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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 Height of a Female Student (inches) f x 60 61 62 63 64 65 66 67 68 69 70 to to to to to to to to to to to less less less less less less less less less less less than than than than than than than than than than than 61 62 63 64 65 66 67 68 69 70 71 Relative Frequency 90 170 460 750 970 760 640 440 320 220 180 .018 .034 .092 .150 .194 .152 .128 .088 .064 .044 .036 N = 5000 Sum = 1.0 2 Figure 6.1 Histogram and polygon for Table 6.1. 3 Figure 6.2 Probability distribution curve for heights. 4 CONTINUOUS PROBABILITY DISTRIBUTION Two characteristics 1. The probability that x assumes a value in any interval lies in the range 0 to 1 2. The total probability of all the (mutually exclusive) intervals within which x can assume a value of 1.0 5 Figure 6.3 Area under a curve between two points. Shaded area is between 0 and 1 x=a x=b x 6 Figure 6.4 Total area under a probability distribution curve. Shaded area is 1.0 or 100% x 7 Figure 6.5 Area under the curve as probability. Shaded area gives the probability P (a ≤ x ≤ b) a b x 8 Figure 6.6 Probability that x lies in the interval 65 to 68 inches. 9 Figure 6.7 The probability of a single value of x is zero. 10 Figure 6.8 Probability “from 65 to 68” and “between 65 and 68”. 11 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. 2. The curve is symmetric about the mean. 3. The two tails of the curve extend indefinitely. 1. 12 Figure 6.11 Normal distribution with mean μ and standard deviation σ. Standard deviation = σ Mean = μ x 13 Figure 6.12 Total area under a normal curve. The shaded area is 1.0 or 100% μ x 14 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 15 Figure 6.14 Areas of the normal curve beyond μ ± 3σ. Each of the two shaded areas is very close to zero μ – 3σ μ μ + 3σ x 16 Figure 6.15 Three normal distribution curves with the same mean but different standard deviations. σ=5 σ = 10 σ = 16 μ = 50 x 17 Figure 6.16 σ=5 µ = 20 Three normal distribution curves with different means but the same standard deviation. σ=5 σ=5 µ = 30 µ = 40 x 18 THE STANDARD NORMAL DISTRIBTUION Definition The normal distribution with μ = 0 and σ = 1 is called the standard normal distribution. 19 Figure 6.17 The standard normal distribution curve. σ=1 µ=0 -3 -2 -1 0 1 2 3 20 z 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. 21 Figure 6.18 Area under the standard normal curve. Each of these two areas is .5 .5 -3 -2 -1 .5 0 1 2 3 z 22 Example 6-1 Find the area under the standard normal curve between z = 0 and z = 1.95. 23 Table 6.2 Area Under the Standard Normal Curve from z = 0 to z = 1.95 z 0.0 0.1 0.2 . 1.9 . . 3.0 .00 .0000 .0398 .0793 . .4713 . . .4987 .01 .0040 .0438 .0832 . .4719 . . .4987 … … … … … … … … … .05 .0199 .0596 .0987 . .4744 . . .4989 … … … … … … … … … .09 .0359 .0753 .1141 . .4767 . . .4990 Required area 24 Figure 6.19 Area between z = 0 to z = 1.95. Shaded area is .4744 0 1.95 z 25 Example 6-2 Find the area under the standard normal curve from z = -2.17 to z = 0. 26 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 27 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 28 Table 6.3 Area Under the Standard Normal Curve from z = 0 to z = 2.17 z 0.0 0.1 0.2 . . 2.1 . 