1 / 41

M ARIO F . T RIOLA

S TATISTICS. E LEMENTARY. Chapter 5 Normal Probability Distributions. M ARIO F . T RIOLA. E IGHTH. E DITION. Continuous random variable Normal distribution. Overview. 5-1. Continuous random variable Normal distribution. Curve is bell shaped and symmetric. µ Score.

zahur
Download Presentation

M ARIO F . T RIOLA

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

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

  2. Continuous random variable Normal distribution Overview 5-1

  3. Continuous random variable Normal distribution Curve is bell shaped and symmetric µ Score Overview 5-1 Figure 5-1

  4. Continuous random variable Normal distribution Curve is bell shaped and symmetric µ Score Overview 5-1 Figure 5-1 x - µ2 ( ) 1  2 y = e Formula 5-1 2 p

  5. The Standard Normal Distribution 5-2

  6. Uniform Distribution a probability distribution in which the continuous random variable values are spread evenly over the range of      possibilities; the graph results in a      rectangular shape. Definitions

  7. Density Curve (or probability density   function)   the graph of a continuous probability   distribution Definitions

  8. Density Curve (or probability density function) : The graph of a continuous probability distribution Definitions 1. The total area under the curve must equal 1. 2. Every point on the curve must have a vertical height that is 0 or greater.

  9. Because the total area under the density curve is equal to 1, there is a correspondence between area and probability.

  10. Times in First or Last Half Hours Figure 5-3

  11. Women: µ = 63.6  = 2.5 Heights of Adult Men and Women Men: µ = 69.0  = 2.8 63.6 69.0 Figure 5-4 Height (inches)

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

  13. Standard Normal Deviation a normal probability distribution that has a mean of 0 and a standard deviation of 1 -3 -2 -1 0 1 2 3 Definition Area found in Table A-2 Area = 0.3413 0.4429 z = 1.58 0 Score (z ) Figure 5-6 Figure 5-5

  14. µ = 0  = 1 Table A-2 Standard Normal Distribution 0 x z

  15. 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 * *

  16. the distance along horizontal scale of the standard normal distribution; refer to the leftmost column and top rowofTable A-2 Area the region under the curve; refer to the values in the body of Table A-2 To find:zScore

  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.

  18. 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. P ( 0 < x < 1.58 ) = 0 1.58

  19. 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 * *

  20. 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. Area = 0.4429 P ( 0 < x < 1.58 ) = 0.4429 0 1.58

  21. The probability that the chosen thermometer will measure freezing water between 0 and 1.58 degrees is 0.4429. 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. Area = 0.4429 P ( 0 < x < 1.58 ) = 0.4429 0 1.58

  22. There is 44.29% 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. Area = 0.4429 P ( 0 < x < 1.58 ) = 0.4429 0 1.58

  23. NOTE: Although a z score can be negative, the area under the curve (or the corresponding probability)can never be negative. Using Symmetry to Find the Area to the Left of the Mean Because of symmetry, these areas are equal. Figure 5-7 (a) (b) 0.4925 0.4925 0 0 z = - 2.43 z = 2.43 Equal distance away from 0

  24. The probability that the chosen thermometer will measure freezing water between -2.43 and 0 degrees is 0.4925. 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 = 0.4925 P ( -2.43 < x < 0 ) = 0.4925 -2.43 0

  25. The Empirical Rule Standard Normal Distribution: µ = 0 and  = 1

  26. The Empirical Rule Standard Normal Distribution: µ = 0 and  = 1 68% within 1 standard deviation 34% 34% x - s x x+s

  27. The Empirical Rule Standard Normal Distribution: µ = 0 and  = 1 95% within 2 standard deviations 68% within 1 standard deviation 34% 34% 13.5% 13.5% x - 2s x - s x x+s x+2s

  28. 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% x - 3s x - 2s x - s x x+s x+2s x+3s

  29. Probability of Half of a Distribution 0.5 0

  30. Finding the Area to the Right of z= 1.27 Value found in Table A-2 0.3980 This area is 0.5 - 0.3980 = 0.1020 z = 1.27 0 Figure 5-8

  31. Finding the Area Between z= 1.20 and z = 2.30 0.4893 (from Table A-2 with z = 2.30) Area A is 0.4893 - 0.3849 = 0.1044 0.3849 A z= 1.20 z= 2.30 0 Figure 5-9

  32. P(a < z < b) denotes the probability that the z score is between a and b P(z > a) denotes the probability that the z score is greater than a P (z < a) denotes the probability that the z score is less than a Notation

  33. Figure 5-10 Interpreting Area Correctly Subtract from 0.5 ‘greater than x’ ‘at least x’ ‘more than x’ ‘not less than x’ Add to 0.5 0.5 x x

  34. Figure 5-10 Interpreting Area Correctly Subtract from 0.5 ‘greater than x’ ‘at least x’ ‘more than x’ ‘not less than x’ Add to 0.5 0.5 x x Add to 0.5 ‘less than x’ ‘at most x’ ‘no more than x’ ‘not greater than x’ Subtract from 0.5 0.5 x x

  35. Figure 5-10 Interpreting Area Correctly Subtract from 0.5 ‘greater than x’ ‘at least x’ ‘more than x’ ‘not less than x’ Add to 0.5 0.5 x x Add to 0.5 ‘less than x’ ‘at most x’ ‘no more than x’ ‘not greater than x’ Subtract from 0.5 0.5 x x Add C Use A = C - B ‘between x1 and x2’ A B x1 x2 x1 x2

  36. 1. Draw a bell-shaped curve, draw the centerline, and identify the region under the curve that corresponds to the given probability. If that region is not bounded by the centerline, work with a known region that is bounded by the centerline. 2. Using the probability representing the area bounded by the centerline, locate the closest probability in the body of Table A-2 and identify the corresponding z score. 3. If the z score is positioned to the left of the centerline, make it a negative. Finding a z - score when given a probabilityUsing Table A-2

  37. Finding z Scores when Given Probabilities 95% 5% 5% or 0.05 0.45 0.50 z 0 ( z score will be positive ) FIGURE 5-11 Finding the 95th Percentile

  38. Finding z Scores when Given Probabilities 95% 5% 5% or 0.05 0.45 0.50 1.645 0 (z score will be positive) FIGURE 5-11 Finding the 95th Percentile

  39. Finding z Scores when Given Probabilities 90% 10% Bottom 10% 0.10 0.40 z 0 (zscore will be negative) FIGURE 5-12 Finding the 10th Percentile

  40. Finding z Scores when Given Probabilities 90% 10% Bottom 10% 0.10 0.40 -1.28 0 (zscore will be negative) FIGURE 5-12 Finding the 10th Percentile

  41. Assignment • Page 240: 1-36 all

More Related