1 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Chapter 6 Continuous Random Variables.

2 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Chapter 6. Continuous Random Variables Reminder: Continuous random variable takes infinitely many values Those values can be associated with measurements on a continuous scale (without gaps or interruptions)

3 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Example: Uniform Distribution A continuous random variable has a uniform distribution if its values are spread evenly over a certain range. pg 251

4 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Example: Uniform Distribution A continuous random variable has a uniform distribution if its values are spread evenly over a certain range. Example: voltage output of an electric generator is between 123 V and 125 V. The actual voltage level may be anywhere in this range.

5 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Using Area to Find Probability Given the uniform distribution illustrated, find the probability that a randomly selected voltage level is greater than 124.5 volts.

6 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Using Area to Find Probability Given the uniform distribution illustrated, find the probability that a randomly selected voltage level is greater than 124.5 volts. Shaded area represents voltage levels greater than 124.5 volts.

7 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Using Area to Find Probability Given the uniform distribution illustrated, find the probability that a randomly selected voltage level is greater than 124.5 volts. Shaded area represents voltage levels greater than 124.5 volts. The area corresponds to probability: P = 0.25.

9 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. A density curve is the graph of a continuous probability distribution. It must satisfy the following properties: Density Curve 1. The total area under the curve must equal 1.

10 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. A density curve is the graph of a continuous probability distribution. It must satisfy the following properties: Density Curve 1. The total area under the curve must equal 1. 2. Every point on the curve must have a vertical height that is 0 or greater. (That is, the curve cannot fall below the x-axis.)

11 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. A density curve is the graph of a continuous probability distribution. It must satisfy the following properties: Density Curve 1. The total area under the curve must equal 1. 2.Every point on the curve must have a vertical height that is 0 or greater. (That is, the curve cannot fall below the x-axis.) 3.Every point on the curve must have a vertical height that is 1 or less

12 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Because the total area under the density curve is equal to 1, there is a correspondence between area and probability. Area and Probability

16 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Standard Normal Distribution Standard normal distribution has the following properties: 1. Its graph is bell-shaped 2. It is symmetric about its center

17 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Standard Normal Distribution Standard normal distribution has the following properties: 1. Its graph is bell-shaped 2. It is symmetric about its center 3. Its mean is equal to 0(  = 0)

18 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Standard Normal Distribution Standard normal distribution has the following properties: 1. Its graph is bell-shaped 2. It is symmetric about its center 3. Its mean is equal to 0(  = 0) 4. Its standard deviation is 1(  = 1)

19 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Standard Normal Distribution The standard normal distribution is a bell- shaped probability distribution with  = 0 and  = 1. The total area under its density curve is equal to 1.

20 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Standard Normal Distribution: Areas and Probabilities Probability that the standard normal random variable takes values less than z is given by the area under the curve from the left up to z. (blue area in the figure)

21 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Thermometers are supposed to give readings of 0ºC at the freezing point of water. Since these instruments are not perfect, some of them give readings below 0ºC and others above 0ºC. Assume the following for the readings: The mean is 0ºC The standard deviation is 1ºC They are normally distributed Examples

22 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. If one thermometer is randomly selected, find the probability that, at the freezing point of water, the reading is less than 1.27º. Example 1

23 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. If one thermometer is randomly selected, find the probability that, at the freezing point of water, the reading is less than 1.27º. Example 1 µ = 0 σ = 1 P (z < 1.27) = ???

27 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. The probability of randomly selecting a thermometer with a reading less than 1.27º is 0.8980. P (z < 1.27) = 0.8980 Example – Thermometers (continued)

28 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Or 89.80% of randomly selected thermometers will have readings below 1.27º. P (z < 1.27) = 0.8980 Example – Thermometers (continued)

29 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. 1.It is designed only for the standard normal distribution, which has a mean of 0 and a standard deviation of 1. 2.It is on two pages, with one page for negative z- scores and the other page for positive z-scores. 3.Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score. Using Table A-2

30 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. 4.When working with a graph, avoid confusion between z-scores and areas. z Score Distance along horizontal scale of the standard normal distribution; refer to the leftmost column and top row of Table A-2. Area Region under the curve; refer to the values in the body of Table A-2. 5.The part of the z-score denoting hundredths is found across the top. Using Table A-2

31 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Working with Excel As before, click the ∑ button, go to Additional Functions, and specify Statistical. Then scroll down to NORMDIST

34 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. If thermometers have an average (mean) reading of 0 degrees and a standard deviation of 1 degree for freezing water, and if one thermometer is randomly selected, find the probability that it reads (at the freezing point of water) above –1.23 degrees. Probability of randomly selecting a thermometer with a reading above –1.23º is 0.8907. P (z > – 1.23) = 0.8907 Example - Thermometers Again

