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random variable x be the number generated when the Sample button is clicked. Explain why the distribution of x is approximately uniform even though x is a discrete rather than continuous random variable. Applying the Concepts Basic NW 238 9 Uranium in the Earth s crust. The American Mineralogist (October 2009) published a study of the evolution of uranium minerals in the Earth s crust. Researchers estimate that the trace amount of uranium x in reservoirs follows a uniform distribution ranging between 1 and 3 parts per million. a. Find E (x) and interpret its value. b. Compute P(2 6 x 6 2.5). c. Compute P(x 6 1.75). 10 Requests to a Web server. According to Brighton Webs LTD, a British company that specialies in data analysis, the arrival time of requests to a Web server within each hour can be modeled by a uniform distribution ( www.brighton-webs. co.uk ). Specifically, the number of seconds x from the start of the hour that the request is made is uniformly distributed between 0 and 3,600 seconds. Find the probability that a request is made to a Web server sometime during the last 15 minutes of the hour. 11 New method for detecting anthrax. Researchers at the University of South Florida Center for Biological Defense have developed a safe method for rapidly detecting anthrax spores in powders and on surfaces ( USF Magaine, Summer 2002). The method has been found to work well even when there are very few anthrax spores in a powder specimen. Consider a powder specimen that has exactly 10 anthrax spores. Suppose that the number of anthrax spores in the sample detected by the new method follows an approximate uniform distribution between 0 and 10. a. Find the probability that 8 or fewer anthrax spores are detected in the powder specimen. b. Find the probability that between 2 and 5 anthrax spores are detected in the powder specimen. 12 Random numbers. The data set listed here was created in the MINITAB random-number generator. Construct a relative frequency histogram for the data (saved in the RANUNI file). Except for the expected variation in relative frequencies among the class intervals, does your histogram suggest that the data are observations on a uniform random variable with c = 0 and d = 100? Explain. 38.8759 98.0716 64.5788 60.8422.8413 99.3734 31.8792 32.9847.7434 93.3017 12.4337 11.7828 87.4506 94.1727 23.0892 47.0121 43.3629 50.7119 88.2612 69.2875 62.6626 55.6267 78.3936 28.6777 71.6829 44.0466 57.8870 71.8318 28.9622 23.0278 35.6438 38.6584 46.7404 11.2159 96.1009 95.3660 21.5478 87.7819 12.0605 75.1015 13 Maintaining pipe wall temperature. Maintaining a constant temperature in a pipe wall in some hot process applications is critical. A new technique that utilies bolt-on trace elements to maintain the temperature was presented in the Journal of Heat Transfer (November 2000). Without bolt-on trace elements, the pipe wall temperature of a switch condenser used to produce plastic has a uniform distribution ranging from 260 t o 290 F. When several bolt-on trace elements are attached to the piping, the wall temperature is uniform from 278 to 285 F. a. Ideally, the pipe wall temperature should range between 280 and 284 F. What is the probability that the temperature will fall into this ideal range when no bolt-on trace elements are used? when bolt-on trace elements are attached to the pipe? b. When the temperature is 268 F or lower, the hot liquid plastic hardens (or plates ), causing a buildup in the piping. What is the probability of the plastic plating when no bolt-on trace elements are used? when bolt-on trace elements are attached to the pipe? Applying the Concepts Intermediate 14 Social network densities. Social networks involve interactions (connections) between members of the network. Sociologists define network density as the ratio of actual network connections to the number of possible one-to-one connections. For example, a network with 10 members has a 10 b = 45 total possible connections. If that network 2 has only 5 connections, the network density is 5/45 =.111. Sociologists at the University of Michigan assumed that the density x of a social network would follow a uniform distribution between 0 and 1 ( Social Networks, 2010). a. On average, what is the density of a randomly selected social network? b. What is the probability that the randomly selected network has a density higher than.7? c. Consider a social network with only 2 members. Explain why the uniform model would not be a good approximation for the distribution of network density. 15 Cycle availability of a system. In the jargon of system maintenance, cycle availability is defined as the probability that a system is functioning at any point in time. The U.S. Department of Defense developed a series of performance measures for assessing system cycle availability ( START, Vol. 11, 2004). Under certain assumptions about the failure time and maintenance time of a system, cycle availability is shown to be uniformly distributed between 0 and 1. Find the following parameters for cycle availability: mean, standard deviation, 10th percentile, lower quartile, and upper quartile. Interpret the results. 16 Time delays at a bus stop. A bus is scheduled to stop at a certain bus stop every half hour on the hour and the half hour. At the end of the day, buses still stop after every 30 minutes, but because delays often occur earlier in the day, the bus is never early and is likely to be late. The director of the bus line claims that the length of time a bus is late is uniformly distributed and the maximum time that a bus is late is 20 minutes. a. If the director s claim is true, what is the expected number of minutes a bus will be late? b. If the director s claim is true, what is the probability that the last bus on a given day will be more than 19 minutes late? c. If you arrive at the bus stop at the end of a day at exactly half-past the hour and must wait more than 19 minutes for the bus, what would you conclude about the director s claim? Why?