3.0 .00 .0000 .0398 .0793 . . .4821 . .4987 .01 .0040 .0438 .0832 . . .4826 . .4987 … … … … … … … … … .07 .0279 .0675 .1064 . . .4850 . .4989 … .09 … .0359 … .0753 … .1141 … . … . … .4857 … . … .4990 Required area 29 Figure 6.21 Area from z = -2.17 to z = 0. Shaded area is .4850 -2.17 0 z 30 Example 6-3 Find the following areas under the standard normal curve. Area to the right of z = 2.32 b) Area to the left of z = -1.54 a) 31 Solution 6-3 a) Area to the right of 2.32 = P (z > 2.32) = .5 - .4898 = .0102 as shown in Figure 6.22 32 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 33 Solution 6-3 b) Area to the left of -1.54 = P (z < -1.54) = .5 - .4382 = .0618 as shown in Figure 6.23 34 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 -1.54 0 z 35 Example 6-4 Find the following probabilities for the standard normal curve. P (1.19 < z < 2.12) b) P (-1.56 < z < 2.31) c) P (z > -.75) a) 36 Solution 6-4 a) P (1.19 < z < 2.12) = Area between 1.19 and 2.12 = .4830 - .3830 = .1000 as shown in Figure 6.24 37 Figure 6.24 Finding P (1.19 < z < 2.12). Shaded area = .4830 - .3830 = .1000 0 1.19 2.12 z 38 Solution 6-4 b) P (-1.56 < z < 2.31) = Area between -1.56 and 2.31 = .4406 + .4896 = .9302 as shown in Figure 6.25 39 Figure 6.25 Finding P (-1.56 < z < 2.31). Total shaded area = .4406 + .4896 = .9302 .4406 -1.56 .4896 0 2.31 z 40 Solution 6-4 c) P (z > -.75) = Area to the right of -.75 = .2734 + .5 = .7734 as shown in Figure 6.26 41 Figure 6.26 Finding P (z > -.75). Total shaded area = .2734 + .5 = .7734 .2734 .5 -.75 0 z 42 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 43 Figure 6.28 Area within two standard deviations of the mean. Total shaded area is .4772 + .4772 = .9544 or 95.44% .4772 -2.0 .4772 0 2.0 z 44 Figure 6.29 Area within three standard deviations of the mean. Total shaded area is .4987 + .4987 = .9974 or 99.74% .4987 -3.0 .4987 0 3.0 z 45 Example 6-5 Find the following probabilities for the standard normal curve. a) b) P (0 < z < 5.67) P (z < -5.35) 46 Solution 6-5 a) P (0 < z < 5.67) = Area between 0 and 5.67 = .5 approximately as shown in Figure 6.30 47 Figure 6.30 Area between z = 0 and z = 5.67. Shaded area is approximately .5 0 5.67 z 48 Solution 6-5 b) P (z < -5.35) = Area to the left of -5.35 = .5 - .5 = .00 approximately as shown in Figure 6.31 49 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 50 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 z x where μ and σ are the mean and standard deviation of the normal distribution of x, respectively. 51 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. a) x = 55 b) x = 35 52 Solution 6-6 a) ajdaj z x 55 50 .50 10 53 Figure 6.32 z value for x = 55. Normal distribution with μ = 50 and σ = 10 μ= 50 x = 55 x Standard normal distribution 0 .50 z z value for x = 55 54 Solution 6-6 b) z x 35 50 1.50 10 55 Figure 6.33 z value for x = 35. σ = 10 35 μ = 50 -1.50 0 x z 56 Example 6-7 Let x be a continuous random variable that is normally distributed with a mean of 25 and a standard deviation of 4. Find the area a) between x = 25 and x = 32 b) between x = 18 and x = 34 57 Solution 6-7 a) The z value for x = 25 is 0 The z value for x = 32 is x 32 25 z 1.75 4 P (25 < x < 32) = P(0 < z < 1.75) = .4599 58 Figure 6.34 Area between x = 25 and x = 32. .4599 25 32 These areas are equal x .4599 0 1.75 z 59 Solution 6-7 b) 18 25 1.75 For x = 18: z 4 For x = 34: z 34 25 2.25 4 P (18 < x < 34) = P (-1.75 < z < 2.25 ) = .4599 + .4878 = .9477 60 Figure 6.35 Area between x = 18 and x = 34. Shaded area = .4599 + .4878 = .9477 .4599 18 -1.75 .4878 25 34 x 0 2.25 z 61 Example 6-8 Let x be a normal random variable with its mean equal to 40 and standard deviation equal to 5. Find the following probabilities for this normal distribution a) b) P (x > 55) P (x < 49) 62 Solution 6-8 a) 55 40 For x = 55: z 5 3.00 P (x > 55) = P (z > 3.00) = .5 - .4987 = .0013 63 Figure 6.36 Finding P (x > 55). Shaded area = .5 - .4987 = .0013 .4987 40 55 x 0 3.00 z 64 Solution 6-8 b) 49 40 z 1.80 5 For x = 49: P (x < 49) = P (z < 1.80) = .5 + .4641 = .9641 65 Figure 6.37 Finding P (x > 49). Shaded area = .5 + .4641 = .9641 .5 .4641 40 49 x 0 1.80 z 66 Example 6-9 Let x be a continuous random variable that has a normal distribution with μ = 50 and σ = 8. Find the probability P (30 ≤ x ≤ 39). 