36 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. A thermometer is randomly selected. Find the probability that it reads (at the freezing point of water) between –2.00 and 1.50 degrees. P (z < –2.00) = 0.0228 P (z < 1.50) = 0.9332 P (–2.00 < z < 1.50) = 0.9332 – 0.0228 = 0.9104 The probability that the chosen thermometer has a reading between – 2.00 and 1.50 degrees is 0.9104. Example - Thermometers III

37 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. If many thermometers are selected and tested at the freezing point of water, then 91.04% of them will read between –2.00 and 1.50 degrees. P (z < –2.00) = 0.0228 P (z < 1.50) = 0.9332 P (–2.00 < z < 1.50) = 0.9332 – 0.0228 = 0.9104 A thermometer is randomly selected. Find the probability that it reads (at the freezing point of water) between –2.00 and 1.50 degrees. Example - continued

39 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Normal Distribution by TI-83/84 Press 2 nd VARS to get the DISTR menu Scroll down to normalcdf( and press ENTER Type in two values: Lower, Upper (separated by commas) and close the parenthesis You see a line like normalcdf(-2.00,1.50) Press ENTER and read the probability.

40 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. P(a < z < b) denotes the probability that the z score is between a and b. P(z > a) denotes the probability that the z score is greater than a. P(z < a) denotes the probability that the z score is less than a. Notation

42 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. P(a < z < b) = P(z < b) – P(z < a) Example: P(-1.5 < z < 0.7) = P(z < 0.7) – P(z < -1.5) P(-1.5 < z < 0.7) = 0.7580364 – 0.0668072 P(-1.5 < z < 0.7) = 0.6912 P(z < b)P(z < a)P(a < z < b)

43 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Normal Distribution by TI-83/84 (continued) If you do not have an upper value, type 999. Example: for P(z>1.2) enter normalcdf(1.2,999) If you do not have a lower value, type -999. Example: for P(z<0.6) enter normalcdf(-999,0.6)

44 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Special Cases in Table A-2  If z-score is above 3.49, then the area=0.9999  If z-score is below -3.49, then the area=0.0001  Some special values are marked by stars: z-score=1.645 corresponds to the area=0.95 z-score=2.575 corresponds to the area=0.995

46 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Finding z Scores When Given Probabilities 5% or 0.05 (z score will be positive) Finding the z-score separating 95% bottom values from 5% top values.

47 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Finding z Scores When Given Probabilities 1.645 5% or 0.05 (z score will be positive) Finding the z-score separating 95% bottom values from 5% top values.

48 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Finding the Bottom 2.5% and Upper 2.5% (One z score will be negative and the other positive) Finding z Scores When Given Probabilities - cont

51 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Notation We use z a to represent the z-score separating the top a from the bottom 1-a. Examples: z 0.025 = 1.96, z 0.05 = 1.645 Area = 1-a Area = a zaza

53 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Normal distributions that are not standard All normal distributions have bell-shaped density curves. A normal distribution is standard if its mean m is 0 and its standard deviation s is 1.

54 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Normal distributions that are not standard All normal distributions have bell-shaped density curves. A normal distribution is standard if its mean m is 0 and its standard deviation s is 1. A normal distribution is not standard if its mean m is not 0, or its standard deviation s is not 1, or both.

55 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Normal distributions that are not standard All normal distributions have bell-shaped density curves. A normal distribution is standard if its mean m is 0 and its standard deviation s is 1. A normal distribution is not standard if its mean m is not 0, or its standard deviation s is not 1, or both. We can use a simple conversion that allows us to standardize any normal distribution so that Table A-2 can be used.

56 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Conversion Formula x – µ σ z = (round z scores to 2 decimal places) Let x be a score for a normal distribution with mean µ and standard deviation σ We convert it to a z score by this formula: Sec 6-3, pg 264

58 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Example – Weights of Passengers Weights of taxi passengers have a normal distribution with mean 172 lb and standard deviation 29 lb. If one passenger is randomly selected, what is the probability he/she weighs less than 174 pounds?

59 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Example – Weights of Passengers Weights of taxi passengers have a normal distribution with mean 172 lb and standard deviation 29 lb. If one passenger is randomly selected, what is the probability he/she weighs less than 174 pounds? µ = 172 and σ = 29 P(x < 174) = ???

