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Chapter 6

Chapter 6. Normal Distributions. Table of Contents. 6.1 Graphs of Normal Probability Distributions. 6.2 Standard Units and Areas Under the Standard Normal Distribution. 6.3 Areas Under Any Normal Curve. 6.4 Normal Approximation to the Binomial Distribution.

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Chapter 6

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  1. Chapter 6 Normal Distributions

  2. Table of Contents 6.1 Graphs of Normal Probability Distributions 6.2 Standard Units and Areas Under the Standard Normal Distribution 6.3 Areas Under Any Normal Curve 6.4 Normal Approximation to the Binomial Distribution

  3. 6.1 Graphs of Normal Probability Distributions Figure 6-1 µ 1. The curve is bell-shaped with the highest point over the mean µ.

  4. Figure 6-1 µ 2. It is symmetrical about a vertical line through µ. 6.1 Graphs of Normal Probability Distributions

  5. Figure 6-1 µ 3. The curve approaches the horizontal axis but never touches or crosses it. 6.1 Graphs of Normal Probability Distributions

  6. Figure 6-1 Downward cup Inflection points Upward cup Upward cup µ – σ µ µ + σ 4. The inflection (transition) points between cupping upward and downward occur above µ + σ and µ – σ. 6.1 Graphs of Normal Probability Distributions

  7. Empirical rule For a distribution that is symmetrical and bell-shaped (in particular, for a normal distribution): Approximately 68% of the data values will lie within one standard deviation on each side of the mean. Approximately 95% of the data values will lie within two standard deviations on each side of the mean. Approximately 99.7% (or almost all) of the data values will lie within three standard deviations on each side of the mean. 6.1 Graphs of Normal Probability Distributions

  8. 2.35% 2.35% 13.5% 34% 34% 13.5% 68% 95% 99.7% Figure 6-3 µ–σ µ µ+σ µ–3σ µ–2σ µ+2σ µ+3σ 6.1 Graphs of Normal Probability Distributions

  9. Example 1 34% 600 µ 700 µ + σ 600 700  µ = 600  700 = 600 + 100 = µ + 100 = µ + σ 6.1 Graphs of Normal Probability Distributions

  10. How to make a control chart for the random variable x A control chart for a random variable x is a plot of observed x values in time sequence order. 1. Find the mean µ and standard deviation σ of the x distribution by (a) using past data from a period during which the process was in “in control” or (b) using specified “target” values for µ and σ. 2. Create a graph in which the vertical axis represents x values and the horizontal axis represents time. 6.1 Graphs of Normal Probability Distributions

  11. How to make a control chart for the random variable x 3. Draw a horizontal line at height µ and horizontal, dashed control-limit lines at µ ± 2σ and µ ± 3σ. 4. Plot the variable x on the graph in time sequence order. Use line segments to connect the points in time sequencing order. 6.1 Graphs of Normal Probability Distributions

  12. Example 2 µ+3σ 33.40 µ+2σ 28.70 Rooms µ 19.30 µ–2σ 9.90 5.20 µ–3σ 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Day 6.1 Graphs of Normal Probability Distributions

  13. µ+3σ µ µ–3σ Out-of-control signals 1.Out-of-Control Signal I: One point falls beyond the 3σ level These points generate an out-of-control signal 6.1 Graphs of Normal Probability Distributions

  14. µ Out-of-control signals 2.Out-of-Control Signal II: A run of nine consecutive points on one side of the center line (the line at target value µ) This sequence generates an out-of-control signal This sequence generates an out-of-control signal 6.1 Graphs of Normal Probability Distributions

  15. µ+2σ µ µ–2σ Out-of-control signals 3.Out-of-Control Signal III: At least two of three consecutive points lie beyond the 2σ level on the same side of the center line Each of these sequences generates an out-of-control signal 6.1 Graphs of Normal Probability Distributions

  16. 6.1 Graphs of Normal Probability Distributions

  17. µ+3σ 33.40 µ+2σ 28.70 Rooms µ 19.30 µ–2σ 9.90 5.20 µ–3σ 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Day Example 3 No type I errors; no type II errors; no type III errors. The x distribution appears to be “in control”. 6.1 Graphs of Normal Probability Distributions