67 Solution 6-9 For x = 30: 30 50 z 2.50 8 For x = 39: 39 50 z 1.38 8 P (30 ≤ x ≤ 39) = P (-2.50 ≤ z ≤ -1.38) = .4938 - .4162 = .0776 68 Figure 6.38 Finding P (30 ≤ x ≤ 39). Shaded area = .4938 - .4162 = .0776 30 39 50 x -2.50 -1.38 0 z 69 Example 6-10 Let x be a continuous random variable that has a normal distribution with a mean of 80 and a standard deviation of 12. Find the area under the normal distribution curve a) from x = 70 to x = 135 b) to the left of 27 70 Solution 6-10 a) For x = 70: 70 80 z .83 12 135 80 For x = 135: z 12 4.58 P (70 ≤ x ≤ 135) = P (-.83 ≤ z ≤ 4.58) = .2967 + .5 = .7967 approximately 71 Figure 6.39 Area between x = 70 and x = 135. Shaded area is approximately .2967 + .5 = .7967 70 80 135 x -.83 0 4.58 z 72 Solution 6-10 b) 27 80 4.42 For x = 27: z 12 P (x < 27) = P (z < -4.42) = .5 - .5 = .00 approximately 73 Figure 6.40 Area to the left of x = 27. Shaded area is approximately .5 - .5 = .00 27 -4.42 80 x 0 z 74 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). 75 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. 76 Solution 6-11 46,000 48,125 1.33 For x = $46,000: z 1600 For x = $49,000: z 49,000 48,125 .55 1600 P ($46,000 < x < $49,000) = P (-1.33 < z < .55) = .4082 + .2088 = .6170 = 61.70% 77 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. 78 Figure 6.41 Area between x = $46,000 and x = $49,000. .4082 .2088 Shaded area is .4082 + .2088 = .6170 $46,000 $48,125 $49,000 -1.33 0 .55 x z 79 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? 80 Solution 6-12 μ = 55 minutes σ = 4 minutes 60 55 z 1.25 4 For x = 60: P (x ≤ 60) = P (z ≤ 1.25) = .5 + .3944 = .8944 81 Solution 6-12 Thus, the probability is .8944 that this worker will finish assembling this racing car before the company closes for the day. 82 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 83 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. 84 Example 6-13 a) b) 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? 85 Solution 6-13 a) For x = 11.97: 11.97 12 z 2.00 .015 For x = 11.99: 11.99 12 z .67 .015 P (11.97 ≤ x ≤ 11.99) = P (-2.00 ≤ z ≤ -.67) = .4772 - .2486 = .2286 86 Figure 6.43 Area between x = 11.97 and x = 11.99. Shaded area = .4772 - .2486 = .2286 11.97 11.99 12 -2.00 -.67 0 x z 87 Solution 6-13 b) For x = 12.02: 12.02 12 z 1.33 .015 For x = 12.07: 12.07 12 z 4.67 .015 P (12.02 ≤ x ≤ 12.07) = P (1.33 ≤ z ≤ 4.67) = .5 - .4082 = .0918 88 Figure 6.44 Area from x = 12.02 to x = 12.07. Shaded area is .5 - .4082 = .0918 μ = 12 0 12.02 1.33 12.07 4.67 x z 89 Example 6-14 The life span of a calculator manufactured by Texas Instruments has a normal distribution with a mean of 54 months and a standard deviation of 8 months. The company guarantees that any calculator that starts malfunctioning within 36 months of the purchase will be replaced by a new one. About what percentage of calculators made by this company are expected to be replaced? 90 Solution 6-14 36 54 z 2.25 8 For x = 36: P (x < 36) = P (z < -2.25) = .5 - .4878 = .0122 Hence, 1.22% of the calculators are expected to be replaced. 91 Figure 6.45 Area to the left of x = 36. Shaded area is .5 - .4878 = .0122 .4878 36 μ = 54 x -2.25 0 z 92 DETERMINING THE z AND x VALUES WHEN AN AREA UNDER THE NORMAL DISTRIBUTION CURVE IS KNOWN Example 6-15 Find a point z such that the area under the standard normal curve between 0 and z is .4251 and the value of z is positive. 93 Figure 6.46 Finding the z value. Shaded area is given to be .4251 0 z z To find this z 94 Table 6.4 Finding the z Value When Area Is Known. Solution 6-15 z 0.0 0.1 0.2 . 1.4 . 