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Chapter 6:

Chapter 6:. CONTINUOUS RANDOM VARIABLES AND THE NORMAL DISTRIBUTION. CONTINUOUS PROBABILITY DISTRIBUTION. Two characteristics The probability that x assumes a value in any interval lies in the range 0 to 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 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

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