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Understanding Continuous Random Variables and Normal Distribution Characteristics

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quyn-case
67 diapositives · 877 Views · Oct 27, 2014

This chapter delves into the fundamentals of continuous random variables and the normal distribution. It explains two key characteristics of probability, illustrating how the area under the probability distribution curve relates to the likelihood of outcomes. The chapter covers the standard normal distribution, including z-scores and their interpretations, as well as practical examples of calculating areas under the curve. Applications of the normal distribution, such as predicting market values, are explored, providing a comprehensive understanding of these statistical concepts.

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

  2. 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

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

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

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

  6. 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.

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

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

  9. 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.

  10. 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

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

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

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

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

  15. 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

  16. 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

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

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

  19. 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

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

  21. 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

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

  23. 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

  24. 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)

  25. 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

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

  27. 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

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

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

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

  31. STANDARDIZING A NORMAL DISTRIBUTION • Converting an x Value to a z Value • For a normal random variable x, a particular value of x can be converted to its corresponding z value by using the formula • where μ and σ are the mean and standard deviation of the normal distribution of x, respectively.

  32. Example 6-6 • Let x be a continuous random variable that has a normal distribution with a mean of 50 and a standard deviation of 10. convert the following x values to z values. • x = 55 • x = 35

  33. Solution 6-6 • ajdaj

  34. Figure 6.32 z value for x= 55. Normal distribution with μ = 50 and σ = 10 x μ= 50 x= 55 Standard normal distribution z 0 .50 zvalue forx= 55

  35. Solution 6-6

  36. Figure 6.33 z value for x= 35. σ = 10 μ = 50 x 35 0 z -1.50

  37. APPLICATIONS OF THE NORMAL DISTRIBUTION • Example 6-11 • According to Automotive Lease Guide, the Porsche 911 sports car is among the vehicles that hold their value best. A Porsche 911 (with price of $87,500 for a new car) is expected to command a price of $48,125 after three years (The Wall Street Journal, August 6, 2002).

  38. Example 6-11 • Suppose the prices of all three-year old Porsche 911 sports cars have a normal distribution with a mean price of $48,125 and a standard deviation of $1600. Find the probability that a randomly selected three-year-old Porsche 911 will sell for a price between $46,000 and $49,000.

  39. Solution 6-11 • For x = $46,000: • For x = $49,000: • P ($46,000 < x < $49,000) = P (-1.33 < z < .55) = .4082 + .2088 = .6170 = 61.70%

  40. Solution 6-11 • Thus, the probability is .6170 that a randomly selected three-year-old Porsche 911 sports car will sell for a price between $46,000 and $49,000.

  41. Figure 6.41 Area between x = $46,000 and x = $49,000. Shaded area is .4082 + .2088 = .6170 .2088 .4082 $46,000 $48,125 $49,000 x z -1.33 0 .55

  42. Example 6-12 • A racing car is one of the many toys manufactured by Mark Corporation. The assembly times for this toy follow a normal distribution with a mean of 55 minutes and a standard deviation of 4 minutes. The company closes at 5 P.M. everyday. If one worker starts to assemble a racing car at 4 P.M., what is the probability that she will finish this job before the company closes for the day?

  43. Solution 6-12 • μ= 55 minutes • σ= 4 minutes • For x = 60: • P (x ≤ 60) = P (z ≤ 1.25) = .5 + .3944 = .8944

  44. Solution 6-12 • Thus, the probability is .8944 that this worker will finish assembling this racing car before the company closes for the day.

  45. Figure 6.42 Area to the left of x = 60. Shaded area is .5 + .3944 = .8944 .5 .3944 55 60 x 0 1.25 z

  46. Example 6-13 • Hupper Corporation produces many types of soft drinks, including Orange Cola. The filling machines are adjusted to pour 12 ounces of soda into each 12-ounce can of Orange Cola. However, the actual amount of soda poured into each can is not exactly 12 ounces; it varies from can to can. It has been observed that the net amount of soda in such a can has a normal distribution with a mean of 12 ounces and a standard deviation of .015 ounce.

  47. Example 6-13 • What is the probability that a randomly selected can of Orange Cola contains 11.97 to 11.99 ounces of soda? • What percentage of the Orange Cola cans contain 12.02 to 12.07 ounces of soda?

  48. Solution 6-13 • For x = 11.97: • For x = 11.99: • P (11.97 ≤ x ≤ 11.99) = P (-2.00 ≤ z ≤ -.67) = .4772 - .2486 = .2286

  49. Figure 6.43 Area between x = 11.97 and x = 11.99. Shaded area = .4772 - .2486 = .2286 11.97 11.99 12 x -2.00 -.67 0 z

  50. Solution 6-13 • For x = 12.02: • For x = 12.07: • P (12.02 ≤ x ≤ 12.07) = P (1.33 ≤ z ≤ 4.67) = .5 - .4082 = .0918