The Exponential Distribution

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Journal of Water and Environment Technology, Vol.4, No.1, 2006 - 61 - Effect of urban emissions on the horizontal distribution of metal concentration in sediments in the vicinity of Asian large cities T. Urase 1* , K. Nadaoka 2 , H. Yagi 2 , T. Iwasa 1 , Y. Suzuki 1 F. Siringan 3 , T. P. Garcia 4 , T. T. Thao 5 1: Dept. of Civil Engineering, Tokyo Institute of Technology, 2-12-1 Ookayama, Meguro, Tokyo, 152-8552 Japan. *: Corresponding author. turase@fluid.cv.titech.ac.jp , +81-3-5734-3548 2: Dept. of Mechanical and Environmental Informatics, Tokyo Institute of Technology, 2-12-1 Ookayama, Meguro, Tokyo, 152-8552 Japan. 3: National Institute of Geological Sciences, University of the Philippines, 1101 Diliman, Quezon City, Philippines 4: Dept. Civil Engineering, College of Engineering, Technological University of the Philippines, Manila, Philippines 5: Department of Analytical Chemistry, Hanoi University of Science, 19- LeThanh Tong street, Hanoi, Vietnam Abstract: Metal contents of sediments in Manila Bay – Laguna Lake watershed in the Philippines were measured and detailed horizontal distribution was obtained. The distribution of zinc and lead concentration in Manila Bay clearly shows the effect of anthropogenic contamination and it was explained by the diffusion of lead and zinc rich anthropogenic particles discharged from Pasig River. The sediments in Laguna Lake were mostly natural particulate matters from surrounding mountains and they contained 20 mgPb/kg and 100 mgZn/kg, while the sediment taken at the heavily polluted branches of the Pasig River contained as high as 88 mgPb/kg and 310 mgZn/kg. The lead and zinc concentrations in the sediments of Manila Bay – Laguna Lake watershed were compared with those in the mouth of the Tama River, Tokyo, where the faster deposition of coarser natural origin particles and slower deposition of lead and zinc rich anthropogenic particles determined the sediment concentration. The comparison was also made with Hanoi City, Vietnam. In spite of the difference in time when leaded gasoline was prohibited, the difference in the lead concentrations of roadside deposits and sediments was not obvious in the vicinity of these three target cities. This is probably due to dilution by a large amount of suspended solids conveyed by the Pasig River in the case of the Philippines. Storm water runoff containing roadside deposits and discharge of untreated wastewater were identified as factors increasing zinc and lead concentrations of sediments in receiving waters based on the measurements on roadside deposits and the estimation of the contribution of untreated wastewater. Keywords: Laguna Lake; lead; Manila Bay; sediment; wastewater; zinc. Introduction Asian cities generally have large populations. Human activities and their impacts on natural environments are concentrated in the vicinity of urban regions. High precipitation in Asian regions results in erosion of land and induces urban runoff during wet weather days. A large amount of particulate matters having natural and anthropogenic sources flows into receiving watersheds. Incomplete sewer problems such as low coverage and The Exponential Distribution The Exponential Distribution By: OpenStaxCollege The exponential distribution is often concerned with the amount of time until some specific event occurs For example, the amount of time (beginning now) until an earthquake occurs has an exponential distribution Other examples include the length, in minutes, of long distance business telephone calls, and the amount of time, in months, a car battery lasts It can be shown, too, that the value of the change that you have in your pocket or purse approximately follows an exponential distribution Values for an exponential random variable occur in the following way There are fewer large values and more small values For example, the amount of money customers spend in one trip to the supermarket follows an exponential distribution There are more people who spend small amounts of money and fewer people who spend large amounts of money The exponential distribution is widely used in the