Continuous Probability Distributions Chapter 7 Chapter Contents 7.1 Describing a Continuous Distribution 7.2 Uniform Continuous Distribution 7.3 Normal Distribution 7.4 Standard Normal Distribution 7.5 Normal Approximations 7.6 Exponential Distribution 7.7 Triangular Distribution (Optional)

Continuous Probability Distributions Chapter 7 Chapter Learning Objectives (LO’s) LO7-1: Define a continuous random variable. LO7-2: Calculate uniform probabilities. LO7-3: Know the form and parameters of the normal distribution. LO7-4: Find the normal probability for given z or x using tables or Excel. LO7-5: Solve for z or x for a given normal probability using tables or Excel.

Continuous Probability Distributions Chapter 7 Chapter Learning Objectives (LO’s) LO6: Use the normal approximation to a binomial or a Poisson distribution. LO7: Find the exponential probability for a given x. LO8: Solve for x for given exponential probability. LO9: Use the triangular distribution for “what-if” analysis (optional).

Events as Intervals 7.1 Describing a Continuous Distribution Chapter 7 LO7-1 Chapter 7 LO7-1: Define a continuous random variable. Events as Intervals Discrete Variable – each value of X has its own probability P(X). Continuous Variable – events are intervals and probabilities are areas under continuous curves. A single point has no probability. 7-4

PDF – Probability Density Function 7.1 Describing a Continuous Distribution LO7-1 Chapter 7 PDF – Probability Density Function Continuous PDF’s: Denoted f(x) Must be nonnegative Total area under curve = 1 Mean, variance and shape depend on the PDF parameters Reveals the shape of the distribution 7-5

7.1 Describing a Continuous Distribution LO7-1 Chapter 7 CDF – Cumulative Distribution Function Continuous CDF’s: Denoted F(x) Shows P(X ≤ x), the cumulative proportion of scores Useful for finding probabilities Describing a Continuous Distribution: For a continuous random variable, the function which is used to determine the cumulative probability for the variable is called the cumulative distribution function, denoted by F(x). F(x) = P(X  x). This function is useful for finding probabilities. The slide shows an example of the cumulative distribution function for a normal distribution. Probability of x being between, say, 65 and 70 is = CDF value at x=70 MINUS CDF value at x=65 7-6

Probabilities as Areas 7.1 Describing a Continuous Distribution LO7-1 Chapter 7 Probability of x being between a and b is = the shaded area under the PDF curve OR CDF value at x=a MINUS CDF value at x=b. So CDF value means area under PDF curve up till that value of x. Probabilities as Areas Continuous probability functions: Unlike discrete distributions, the probability at any single point = 0. The entire area under any PDF, by definition, is set to 1. Mean is the balance point of the distribution. 7-7

Expected Value and Variance 7.1 Describing a Continuous Distribution LO7-1 Chapter 7 Expected Value and Variance The mean and variance of a continuous random variable are analogous to E(X) and Var(X ) for a discrete random variable, Here the integral sign replaces the summation sign. Calculus is required to compute the integrals. In class Examples: 7.1, 7.4 7-8

Characteristics of the Uniform Distribution 7.2 Uniform Continuous Distribution LO7-2 Chapter 7 LO7-2: Calculate uniform probabilities. If X is a random variable that is uniformly distributed between a and b, its PDF has constant height. Characteristics of the Uniform Distribution Denoted U(a, b) Area = base x height = (b-a) x 1/(b-a) = 1

7.2 Uniform Continuous Distribution LO7-2 Chapter 7 Characteristics of the Uniform Distribution

Example 7.1 (Text): Anesthesia Effectiveness 7.2 Uniform Continuous Distribution LO7-2 Chapter 7 Example 7.1 (Text): Anesthesia Effectiveness An oral surgeon injects a painkiller prior to extracting a tooth. Given the varying characteristics of patients, the dentist views the time for anesthesia effectiveness as a uniform random variable that takes between 15 minutes and 30 minutes. X is U(15, 30) a = 15, b = 30, find the mean and standard deviation. Find the probability that the effectiveness anesthetic takes between 20 and 25 minutes.