60 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Example – Weights of Passengers Weights of taxi passengers have a normal distribution with mean 172 lb and standard deviation 29 lb. If one passenger is randomly selected, what is the probability he/she weighs less than 174 pounds? P(x < 174) = ??? µ = 172 and σ = 29

61 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Example – Weights of Passengers Weights of taxi passengers have a normal distribution with mean 172 lb and standard deviation 29 lb. If one passenger is randomly selected, what is the probability he/she weighs less than 174 pounds? P(z < 0.069) = 0.5279 P(x < 174) = ??? µ = 172 and σ = 29

62 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Example – Weights of Passengers Weights of taxi passengers have a normal distribution with mean 172 lb and standard deviation 29 lb. If one passenger is randomly selected, what is the probability he/she weighs less than 174 pounds? P(x < 174) = 0.5279 P(z < 0.069) = 0.5279 P(x < 174) = ??? µ = 172 and σ = 29

63 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. 1. Find the z score corresponding to the given probability (the area to the left). Use Table A-2, your calculator, or Excel 2. Use the values for µ, σ, and the z score found in step 1, to find x: x = µ + (z σ) (If z is located to the left of the mean, be sure that it is a negative number.) Finding x Scores When Given Probabilities pg 267

64 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Example – Lightest and Heaviest The weights of taxi passengers have a normal distribution with mean 172 lb and standard deviation 29 lb. Determine what weight separates the lightest 99.5% from the heaviest 0.5%?

65 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Example – Lightest and Heaviest The weights of taxi passengers have a normal distribution with mean 172 lb and standard deviation 29 lb. Determine what weight separates the lightest 99.5% from the heaviest 0.5%. P(x < ???) = 0.995 µ = 172 and σ = 29

66 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Example – Lightest and Heaviest P(x < ???) = 0.995 P(z < 0.995) = 2.575 µ = 172 and σ = 29 The weights of taxi passengers have a normal distribution with mean 172 lb and standard deviation 29 lb. Determine what weight separates the lightest 99.5% from the heaviest 0.5%.

67 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Example – Lightest and Heaviest P(x < ???) = 0.995 P(z < 0.995) = 2.575 x = µ + σ*z = 172 + 2.575*29 = 246.7 µ = 172 and σ = 29 The weights of taxi passengers have a normal distribution with mean 172 lb and standard deviation 29 lb. Determine what weight separates the lightest 99.5% from the heaviest 0.5%.

68 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Example – Lightest and Heaviest P(x < 247) = 0.995 P(z < 0.995) = 2.575 x = µ + σ*z = 172 + 2.575*29 = 246.7 Separating weight: 247 lb µ = 172 and σ = 29 The weights of taxi passengers have a normal distribution with mean 172 lb and standard deviation 29 lb. Determine what weight separates the lightest 99.5% from the heaviest 0.5%.

69 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Inverse Normal by TI-83/84 Press 2 nd VARS to get the DISTR menu Scroll down to invNnorm( and press ENTER Type in the desired area, mean, st.deviation and close the parenthesis You see a line like invNorm(0.995,172,29) Press ENTER and read the x-score.

71 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Central Limit Theorem The Central Limit Theorem tells us that the distribution of the sample mean x for a sample of size n approaches a normal distribution, as the sample size n increases. pg 287

72 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Central Limit Theorem 1. The random variable x has a distribution (which may or may not be normal) with mean µ and standard deviation ∑. 2. A random sample of size n is selected from the population. Given:

74 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. 1. The distribution of the sample mean x will, as the sample size increases, approach a normal distribution. Conclusions: Central Limit Theorem – cont.

75 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. 1. The distribution of the sample mean x will, as the sample size increases, approach a normal distribution. 2. The mean of that normal distribution is the same as the population mean µ. Conclusions: Central Limit Theorem – cont.

76 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. 1. The distribution of the sample mean x will, as the sample size increases, approach a normal distribution. 2. The mean of that normal distribution is the same as the population mean µ. 3. The standard deviation of that normal distribution is (So it is smaller than the standard deviation of the population.) Conclusions: Central Limit Theorem – cont.

78 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Practical Rules: 1. For samples of size n larger than 30, the distribution of the sample mean can be approximated by a normal distribution. 2. If the original population is normally distributed, then for any sample size n, the sample means will be normally distributed. 3. We can apply Central Limit Theorem if either n>30 or the original population is normal.

79 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Assume the population of taxi passengers is normally distributed with a mean of 172 lb and a standard deviation of 29 lb Example: Water Taxi Passengers a)Find the probability that if an individual passenger is randomly selected, his weight is greater than 175 lb. b)Find the probability that 20 randomly selected passengers will have a mean weight that is greater than 175 lb. pg 290

80 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Assume the population of taxi passengers is normally distributed with a mean of 172 lb and a standard deviation of 29 lb. Example: Water Taxi Passengers a)Find the probability that if an individual passenger is randomly selected, his weight is greater than 175 lb. b)Find the probability that 20 randomly selected passengers will have a mean weight that is greater than 175 lb.