field of reliability Reliability deals with the amount of time a product lasts Let X = amount of time (in minutes) a postal clerk spends with his or her customer The time is known to have an exponential distribution with the average amount of time equal to four minutes X is a continuous random variable since time is measured It is given that μ = minutes To any calculations, you must know m, the decay parameter m = μ Therefore, m = = 0.25 The standard deviation, σ, is the same as the mean μ = σ The distribution notation is X ~ Exp(m) Therefore, X ~ Exp(0.25) The probability density function is f(x) = me-mx The number e = 2.71828182846 It is a number that is used often in mathematics Scientific calculators have the key "ex." If you enter one for x, the calculator will display the value e 1/26 The Exponential Distribution The curve is: f(x) = 0.25e–0.25x where x is at least zero and m = 0.25 For example, f(5) = 0.25e−(0.25)(5) = 0.072 The postal clerk spends five minutes with the customers The graph is as follows: Notice the graph is a declining curve When x = 0, f(x) = 0.25e(−0.25)(0) = (0.25)(1) = 0.25 = m The maximum value on the y-axis is m Try It The amount of time spouses shop for anniversary cards can be modeled by an exponential distribution with the average amount of time equal to eight minutes Write the distribution, state the probability density function, and graph the distribution X ~ Exp(0.125); f(x) = 0.125e–0.125x; 2/26 The Exponential Distribution a Using the information in [link], find the probability that a clerk spends four to five minutes with a randomly selected customer a Find P(4 < x < 5) The cumulative distribution function (CDF) gives the area to the left P(x < x) = – e–mx P(x < 5) = – e(–0.25)(5) = 0.7135 and P(x < 4) = – e(–0.25)(4) = 0.6321 NOTE You can these calculations easily on a calculator The probability that a postal clerk spends four to five minutes with a randomly selected customer is P(4 < x < 5) = P(x < 5) – P(x < 4) = 0.7135 − 0.6321 = 0.0814 On the home screen, enter (1 – e^(–0.25*5))–(1–e^(–0.25*4)) or enter e^(–0.25*4) – e^(–0.25*5) b Half of all customers are finished within how long? (Find the 50th percentile) b Find the 50th percentile P(x < k) = 0.50, k = 2.8 minutes (calculator or computer) 3/26 The Exponential Distribution Half of all customers are finished within 2.8 minutes You can also the calculation as follows: P(x < k) = 0.50 and P(x < k) = –e–0.25k Therefore, 0.50 = − e−0.25k and e−0.25k = − 0.50 = 0.5 Take natural logs: ln(e–0.25k) = ln(0.50) So, –0.25k = ln(0.50) ln(0.50) Solve for k:k = −0.25 = 2.8 minutes The calculator simplifies the calculation for percentile k See the following two notes Note A formula for the percentile k is k = ln(1 − AreaToTheLeft) −m where ln is the natural log Collaborative Exercise On the home screen, enter ln(1 – 0.50)/–0.25 Press the (-) for the negative c Which is larger, the mean or the median? c From part b, the median or 50th percentile is 2.8 minutes The theoretical mean is four minutes The mean is larger Try It The number of days ahead travelers purchase their airline tickets can be modeled by an exponential distribution with the average amount of time equal to 15 days Find the probability that a traveler will purchase a ticket fewer than ten days in advance How many days half of all travelers wait? P(x < 10) = 0.4866 50th percentile = 10.40 Collaborative Exercise Have each class member count the change he or she has in his or her pocket or purse Your instructor will record the amounts in dollars and cents Construct a histogram of 4/26 The Exponential Distribution the data taken by the class Use five intervals Draw a smooth curve through the bars The graph should look approximately exponential Then calculate the mean Let X = the amount of money a student in your class has in his or her pocket or purse The distribution for X is approximately exponential with mean, μ = _ and m = _ The standard deviation, σ = Draw the appropriate exponential graph You should ... 