Continuous Probability Distributions Chapter Contents Uniform Continuous Distribution Exponential Distribution Describing a Continuous Distribution Normal Distribution Standard Normal Distribution Normal Approximations

Continuous Probability Distributions Chapter Learning Objectives Define a continuous random variable. Calculate uniform probabilities. Know the form and parameters of the normal distribution. Find the normal probability for given z or x using tables or Excel. Solve for z or x for a given normal probability using tables or Excel. Use the normal approximation to a binomial or a Poisson distribution. Find the exponential probability for a given x. Solve for x for given exponential probability.

Characteristics of the Uniform Distribution Uniform Continuous Distribution LO7-2 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

Characteristics of the Uniform Distribution

Example: 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 P(20 < X < 25) = (25 – 20)/(30 – 15) = 5/15 = 0.3333 = 33.33%

1. The time that the customers at the “self serve” check out stations at the Mejers store spend checking out follows a uniform distribution between 0 and 3 minutes. Determine the height and draw this uniform distribution. b. How long does the typical customer wait to check out? c. Determine the standard deviation of the wait time. d. What is the probability a particular customer will wait less than one minute? e. What is the probability a particular customer will wait between 1.5 and 2 minutes?

A statistics instructor collected data on the time it takes the students to complete a test. The test taking time is uniformly distributed within a range of 35 minutes to 55 minutes. a. Determine the height. b. How long does the typical test taking time? c. Determine the standard deviation of the test taking time. d. What is the probability a particular student will take less than 45 minutes? e. What is the probability a particular student will take between 45 and 50 minutes?

3. A tube of Listerine Tartar Control toothpaste contains 4. 2 ounces 3. A tube of Listerine Tartar Control toothpaste contains 4.2 ounces. As people use the toothpaste, the amount remaining in any tube is random. Assume the amount of toothpaste left in the tube follows a uniform distribution. From this information, we can determine the following information about the amount remaining in a toothpaste tube without invading anyone’s privacy. How much toothpaste would you expect to be remaining in the tube? b. What is the standard deviation of the amount remaining in the tube? c. What is the likelihood there is less than 3.0 ounces remaining in the tube? d. What is the probability there is more than 1.5 ounces remaining in the tube?

Characteristics of the Exponential Distribution If events per unit of time follow a Poisson distribution, the time until the next event follows the Exponential distribution. The time until the next event is a continuous variable. NOTE: Here we will find probabilities > x or ≤ x.

Characteristics of the Exponential Distribution Probability of waiting less than or equal to x Probability of waiting more than x

Example Customer Waiting Time Between 2P.M. and 4P.M. on Wednesday, patient insurance inquiries arrive at Blue Choice insurance at a mean rate of 2.2 calls per minute. What is the probability of waiting more than 30 seconds (i.e., 0.50 minutes) for the next call? Set  = 2.2 events/min and x = 0.50 min P(X > 0.50) = e–x = e–(2.2)(0.5) = .3329 or 33.29% chance of waiting more than 30 seconds for the next call.

Example Customer Waiting Time P(X > 0.50) P(X ≤ 0.50)

Inverse Exponential If the mean arrival rate is 2.2 calls per minute, we want the 90th percentile for waiting time (the top 10% of waiting time). Find the x-value that defines the upper 10%.

Inverse Exponential

Mean Time Between Events

5. If arrivals occur at a mean rate of 3 5. If arrivals occur at a mean rate of 3.6 events per hour, the exponential probability of: waiting more than 0.5 hour for the next arrival is:  P(X > .50) = exp(-3.6 × 0.50) = .1653. 6. waiting less than 0.5 hour for the next arrival is: P(X < .50) = 1 - exp(-3.6 × 0.50) = 1 - .1653 = .8347 7. If arrivals occur at a mean rate of 2.6 events per minute, the exponential probability of waiting more than 1.5 minutes for the next arrival is:  P(X > 1.5) = exp(-2.6 × 1.50) = .0202

8. If arrivals occur at a mean rate of 1 8. If arrivals occur at a mean rate of 1.6 events per minute, the exponential probability of waiting less than 1 minute for the next arrival is:  (X < 1) = 1 - exp(-1.6 × 1) = 1 - .2019 = .7981. 9. On average, a major earthquake (Richter scale 6.0 or above) occurs 3 times a decade in a certain California county. What is the probability that less than six months will pass before the next earthquake?  Set λ = 3/120 = 0.025 earthquake/month so P(X < 6) = 1 - exp(-0.025 × 6) = 1 - .8607 = .1393

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. 7-19

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 7-20

Continuous CDF’s: CDF – Cumulative Distribution Function 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. 7-21

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. 7-22

Characteristics of the Normal Distribution Defined by two parameters, µ and . 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. Symmetric and unimodal about the mean.

Characteristics of the Normal Distribution

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 Normal CDF

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.

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 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 Table 1 The table allows you to find the area under the curve from 0 to z. For example, find P(0 < Z < 1.96):

Normal Areas from Table 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 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 Cumulative Table 2 Cumulative table 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 Tables Both tables yield identical results. Use whichever table is easiest. Finding z for a Given Area Both tables 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 Table1.

Finding Areas by using Standardized Variables 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.

The net sales and the number of employees for aluminum fabricators with similar characteristics are organized into frequency distributions. Both are normally distributed. For the net sales, the mean is $180 million and the standard deviation is $25 million. For the number of employees, the mean is 1,500 and the standard deviation is 120. Clarion Fabricators had sales of $170 million and 1,850 employees. a. Convert Clarion’s sales and number of employees to z values. b. Locate the two z values. c. Compare Clarion’s sales and number of employees with those of the other fabricators.

Finding Areas by using Standardized Variables NOTE: You can use Excel, Minitab, TI83/84 etc. to compute these probabilities directly.

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

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.

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.

Inverse Normal

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 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 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 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.

Example Coin Flips P(X > 17) = P(X ≥ 18)  P(X ≥ 17.5) = P(Z > 0.53) = 0.2981