Example: Anesthesia Effectiveness 7.2 Uniform Continuous Distribution LO7-2 Chapter 7 Example: Anesthesia Effectiveness P(c < X < d) = P(X<=d) MINUS P(X<=c) = (d – c)/(b – a) P(20 < X < 25) = (25 – 20)/(30 – 15) = 5/15 = 0.3333 = 33.33% Uniform distribution useful if we have no values more likely than others. Leads to a higher standard deviation than if some values were more likely. In class Examples: 7.5, 7.6

Characteristics of the Normal Distribution LO7-3 Chapter 7 LO7-3: Know the form and parameters of the normal distribution. Characteristics of the Normal Distribution Normal or Gaussian (or bell shaped) distribution was named for German mathematician Karl Gauss (1777 – 1855). Defined by two parameters, µ and . Denoted N(µ, ). Domain is –  < X < +  (continuous scale). Almost all (99.7%) of the area under the normal curve is included in the range µ – 3 < X < µ + 3. Recall Empirical Rule in chapter 4. Symmetric and unimodal about the mean.

Characteristics of the Normal Distribution LO7-3 Chapter 7 Characteristics of the Normal Distribution

Characteristics of the Normal Distribution LO7-3 Chapter 7 Characteristics of the Normal Distribution Normal PDF f(x) reaches a maximum at µ and has points of inflection at µ ±  Bell-shaped curve NOTE: All normal distributions have the same shape but differ in the axis scales.

Characteristics of the Normal Distribution LO7-3 Chapter 7 Characteristics of the Normal Distribution Normal CDF To be considered normal, a random variable must -be considered continuous, uni-modal (1 peak), have tapering tails, be symmetric about the mean. In class Examples: 7.10, 7.12

Characteristics of the Standard Normal Distribution LO7-3 Chapter 7 Characteristics of the Standard Normal Distribution Since for every value of µ and , there is a different normal distribution, we transform a normal random variable to a standard normal distribution with µ = 0 and  = 1 using the formula. We use the Z standardized variable to do this, just like in Chapter 4.

Characteristics of the Standard Normal 7.4 Standard Normal Distribution LO7-3 Chapter 7 Characteristics of the Standard Normal Standard normal PDF f(x) reaches a maximum at z = 0 and has points of inflection at +1. Shape is unaffected by the transformation. It is still a bell-shaped curve. Figure 7.11

Characteristics of the Standard Normal 7.4 Standard Normal Distribution LO7-3 Chapter 7 Characteristics of the Standard Normal Standard normal CDF A common scale from -3 to +3 is used. Entire area under the curve is unity. The probability of an event P(z1 < Z < z2) is a definite integral of f(z). However, standard normal tables or Excel functions can be used to find the desired probabilities.

Normal Areas from Appendix C-1 7.4 Standard Normal Distribution LO7-3 Chapter 7 Normal Areas from Appendix C-1 Appendix C-1 allows you to find the area under the curve from 0 to z. For example, find P(0 < Z < 1.96):

Normal Areas from Appendix C-1 7.4 Standard Normal Distribution LO7-3 Chapter 7 Normal Areas from Appendix C-1 Now find P(-1.96 < Z < 1.96). Due to symmetry, P(-1.96 < Z) is the same as P(Z < 1.96). So, P(-1.96 < Z < 1.96) = .4750 + .4750 = .9500 or 95% of the area under the curve.

Basis for the Empirical Rule 7.4 Standard Normal Distribution LO7-3 Chapter 7 Basis for the Empirical Rule Approximately 68% of the area under the curve is between + 1 Approximately 95% of the area under the curve is between + 2 Approximately 99.7% of the area under the curve is between + 3

Normal Areas from Appendix C-2 7.4 Standard Normal Distribution LO7-4 Chapter 7 LO7-4: Find the normal probability for given z or x using tables or Excel. Normal Areas from Appendix C-2 Appendix C-2 allows you to find the area under the curve from the left of z (similar to Excel). For example, P(Z < 1.96) P(Z < -1.96) P(-1.96 < Z < 1.96)

Normal Areas from Appendices C-1 or C-2 7.4 Standard Normal Distribution LO7-4 Chapter 7 Normal Areas from Appendices C-1 or C-2 Appendices C-1 and C-2 yield identical results. Use whichever table is easiest. In class examples: 7.13, 7.15, 7.19 Finding z for a Given Area Appendices C-1 and C-2 can be used to find the z-value corresponding to a given probability. For example, what z-value defines the top 1% of a normal distribution? This implies that 49% of the area lies between 0 and z which gives z = 2.33 by looking for an area of 0.4900 in Appendix C-1. In class examples: 7.21, 7.22, 7.25

7.4 Standard Normal Distribution LO7-4 Chapter 7 Finding Areas by using Standardized Variables (Z) Suppose John took an economics exam and scored 86 points. The class mean was 75 with a standard deviation of 7. What percentile is John in? That is, what is P(X < 86) where X represents the exam scores? So John’s score is 1.57 standard deviations about the mean. P(X < 86) = P(Z < 1.57) = .9418 (from Appendix C-2) So, John is approximately in the 94th percentile.