7 CHAPTER The Binomial Distribution Introduction Many probability problems involve assigning probabilities to the outcomes of a probability experiment. These probabilities and the corresponding outcomes make up a probability distribution. There are many different probability distributions. One special probability distribution is called the binomial distribution. The binomial distribution has many uses such as in gambling, in inspecting parts, and in other areas. 114 Copyright © 2005 by The McGraw-Hill Companies, Inc. Click here for terms of use. Discrete Probability Distributions In mathematics, a variable can assume different values. For example, if one records the temperature outside every hour for a 24-hour period, temperature is considered a variable since it assumes different values. Variables whose values are due to chance are called random variables. When a die is rolled, the value of the spots on the face up occurs by chance; hence, the number of spots on the face up on the die is considered to be a random variable. The outcomes of a die are 1, 2, 3, 4, 5, and 6, and the probability of each outcome occurring is 1 6 . The outcomes and their corresponding probabilities can be written in a table, as shown, and make up what is called a probability distribution. Value, x 123456 Probability, P(x) 1 6 1 6 1 6 1 6 1 6 1 6 A probability distribution consists of the values of a random variable and their corresponding probabilities. There are two kinds of probability distributions. They are discrete and continuous.Adiscrete variable has a countable number of values (countable means values of zero, one, two, three, etc.). For example, when four coins are tossed, the outcomes for the number of heads obtained are zero, one, two, three, and four. When a single die is rolled, the outcomes are one, two, three, four, five, and six. These are examples of discrete variables. A continuous variable has an infinite number of values between any two values. Continuous variables are measured. For example, temperature is a continuous variable since the variable can assume any value between 108 and 208 or any other two temperatures or values for that matter. Height and weight are continuous variables. Of course, we are limited by our measuring devices and values of continuous variables are usually ‘‘rounded off.’’ EXAMPLE: Construct a discrete probability distribution for the number of heads when three coins are tossed. SOLUTION: Recall that the sample space for tossing three coins is TTT, TTH, THT, HTT, HHT, HTH, THH, and HHH. CHAPTER 7 The Binomial Distribution 115 The outcomes can be arranged according to the number of heads, as shown. 0 heads TTT 1 head TTH, THT, HTT 2 heads THH, HTH, HHT 3 heads HHH Finally, the outcomes and corresponding probabilities can be written in a table, as shown. Outcome, x 0123 Probability, P(x) 1 8 3 8 3 8 1 8 The sum of the probabilities of a probability distribution must be 1. A discrete probability distribution can also be shown graphically by labeling the x axis with the values of the outcomes and letting the values on the y axis represent the probabilities for the outcomes. The graph for the discrete probability distribution of the number of heads occurring when three coins are tossed is shown in Figure 7-1. There are many kinds of discrete probability distributions; however, the distribution of the number of heads when three coins are tossed is a special kind of CHAPTER 9 The Normal Distribution Introduction A branch of mathematics that uses probability is called statistics. Statistics is the branch of mathematics that uses observations and measurements called data to analyze, summarize, make inferences, and draw conclusions based on the data gathered. This chapter will explain some basic concepts of statistics such as measures of average and measures of variation. Finally, the relationship between probability and normal distribution will be explained in the last two sections. 