3.0 .00 .0000 .0398 .0793 . . .4987 .01 .0040 .0438 .0832 . . .4987 Therefore, z = 1.44 … … … … … … … .04 .4251 . .4988 … … … … … … … … .09 .0359 .0753 .1141 . . … .4990 We locate this value in Table VII of Appendix C 95 Example 6-16 Find the value of z such that the area under the standard normal curve in the right tail is .0050. 96 Solution 6-16 Area between 0 and z = .5 - .005 = .4950 Look for .4950 in the normal distribution table Find the value closest to .4950 Table VII does not contain .4950 It is either .4949 or .4951 z = 2.57 or 2.58. 97 Figure 6.47 Finding the z value. Shaded area is given to be .0050 .4950 0 z z To find this z value 98 Example 6-17 Find the value of z such that the area under the standard normal curve in the left tail is .05. 99 Solution 6-17 Area between 0 and z = .5 - .05 = .4500 z = -1.65. 100 Figure 6.48 Finding the z value. Shaded area is given to be .05 .4500 z 0 z To find this z value 101 Finding an x Value for a Normal Distribution For a normal curve, with known values of μ and σ and for a given area under the curve between the mean and x, the x value is calculated as x = μ + zσ 102 Example 6-18 It is known that the life of a calculator manufactured by Texas Instruments has a normal distribution with a mean of 54 months and a standard deviation of 8 months. What should the warranty period be to replace a malfunctioning calculator if the company does not want to replace more than 1% of all the calculators sold? 103 Solution 6-18 Area between the mean and the x value = .5 - .01 = .4900 z = -2.33 x = μ + zσ = 54 + (-2.33)(8) = 54 – 18.64 = 35.36 104 Solution 6-18 Thus, the company should replace all calculators that start to malfunction within 35.36 months (which can be rounded to 35 months) of the date of purchase so that they will not have to replace more than 1% of the calculators. 105 Figure 6.49 Finding an x value. Shaded area is given as.01 .5 - .01 = .4900 To find this x value x μ = 54 x -2.33 0 z From Table VII, this value of z is approximately -2.33 106 Example 6-19 Almost all high school students who intend to go to college take the SAT test. In 2002, the mean SAT score (in verbal and mathematics) of all students was 1020. Debbie is planning to take this test soon. Suppose the SAT scores of all students who take this test with Debbie will have a normal distribution with a mean of 1020 and a standard deviation of 153. What should her score be on this test so that only 10% of all examinees score higher than she does? 107 Solution 6-19 Area between μ and the x value = .5 - .10 = .4000 x = μ + zσ = 1020 + 1.28(153) = 1020 + 195.84 = 1215.84 ≈ 1216 Thus, if Debbie scores 1216 on the SAT, only about 10% of the examinees are expected to score higher than she does. 108 Figure 6.50 Finding an x value. Shaded area is given to be .10 .4000 μ = 1020 0 From normal distribution table, this value of z is approximately 1.28 x x 1.28 z To find this x value 109 THE NORMAL APPROXIMATION OF THE BINOMIAL DISTRIBUTION 1. 2. 3. 4. The binomial distribution is applied to a discrete random variable. Each repetition, called a trial, of a binomial experiment results in one of two possible outcomes, either a success or a failure. The probabilities of the two outcomes remain the same for each repetition of the experiment. The trials are independent. 110 THE NORMAL APPROXIMATION OF THE BINOMIAL DISTRIBUTION cont. The binomial formula, which gives the probability of x successes in n trials, is P( x) n C x p q x n x 111 THE NORMAL APPROXIMATION OF THE BINOMIAL DISTRIBUTION cont. Normal Distribution as an Approximation to Binomial Distribution Usually, the normal distribution is used as an approximation to the binomial distribution when np and nq are both greater than 5 - that is , when np > 5 and nq > 5 112 Table 6.5 x 0 1 2 3 4 5 6 7 8 9 10 11 12 The Binomial Probability Distribution for n = 12 and p = .50 P(x) .0002 .0029 .0161 .0537 .1208 .1934 .2256 .1934 .1208 .0537 .0161 .0029 .0002 113 Figure 6.51 Histogram for the probability distribution of Table 6.5. 