7.4 Standard Normal Distribution LO7-4 Chapter 7 Finding Areas by using Standardized Variables NOTE: You can use Excel, Minitab, TI83/84 etc. to compute these probabilities directly. Discussion: What is the probability that a test taker scores at least 65? In class examples: 7.26

7.4 Standard Normal Distribution LO7-5 Chapter 7 LO7-5: Solve for z or x for a normal probability using tables or Excel. Inverse Normal How can we find the various normal percentiles (5th, 10th, 25th, 75th, 90th, 95th, etc.) known as the inverse normal? That is, how can we find X for a given area? We simply turn the standardizing transformation around: Solving for x in z = (x − μ)/ gives x = μ + zσ

7.4 Standard Normal Distribution LO7-5 Chapter 7 Inverse Normal For example, suppose that John’s economics professor has decided that any student who scores below the 10th percentile must retake the exam. The exam scores are normal with μ = 75 and σ = 7. What is the score that would require a student to retake the exam? We need to find the value of x that satisfies P(X < x) = .10. The z-score for with the 10th percentile is z = −1.28.

7.4 Standard Normal Distribution LO7-5 Chapter 7 Inverse Normal The steps to solve the problem are: Use Appendix C or Excel to find z = −1.28 to satisfy P(Z < −1.28) = .10. Substitute the given information into z = (x − μ)/σ to get −1.28 = (x − 75)/7 Solve for x to get x = 75 − (1.28)(7) = 66.03 (or 66 after rounding) Students who score below 66 points on the economics exam will be required to retake the exam.

7.4 Standard Normal Distribution LO7-5 Chapter 7 Inverse Normal In class examples: 7.33, 7.37

Normal Approximation to the Binomial 7.5 Normal Approximations LO7-6 Chapter 7 LO7-6: Use the normal approximation to a binomial or a Poisson. Normal Approximation to the Binomial Binomial probabilities are difficult to calculate when n is large. Use a normal approximation to the binomial distribution. As n becomes large, the binomial bars become smaller and continuity is approached.

Example Coin Flips 7.5 Normal Approximations Chapter 7 LO7-6 Normal Approximation to the Binomial Rule of thumb: when n ≥ 10 and n(1- ) ≥ 10, then it is appropriate to use the normal approximation to the binomial distribution. In this case, the mean and standard deviation for the binomial distribution will be equal to the normal µ and , respectively. Example Coin Flips If we were to flip a coin n = 32 times and  = .50, are the requirements for a normal approximation to the binomial distribution met?

Example Coin Flips 7.5 Normal Approximations Chapter 7 LO7-6 n = 32 x .50 = 16 n(1- ) = 32 x (1 - .50) = 16 So, a normal approximation can be used. When translating a discrete scale into a continuous scale, care must be taken about individual points. For example, find the probability of more than 17 heads in 32 flips of a fair coin. This can be written as P(X  18). However, “more than 17” actually falls between 17 and 18 on a discrete scale.

Example Coin Flips 7.5 Normal Approximations Chapter 7 LO7-6 Since the cutoff point for “more than 17” is halfway between 17 and 18, we add 0.5 to the lower limit and find P(X > 17.5). This addition to X is called the Continuity Correction. At this point, the problem can be completed as any normal distribution problem.

7.5 Normal Approximations LO7-6 Chapter 7 Example Coin Flips P(X > 17) = P(X ≥ 18)  P(X ≥ 17.5) = P(Z > 0.53) = 0.2981 In class exercises: 7.47, 7.49

Normal Approximation to the Poisson 7.5 Normal Approximations LO7-6 Chapter 7 Normal Approximation to the Poisson The normal approximation to the Poisson distribution works best when  is large (e.g., when  exceeds the values in Appendix B). Set the normal µ and  equal to the mean and standard deviation for the Poisson distribution. Example Utility Bills On Wednesday between 10A.M. and noon customer billing inquiries arrive at a mean rate of 42 inquiries per hour at Consumers Energy. What is the probability of receiving more than 50 calls in an hour?  = 42 which is too big to use the Poisson table. Use the normal approximation with  = 42 and  = 6.48074

Example Utility Bills 7.5 Normal Approximations Chapter 7 LO7-6 To find P(X > 50) calls, use the continuity-corrected cutoff point halfway between 50 and 51 (i.e., X = 50.5). At this point, the problem can be completed as any normal distribution problem. In class example: 7.51

Homework 7.2 7.8 Above due Tuesday June 13th by 5 pm 7.14 7.18 7.20 7.24 7.30 7.36 7.14 – 36 Home work due Friday June 16th at noon.