  18. 6.2 Standard Units and Areas Under the Standard Normal Distribution Figure 6-16 64 72 82 74

  19. x – µ z = ──── σ Definition The z value or z score gives the number of standard deviations between the original measurement x and the mean µ of the x distribution. 6.2 Standard Units and Areas Under the Standard Normal Distribution

  20. 6.2 Standard Units and Areas Under the Standard Normal Distribution

  21. Unless otherwise stated, in the remainder of the book we will take the word average to be either the sample arithmetic mean x or the population mean µ. ¯ Note 6.2 Standard Units and Areas Under the Standard Normal Distribution

  22. Definition Given an x distribution with mean µ and standard deviation σ, the raw score x corresponding to a z score is x = zσ + µ 6.2 Standard Units and Areas Under the Standard Normal Distribution

  23. Definition The standard normal distribution is a normal distribution with mean µ = 0 and standard deviation σ = 1. 6.2 Standard Units and Areas Under the Standard Normal Distribution

  24. 68% 95% 99.7% Figure 6-15 –1 0 1 –3 –2 2 3 6.2 Standard Units and Areas Under the Standard Normal Distribution

  25. Calculator function The TI-83/84 calculator has a function to compute the area under the standard normal curve. DISTR|DISTR|normalcdf(low z,high z) [2nd][VARS] 6.2 Standard Units and Areas Under the Standard Normal Distribution

  26. Calculator function The TI-83/84 calculator has a function to compute the area under the standard normal curve. DISTR|DISTR|normalcdf(low z,high z) Calculates the area under the standard normal distribution for a z-interval ranging from low z to high z. 6.2 Standard Units and Areas Under the Standard Normal Distribution

  27. 0 z Calculator function z-interval is (–∞,z] normalcdf(-1E99,z) Use [(-)], not [–]. [2nd][,] 6.2 Standard Units and Areas Under the Standard Normal Distribution

  28. 0 z Calculator function z-interval is (–∞,z] normalcdf(-1E99,z) Since the calculator does not have a representation for –∞, we approximate it with -1E99, which stands for –1 × 1099. 6.2 Standard Units and Areas Under the Standard Normal Distribution

  29. Example 5 (a) The z-interval is (–∞,–1]. normalcdf(-1E99,-1) .1586552596 (b) The z-interval is (–∞,1.18]. normalcdf(-1E99,1.18) .880999834 6.2 Standard Units and Areas Under the Standard Normal Distribution

  30. 0 z Calculator function z-interval is [z,∞) normalcdf(z,1E99) 6.2 Standard Units and Areas Under the Standard Normal Distribution

  31. z1 0 z2 Calculator function z-interval is [z1,z2] normalcdf(z1,z2) 6.2 Standard Units and Areas Under the Standard Normal Distribution

  32. Example 6 (a) The z-interval is [1,2.7]. normalcdf(1,2.7) .1551882365 (b) The z-interval is [0.94,∞). normalcdf(0.94,1E99) .1736087621 6.2 Standard Units and Areas Under the Standard Normal Distribution

  33. 6.3 Areas Under Any Normal Curve Calculator function The TI-83/84 calculator has a function to compute the area under any normal curve. DISTR|DISTR|normalcdf(low x,high x,µ,σ) Calculates the area under any normal distribution with mean µ and standard deviation σ for an x-interval ranging from low x to high x.

  34. Example 7 Let x have a normal distribution with µ = 10 and σ = 2. Find the probability that an x value selected at random from this distribution is between 11 and 14. In symbols, find P(11 ≤ x ≤ 14). normalcdf(11,14,10,2) .2857874702 6.3 Areas Under Any Normal Curve

  35. So far, we’ve talked about finding the probability A given a z- or x-interval. Now we’ll be discussing how to find the z- or x-interval given a probability A. 6.3 Areas Under Any Normal Curve

  36. 0 z Calculator function A Left-tail case DISTR|DISTR|invNorm(A) Returns the z value so that the probability corresponding to the z-interval (–∞,z] for the standard normal distribution is A. 6.3 Areas Under Any Normal Curve