81 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Assume the population of taxi passengers is normally distributed with a mean of 172 lb and a standard deviation of 29 lb Example: Water Taxi Passengers a)Find the probability that if an individual passenger is randomly selected, his weight is greater than 175 lb. b)Find the probability that 20 randomly selected passengers will have a mean weight that is greater than 175 lb. µ = 172 σ = 29 P(x > 175) = ???

82 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Assume the population of taxi passengers is normally distributed with a mean of 172 lb and a standard deviation of 29 lb Example: Water Taxi Passengers a)Find the probability that if an individual passenger is randomly selected, his weight is greater than 175 lb. b)Find the probability that 20 randomly selected passengers will have a mean weight that is greater than 175 lb. µ = 172 σ = 29 P(x > 175) = ???

83 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. a)Find the probability that if an individual man is randomly selected, his weight is greater than 175 lb. Example – cont µ = 172 σ = 29 P(x > 175) = ???

84 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. a)Find the probability that if an individual man is randomly selected, his weight is greater than 175 lb. Example – cont µ = 172 σ = 29 P(x > 175) = ???

85 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. a)Find the probability that if an individual man is randomly selected, his weight is greater than 175 lb. Example – cont µ = 172 σ = 29 P(z > 0.10) = 0.4602 P(x > 175) = ???

86 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. a)Find the probability that if an individual man is randomly selected, his weight is greater than 175 lb. Example – cont µ = 172 σ = 29 P(z > 0.10) = 0.4602 P(x > 175) = 0.4602

87 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Example – cont b) Find the probability that 20 randomly selected men will have a mean weight that is greater than 175 lb (so that their total weight exceeds the safe capacity of 3500 pounds). µ = 172 σ = 29 n = 20 P(x > 175) = ???

88 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Example – cont b) Find the probability that 20 randomly selected men will have a mean weight that is greater than 175 lb (so that their total weight exceeds the safe capacity of 3500 pounds). µ = 172 σ = 29 n = 20 P(x > 175) = ???

89 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Example – cont P(z > 0.4626) = 0.3228 b) Find the probability that 20 randomly selected men will have a mean weight that is greater than 175 lb (so that their total weight exceeds the safe capacity of 3500 pounds). µ = 172 σ = 29 n = 20 P(x > 175) = ???

90 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Example – cont P(z > 0.4626) = 0.3228 P(x > 175) = 0.3228 b) Find the probability that 20 randomly selected men will have a mean weight that is greater than 175 lb (so that their total weight exceeds the safe capacity of 3500 pounds). µ = 172 σ = 29 n = 20

91 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. b)Find the probability that 20 randomly selected passengers will have a mean weight that is greater than 175 lb. It is much easier for an individual to deviate from the mean than it is for a group of 20 to deviate from the mean. a)Find the probability that if an individual passenger is randomly selected, his weight is greater than 175 lb. Example - cont P(x > 175) = 0.4602 P(x > 175) = 0.3228

93 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Review Binomial Probability Distribution 1. The procedure must have a fixed number of trials, n. 2. The trials must be independent. 3. Each trial must have all outcomes classified into two categories (commonly, success and failure). 4.The probability of success p remains the same in all trials (the probability of failure is q = 1-p) Solve by binomial probability formula, Excel, or calculator

94 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Approximation of a Binomial Distribution with a Normal Distribution If np  5 and nq  5 Then µ = np and  = npq and the random variable has distribution. (normal) a pg 299

95 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Procedure for Using a Normal Distribution to Approximate a Binomial Distribution 1. Verify that both np  5 and nq  5. If not, you cannot use normal approximation to binomial. 2. Find the values of the parameters µ and  by calculating µ = np and  = npq. 3. Identify the discrete whole number x that is relevant to the binomial probability problem. Use the continuity correction (see next). pg 301

96 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Continuity Correction When we use the normal distribution (which is a continuous probability distribution) as an approximation to the binomial distribution (which is discrete), a continuity correction is made to a whole number x in the binomial distribution by representing the number x by the interval from x – 0.5 to x + 0.5 (that is, adding and subtracting 0.5).

97 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. The value 122 is represented by the interval (121.5,122.5) Finding the Probability of “At Least 122 Men” Among 213 Passengers Example: The values “at least 122 men” are represented by the interval starting at 121.5.

98 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. at least 8 (includes 8 and above) more than 8 (doesn’t include 8) at most 8 (includes 8 and below) fewer than 8 (doesn’t include 8) exactly 8

1 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Chapter 6 Continuous Random Variables.

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1 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Chapter 6 Continuous Random Variables.

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