114 Example 6-20 According to an estimate, 50% of the people in America have at least one credit card. If a random sample of 30 persons is selected, what is the probability that 19 of them will have at least one credit card? 115 Solution 6-20 n = 30 p = .50, q = 1 – p = .50 x = 19, n – x = 30 – 19 = 11 From the binomial formula, P(19) 30 C19 (.5) (.5) .0509 19 11 116 Solution 6-20 np 30(.50) 15 npq 30(.50)(.50) 2.73861279 117 Continuity Correction Factor Definition The addition of .5 and/or subtraction of .5 from the value(s) of x when the normal distribution is used as an approximation to the binomial distribution, where x is the number of successes in n trials, is called the continuity correction factor. 118 Figure 6.52 Correction for continuity x 119 Solution 6-20 For x = 18.5: 18.5 15 z 1.28 2.73861279 19.5 15 z 1.64 2.73861279 For x = 19.5: P (18.5 ≤ x ≤ 19.5) = P (1.28 ≤ z ≤ 1.64) = .4495 - .3997 = .0498 120 Solution 6-20 Thus, based on the normal approximation, the probability that 19 persons in a sample of 30 will have at lease on credit card is approximately .0498. Using the binomial formula, we obtain the exact probability .0509. The error due to using the normal approximation is .0509 - .0498 = .0011. 121 Figure 6.53 Area between x = 18.5 and x = 19.5. Shaded area is .4495 - .3997 = .0498 15 18.5 19.5 x z 0 1.28 1.64 122 Example 6-21 In a recent survey conducted for Money magazine, 80% of the women (married or single) surveyed said that they are more knowledgeable about investing now than they were just five years ago (Money, June 2002). Suppose this result is true for the current population of all women. What is the probability that in a random sample of 100 women, 72 to 76 will say that they know more about investing now than just five years ago? 123 Solution 6-21 np 100(.80) 80 npq 100(.80)(.20) 4.0 124 Solution 6-21 To make the continuity correction, we subtract .5 from 72 and add .5 to 76 to obtain the interval 71.5 to 76.5. 125 Solution 6-21 For x = 71.5: 71.5 80 z 2.13 4.0 For x = 76.5: 76.5 80 z .88 4.0 P (71.5 ≤ x ≤ 76.5) = P (-2.13 ≤ z ≤ -.88) = .4834 - .3106 = .1728 126 Solution 6-21 Thus, the probability that 72 to 76 women in a random sample of 100 say that they are more knowledgeable about investing now than they were just five years ago is approximately .1728. 127 Figure 6.54 Area from x = 71.5 to x = 76.5. Shaded area is .4834 - .3106 = .1728 71.5 76.5 80 x -2.13 -.88 0 z 128 Example 6-22 According to the 2001 Youth Risk Behavior Surveillance by the Centers for Disease Control and Prevention, 39% of the 10th-graders surveyed said that they watch three or more hours of television on a typical school day. Assume that this percentage is true for the current population of all 10th-graders. What is the probability that 86 or more of the 10thgraders in a random sample of 200 watch three or more hours of television on a typical school day? 129 Solution 6-22 np 200(.39) 78 npq 200(.39)(.61) 6.89782574 130 Solution 6-22 For the continuity correction, we subtract .5 from 86, which gives 85.5. 131 Solution 6-22 85.5 78 z 1.09 6.89782574 For x = 85.5: P (x ≥ 85.5) = P (z ≥ 1.09) = .5 - .3621 = .1379 132 Solution 6-22 Thus, the probability that 86 or more of the 10th-graders in a random sample of 200 watch three or more hours of television on a typical school say is approximately .1379. 133 Figure 6.55 Area to the right of x = 85.5. The required probability = .5 .3621 = .1379 78 0 85.5 1.09 x z 134