .4952 .4964 .4974 .4981 .4986 .4990 * *

  19. Example: If thermometers have an average (mean) reading of 0 degrees and a standard deviation of 1 degree for freezing water and if one thermometer is randomly selected, find the probability that it reads freezing water between 0 degrees and 1.58 degrees. Since  = 0 and  =1, the values are standard or z-scores. P ( 0 < x < 1.58 ) = Normalcdf(0, 1.58) = .443 0 1.58

  20. The probability that the chosen thermometer will measure freezing water between 0 and 1.58 degrees is 0.443. Example: If thermometers have an average (mean) reading of 0 degrees and a standard deviation of 1 degree for freezing water and if one thermometer is randomly selected, find the probability that it reads freezing water between 0 degrees and 1.58 degrees. Since  = 0 and  =1, the values are standard or z-scores. Area = 0.443 P ( 0 < x < 1.58 ) = 0.443 0 1.58

  21. There is 44.3% of the thermometers with readings between 0 and 1.58 degrees. Example: If thermometers have an average (mean) reading of 0 degrees and a standard deviation of 1 degree for freezing water and if one thermometer is randomly selected, find the probability that it reads freezing water between 0 degrees and 1.58 degrees. Since  = 0 and  =1, the values are standard or z-scores. Area = 0.443 P ( 0 < x < 1.58 ) = 0.443 0 1.58

  22. Example: If thermometers have an average (mean) reading of 0 degrees and a standard deviation of 1 degree for freezing water, and if one thermometer is randomly selected, find the probability that it reads freezing water between -2.43 degrees and 0degrees. -2.43 0

  23. Example: If thermometers have an average (mean) reading of 0 degrees and a standard deviation of 1 degree for freezing water, and if one thermometer is randomly selected, find the probability that it reads freezing water between -2.43 degrees and 0degrees. Area = 0.493 P ( -2.43 < x < 0 ) = Normalcdf(-2.43, 0)= 0.493 -2.43 0 NOTE: Although a z score can be negative, the area under the curve (or the corresponding probability)can never be negative.

  24. P(a < z < b) → Normalcdf(a,b) denotes the probability that the z score is between a and b P(z > a) → Normalcdf(a,9999) denotes the probability that the z score is greater than a P (z < a) → Normalcdf(-9999,a) denotes the probability that the z score is less than a Note: 9999 is used to represent infinity Notation

  25. Probability of Half of a Distribution Remember the graph represents the entire distribution with its area of 1. 0.5 0.5 =0

  26. Finding the Area to the Right of z= 1.27 P(z>1.27) = normalcdf(1.27, 9999) = 0.1020 z = 1.27 0

  27. Finding the Area Between z= 1.20 and z = 2.30 P(1.20 <z< 2.30) = normalcdf(1.20, 2.30) = 0.104 Area = .104 z= 1.20 z= 2.30 0

  28. Figure 5-10 Interpreting Area Correctly ‘greater than x’ ‘at least x’ ‘more than x’ ‘not less than x’ x x

  29. Interpreting Area Correctly ‘greater than x’ ‘at least x’ ‘more than x’ ‘not less than x’ x x ‘less than x’ ‘at most x’ ‘no more than x’ ‘not greater than x’ x x

  30. 1. Draw a bell-shaped curve, draw the centerline, and shade the region under the curve that corresponds to the given probability. The line(s) that bound the region denotes the z-score(s) you are trying to find. Using the probability representing the area bounded by the z-score, convert it to area to the left of the z-score and enter that value into invNorm() function in your calculator. z = invNorm(area to the left) This command finds the z-score that has area to the left of the boundary under the normal curve. Since the area represent probability the argument must be between 0 and 1 inclusive. 3. If the z score is positioned to the left of the centerline, make sure it’s negative. Finding a z - score when given a probability

  31. Finding z Scores when Given Probabilities Find the z-score that denotes the 95th Percentile. 95% 0.95 z 0 z = invNorm(.95) =1.65 ( z score will be positive )

  32. Finding z Scores when Given Probabilities Find the z-score that separate the bottom 10% and upper 90%. 90% 10% Bottom 10% 0.90 0.10 z 0 z = invNorm(.10) = -1.28 (zscore will be negative)

  33. Finding z Scores when Given Probabilities Find the z-scores that denotes the middle 60% 20% 20% 60% Bottom 20% 0.60 0.20 0.20 0 z = invNorm(.20) = -0.84 z = invNorm(.80) = 0.84

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