147 Copyright © 2005 by The McGraw-Hill Companies, Inc. Click here for terms of use. Measures of Average There are three statistical measures that are commonly used for average. They are the mean, median, and mode. The mean is found by adding the data values and dividing by the number of values. EXAMPLE: Find the mean of 18, 24, 16, 15, and 12. SOLUTION: Add the values: 18 þ 24 þ 16 þ 15 þ 12 ¼ 85 Divide by the number of values, 5: 85 Ä 5 ¼ 17 Hence the mean is 17. EXAMPLE: The ages of 6 executives are 48, 56, 42, 52, 53 and 52. Find the mean. SOLUTION: Add: 48 þ 56 þ 42 þ 52 þ 53 þ 52 ¼ 303 Divide by 6: 303 Ä 6 ¼ 50.5 Hence the mean age is 50.5. The median is the middle data value if there is an odd number of data values or the number halfway between the two data values at the center, if there is an even number of data values, when the data values are arranged in order. EXAMPLE: Find the median of 18, 24, 16, 15, and 12. SOLUTION: Arrange the data in order: 12, 15, 16, 18, 24 Find the middle value: 12, 15, 16, 18, 24 The median is 16. EXAMPLE: Find the median of the number of minutes 10 people had to wait in a checkout line at a local supermarket: 3, 0, 8, 2, 5, 6, 1, 4, 1, and 0. SOLUTION: Arrange the data in order: 0, 0, 1, 1, 2, 3, 4, 5, 6, 8 The middle falls between 2 and 3; hence, the median is (2 þ 3) Ä 2 ¼ 2.5. CHAPTER 9 The Normal Distribution 148 The third measure of average is called the mode. The mode is the data value that occurs most frequently. EXAMPLE: Find the mode for 22, 27, 30, 42, 16, 30, and 18. SOLUTION: Since 30 occurs twice and more frequently than any other value, the mode is 30. EXAMPLE: Find the mode for 2, 3, 3, 3, 4, 4, 6, 6, 6, 8, 9, and 10. SOLUTION: In this example, 3 and 6 occur most often; hence, 3 and 6 are used as the mode. In this case, we say that the distribution is bimodal. EXAMPLE: Find the mode for 18, 24, 16, 15, and 12. SOLUTION: Since no value occurs more than any other value, there is no mode. A distribution can have one mode, more than one mode, or no mode. Also, the mean, median, and mode for a set of values most often differ somewhat. PRACTICE 1. Find the mean, median, and mode for the number of sick days nine employees used last year. The data are 3, 6, 8, 2, 0, 5, 7, 8, and 5. 2. Find the mean, median, and mode for the number of rooms seven hotels in a large city have. The data are 332, 256, 300, 275, 216, 314, and 192. 3. Find the mean, median, and mode for the number of tornadoes that occurred in a specific state over the last 5 years. The data are 18, 6, 3, 9, and 10. 4. Find the mean, median, and mode for the number of items 9 people purchased at the express checkout register. The data are 12, 8, 6, 1, 5, 4, 6, 2, and 6. 5. Find the mean, median, and mode for the ages of 10 children who participated in a field trip to the zoo. The ages are 7, 12, 11, 11, 5, 8, 11, 7, 8, and 6. CHAPTER 9 The Normal Distribution 149 ANSWERS 1. Mean ¼ 3 þ 6 þ 8 þ 2 þ 0 þ 5 þ OptiX OSN 3500 Installation Manual Contents Contents 6 Installing the Cable Distribution Plate 1 6.1 Installation Position 1 6.2 Installing the Cable Distribution Plate in the Cabinet 2 Issue 05 (2006-11-20) Huawei Technologies Proprietary i Figures Installing the cable distribution plate 3 T2-0416xx- 20050330-C- 1.20 Huawei Technologies Proprietary iii OptiX OSN 3500 IM 6 Installing the Cable Distribution Plate About This Chapter This chapter guides you to install the cable distribution plate. One cable distribution plate is delivered with one subrack. The cable distribution plate is installed over the subrack. The following table lists the contents of this chapter. Section Description 6.1Installation Position Describes the position of the cable distribution plate in the cabinet. 6.2Installing the Cable Distribution Plate in the Cabinet Describes the steps to install the cable distribution plate 6.1 Installation Position Figure 1.1 shows the position of the cable distribution plate in the 2200mm-high 1 6Installing the Cable Distribution Plate OptiX OSN 3500 Installation Manual cabinet. Figure 1.1 Position for mounting ears Cable distribution plate Installation position Cable distribution plate for the lower subrack 37, 38 Cable distribution plate for the upper subrack 62, 63 Note: The holes are numbered from bottom to top. The hole at the bottom is numbered 1. 6.2 Installing the Cable Distribution Plate in the Cabinet Purpose This procedure guides you to install the cable distribution plate in the cabinet. Tools /Materials Cross screwdriver Cable distribution plate Prerequisites The cabinet and the subrack have been installed. Required/As needed Required Step 1 Secure the cable distribution plate into the cabinet over the subrack. See Figure 1.1. 