  37. µ x Calculator function A Left-tail case DISTR|DISTR|invNorm(A,µ,σ) Returns the x value so that the probability corresponding to the x-interval (–∞,x] for a normal distribution defined by µ and σ is A. 6.3 Areas Under Any Normal Curve

  38. Example 8 Computer life is normally distributed with mean µ = 30 months and standard deviation σ = 4 months. Find x so that the probability that a randomly selected computer fails in x or fewer months does not exceed A = 7%. invNorm(0.07,30,4) 24.09683589 6.3 Areas Under Any Normal Curve

  39. 0 z Calculator function A Right-tail case DISTR|DISTR|invNorm(1–A) Returns the z value so that the probability corresponding to the z-interval [z,∞) for the standard normal distribution is A. 6.3 Areas Under Any Normal Curve

  40. µ x Calculator function A Right-tail case DISTR|DISTR|invNorm(1–A,µ,σ) Returns the x value so that the probability corresponding to the x-interval [x,∞) for a normal distribution defined by µ and σ is A. 6.3 Areas Under Any Normal Curve

  41. –z 0 z Calculator function A Center case DISTR|DISTR|invNorm((1–A)/2) Returns the –z value so that the probability corresponding to the z-interval [–z,z] for the standard normal distribution is A. 6.3 Areas Under Any Normal Curve

  42. We will not be dealing with the center case for x-intervals in this course. 6.3 Areas Under Any Normal Curve

  43. Example 9 Find the z value such that 90% of the area under the standard normal curve lies between –z and z. invNorm((1–0.9)/2) -1.644853626 90% of the area under the standard normal curve lies between the z values –1.645 and 1.645. 6.3 Areas Under Any Normal Curve

  44. ─── µ = np and σ = √npq 6.4 Normal Approximation to the Binomial Distribution Consider a binomial distribution where n = number of trials r = number of successes p = probability of success on a single trial q = 1 – p = probability of failure on a single trial If np > 5 and nq > 5, then r has a binomial distribution that is approximated by a normal distribution with

  45. Example 11 This example is just a recap of what we’ve already done in chapter 5. As a means to refresh your memory, let’s do the first case asked for in the text, where n = 3. binompdf(3,0.25,0) binompdf(3,0.25,1) binompdf(3,0.25,2) binompdf(3,0.25,3) 6.4 Normal Approximation to the Binomial Distribution

  46. P(r) 0.40 0.35 0.30 0.25 0.20 0.15 0.10 0.05 0 1 2 3 r Example 11 Figure 6-34 6.4 Normal Approximation to the Binomial Distribution

  47. How to make the continuity correction Convert the discrete random variable r (number of successes) to the continuous normal random number variable x by doing the following: 1. If r is a left-point of an interval, subtract 0.5 to obtain the corresponding normal variable x; that is, x = r – 0.5. 2. If r is a right-point of an interval, add 0.5 to obtain the corresponding normal variable x; that is, x = r + 0.5. 6.4 Normal Approximation to the Binomial Distribution

  48. 6 7 8 9 10 P(5.5 ≤ x ≤ 10.5) E.g., P(6 ≤ r ≤ 10) [5.5,10.5] 5.5 and 10.5 are like class boundaries in histograms. 6.4 Normal Approximation to the Binomial Distribution

  49. 6 0 1 7 8 2 9 3 4 10 P(5.5 ≤ x ≤ 10.5) P(x ≤ 4.5) E.g., P(6 ≤ r ≤ 10) [5.5,10.5] E.g., P(r ≤ 4) (–∞,4.5] When using the normal approximation to the binomial distribution, we figure a single inequality as going to –∞ or ∞. 6.4 Normal Approximation to the Binomial Distribution

  50. … 8 9 10 P(x ≥ 7.5) Example 12 (a) The x-interval is [7.5,∞). normalcdf(7.5,1E99,25*0.25,√(25*0.25*0.75)) .2818513864 (b) 1-binomcdf(25,0.25,8–1) .2734937886 (c) 0.2819 – 0.2735 = 0.0084 6.4 Normal Approximation to the Binomial Distribution

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