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Chapter 6:

Chapter 6:. CONTINUOUS RANDOM VARIABLES AND THE NORMAL DISTRIBUTION. CONTINUOUS PROBABILITY DISTRIBUTION. Table 6.1 Frequency and Relative Frequency Distributions of Heights of Female Students. Figure 6.1 Histogram and polygon for Table 6.1.

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Chapter 6:

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  1. Chapter 6: CONTINUOUS RANDOM VARIABLES AND THE NORMAL DISTRIBUTION

  2. CONTINUOUS PROBABILITY DISTRIBUTION Table 6.1Frequency and Relative Frequency Distributions of Heights of Female Students

  3. Figure 6.1 Histogram and polygon for Table 6.1.

  4. Figure 6.2 Probability distribution curve for heights.

  5. CONTINUOUS PROBABILITY DISTRIBUTION Two characteristics • The probability that x assumes a value in any interval lies in the range 0 to 1 • The total probability of all the (mutually exclusive) intervals within which x can assume a value of 1.0

  6. Figure 6.3 Area under a curve between two points. Shaded area is between 0 and 1 x = a x = b x

  7. Figure 6.4 Total area under a probability distribution curve. Shaded area is 1.0 or 100% x

  8. Figure 6.5 Area under the curve as probability. Shaded area gives the probability P (a ≤x ≤b) a b x

  9. Figure 6.6 Probability that x lies in the interval 65 to 68 inches.

  10. Figure 6.7 The probability of a single value of x is zero.

  11. Figure 6.8 Probability “from 65 to 68” and “between 65 and 68”.

  12. THE NORMAL DISTRIBUTION • Normal Probability Distribution • A normal probability distribution , when plotted, gives a bell-shaped curve such that • The total area under the curve is 1.0. • The curve is symmetric about the mean. • The two tails of the curve extend indefinitely.

  13. Figure 6.11 Normal distribution with mean μand standard deviation σ. Standard deviation = σ Mean =μ x

  14. Figure 6.12 Total area under a normal curve. The shaded area is 1.0 or 100% μ x

  15. Figure 6.13 A normal curve is symmetric about the mean. Each of the two shaded areas is .5 or 50% .5 .5 μ x

  16. Figure 6.14 Areas of the normal curve beyond μ± 3σ. Each of the two shaded areas is very close to zero μ μ–3σ μ + 3σ x

  17. Figure 6.15 Three normal distribution curves with the same mean but different standard deviations. σ = 5 σ = 10 σ = 16 x μ = 50

  18. Figure 6.16 Three normal distribution curves with different means but the same standard deviation. σ = 5 σ = 5 σ = 5 µ = 20 µ = 30 µ = 40 x

  19. THE STANDARD NORMAL DISTRIBTUION • Definition • The normal distribution with μ = 0 and σ = 1 is called the standard normal distribution.

  20. Figure 6.17 The standard normal distribution curve. σ = 1 µ = 0 z -3 -2 -1 0 1 2 3

  21. THE STANDARD NORMAL DISTRIBTUION • Definition • The units marked on the horizontal axis of the standard normal curve are denoted by z and are called the z values or z scores. A specific value of z gives the distance between the mean and the point represented by z in terms of the standard deviation.

  22. Figure 6.18 Area under the standard normal curve. Each of these two areas is .5 . 5 . 5 -3 -2 -1 0 1 2 3 z

  23. Example 6-1 • Find the area under the standard normal curve between z = 0 and z = 1.95.

  24. Table 6.2 Area Under the Standard Normal Curve from z = 0 to z = 1.95 Required area

  25. Figure 6.19 Area between z = 0 to z = 1.95. Shaded area is .4744 0 1.95 z

  26. Example 6-2 • Find the area under the standard normal curve from z = -2.17 to z = 0.

  27. Solution 6-2 • Because the normal distribution is symmetric about the mean, the area from z = -2.17 to z = 0 is the same as the area from z = 0 to z = 2.17. • Area from -2.17 to 0 = P(-2.17≤ z ≤ 0) = .4850

  28. Figure 6.20 Area from z = 0 to z = 2.17 equals area from z = -2.17 to z = 0. Because the symmetry the shaded areas are equal -2.17 0 z 0 2.17 z

  29. Table 6.3 Area Under the Standard Normal Curve from z = 0 to z = 2.17 Required area

  30. Figure 6.21 Area from z = -2.17 to z = 0. Shaded area is .4850 -2.17 0 z

  31. Example 6-3 • Find the following areas under the standard normal curve. • Area to the right of z = 2.32 • Area to the left of z = -1.54

  32. Solution 6-3 • Area to the right of 2.32 = P (z > 2.32) = .5 - .4898 = .0102 as shown in Figure 6.22

  33. Figure 6.22 Area to the right of z = 2.32. This area is .4898 from Table VII Shaded area is .5 - .4898 - .0102 .4898 0 2.32 z

  34. Solution 6-3 • Area to the left of -1.54 = P (z < -1.54) = .5 - .4382 = .0618 as shown in Figure 6.23

  35. Figure 6.23 Area to the left of z = -1.54. This area is .4382 from Table VII Shaded area is .5 - .4382 = .0618 .4382 0 z -1.54

  36. Example 6-4 • Find the following probabilities for the standard normal curve. • P (1.19 < z < 2.12) • P (-1.56 < z < 2.31) • P (z > -.75)

  37. Solution 6-4 • P (1.19 < z < 2.12) = Area between 1.19 and 2.12 = .4830 - .3830 = .1000 as shown in Figure 6.24

  38. Figure 6.24 Finding P (1.19 <z < 2.12). Shaded area = .4830 - .3830 = .1000 z 0 1.19 2.12

  39. Solution 6-4 • P (-1.56 < z < 2.31) = Area between -1.56 and 2.31 = .4406 + .4896 = .9302 as shown in Figure 6.25

  40. Figure 6.25 Finding P (-1.56 < z < 2.31). Total shaded area = .4406 + .4896 = .9302 .4406 .4896 z -1.56 0 2.31

  41. Solution 6-4 • P (z > -.75) = Area to the right of -.75 = .2734 + .5 = .7734 as shown in Figure 6.26

  42. Figure 6.26 Finding P (z > -.75). Total shaded area = .2734 + .5 = .7734 .2734 .5 -.75 0 z

  43. Figure 6.27 Area within one standard deviation of the mean. Total shaded area is .3413 + .3413 = .6826 or 68.26% .3413 .3413 -1.0 0 1.0 z

  44. Figure 6.28 Area within two standard deviations of the mean. Total shaded area is .4772 + .4772 = .9544 or 95.44% .4772 .4772 -2.0 0 2.0 z

  45. Figure 6.29 Area within three standard deviations of the mean. Total shaded area is .4987 + .4987 = .9974 or 99.74% .4987 .4987 z -3.0 0 3.0

  46. Example 6-5 • Find the following probabilities for the standard normal curve. • P (0 < z < 5.67) • P (z < -5.35)

  47. Solution 6-5 • P (0 < z < 5.67) = Area between 0 and 5.67 = .5 approximately as shown in Figure 6.30

  48. Figure 6.30 Area between z = 0 and z = 5.67. Shaded area is approximately .5 0 z 5.67

  49. Solution 6-5 • P (z < -5.35) = Area to the left of -5.35 = .5 - .5 = .00 approximately as shown in Figure 6.31

  50. Figure 6.31 Area to the left of z = -5.35. This area is approximately .5 Shaded area is approximately .00 -5.35 0 z

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