2 Huawei Technologies Proprietary Issue 05 (2006-11-20) OptiX OSN 3500 Installation Manual 6Installing the Cable Distribution Plate Figure 1.1 Installing the cable distribution plate End Issue 05 (2006-11-20) Huawei Technologies Proprietary 3 FEDERAL RESERVE BANK OF ST . LOUIS RE V I EW SEPTEMBER / OCTOBER 201 0 395 The Geographic Distribution and Characteristics of U.S. Bank Failures, 2007-2010: Do Bank Failures Still Reflect Local Economic Conditions? Craig P. Aubuchon and David C. Wheelock The financial crisis and recession that began in 2007 brought a sharp increase in the number of bank failures in the United States. This article investigates characteristics of banks that failed and regional patterns in bank failure rates during 2007-10. The article compares the recent experience with that of 1987-92, when the United States last experienced a high number of bank failures. As during the 1987-92 and prior episodes, bank failures during 2007-10 were concentrated in regions of the country that experienced the most serious distress in real estate markets and the largest declines in economic activity. Although most legal restrictions on branch banking were eliminated in the 1990s, the authors find that many banks continue to operate in a small number of markets and are vulnerable to localized economic shocks. (JEL E32, G21, G28, R11) Federal Reserve Bank of St. Louis Review, September/October 2010, 92(5), pp. 395-415. fewer than four banks failed per year. Bank fail- ures were much more common in the 1980s and early 1990s, however, including more than 100 commercial bank failures each year from 1987 to 1992. As percentages of the total number of U.S. banks and volume of bank deposits, the failures of 2007-10 approach the failures of the 1980s and early 1990s (Figures 1 and 2). 2 The bank failures of the 1980s and early 1990s were concentrated in regions of the country that T he financial crisis and recession that began in 2007 brought a sharp increase in the number of failures of banks and other financial firms in the United States. The failures and near-failures of very large financial firms, such as Bear Stearns, Lehman Brothers, and American International Group (AIG), grabbed the headlines. However, 206 fed- erally insured banks (commercial banks, savings banks, and savings and loan associations, here- after “banks”)—or 2.4 percent of all banks in oper- ation on December 31, 2006—failed between January 1, 2007, and March 31, 2010. 1 Failed banks held $373 billion of deposits (6.5 percent of total U.S. bank deposits) as of June 30, 2006; Washington Mutual Bank alone accounted for $211 billion of deposits in failed banks. The recent spike in bank failures followed a period of relative tranquility in the U.S. banking industry. Between 1995 and 2007, on average 1 The 206 failures include only banks that were declared insolvent by their primary regulator and were either liquidated or sold, in whole or in part, to another financial institution by the Federal Deposit Insurance Corporation (FDIC). This total does not include banks, bank holding companies, or other firms that received govern- ment assistance but remained going concerns, such as the Federal National Mortgage Association (Fannie Mae), Federal Home Loan Mortgage Corporation (Freddie Mac), Citigroup, and GMAC. 2 Figures 1 and 2 include data for both commercial banks and savings institutions but exclude another 747 savings institutions (with $394 billion of total assets) that were resolved by ... Relationship between the Poisson and the Exponential Distribution There is an interesting relationship between the exponential distribution and the Poisson distribution Suppose that the time that elapses... cumulative distribution function? Draw the distribution 16/26 The Exponential Distribution Find P(x < 4) Find the 30th percentile 0.4756 Find the median Which is larger, the mean or the median? The. .. Poisson distribution, the time between accidents follows the exponential distribution If there are an average of three per week, then on average there is μ = of a week between accidents, and the