17 Soft-drink dispenser. The manager of a local soft-drink bottling company believes that when a new beveragedispensing machine is set to dispense 7 ounces, it in fact dispenses an amount x at random anywhere between 6.5 and 7.5 ounces, inclusive. Suppose x has a uniform probability distribution. a. Is the amount dispensed by the beverage machine a discrete or a continuous random variable? Explain. b. Graph the frequency function for x, the amount of beverage the manager believes is dispensed by the new machine when it is set to dispense 7 ounces. c. Find the mean and standard deviation for the distribution graphed in part b, and locate the mean and the interval m { 2s on the graph. d. Find P(x Ú 7 ). e. Find P(x 6 6 ). f. Find P(6.5 x 7.25). g. What is the probability that each of the next six bottles filled by the new machine will contain more than 7.25 ounces of beverage? Assume that the amount of beverage dispensed in one bottle is independent of the amount dispensed in another bottle. Applying the Concepts Advanced 18 Reliability of a robotic device. The reliability of a piece of equipment is frequently defined to be the probability p that the equipment performs its intended function successfully for a given period under specific conditions (Render and Heier, Principles of Operations Management, 1995). Because p varies from one point in time to another, some reliability analysts treat p as if it were a random variable. Suppose an analyst characteries the uncertainty about the reliability of a particular robotic device used in an automobile assembly line by means of the following distribution: f(p) = e 1 0 p 1 0 otherwise a. Graph the analyst s probability distribution for p. b. Find the mean and variance of p. c. According to the analyst s probability distribution for p, what is the probability that p is greater than.95? Less than.95? d. Suppose the analyst receives the additional information that p is definitely between.90 and.95, but that there is complete uncertainty about where it lies between these values. Describe the probability distribution the analyst should now use to describe p. 19 Gouges on a spindle. A tool-and-die machine shop produces extremely high-tolerance spindles. The spindles are 18-inch slender rods used in a variety of military equipment. A piece of equipment used in the manufacture of the spindles malfunctions on occasion and places a single gouge somewhere on the spindle. However, if the spindle can be cut so that it has 14 consecutive inches without a gouge, then the spindle can be salvaged for other purposes. Assuming that the location of the gouge along the spindle is best described by a uniform distribution, what is the probability that a defective spindle can be salvaged? 3 The Normal Distribution One of the most commonly observed continuous random variables has a bell-shaped probability distribution (or bell curve ), as shown in Figure 6. It is known as a normal random variable and its probability distribution is called a normal distribution. f(x) Figure 6 A normal probability distribution x The normal distribution plays a very important role in the science of statistical inference. Moreover, many phenomena generate random variables with probability distributions that are very well approximated by a normal distribution. For example, the error made in measuring a person s blood pressure may be a normal random variable, and the probability distribution for the yearly rainfall in a certain region might be approximated by a normal probability distribution. You can determine the adequacy of the normal approximation to an existing population of data by comparing the relative frequency distribution of a large sample of the data with the normal probability distribution. Methods for detecting disagreement between a set of data and the assumption of normality are presented in Section 4. The normal distribution is perfectly symmetric about its mean m, as can be seen in the examples in Figure 7. Its spread is determined by the value of its standard deviation 239
BIOGRAPHY CARL F. GAUSS (1777 1855) The Gaussian Distribution The normal distribution began in the 18th century as a theoretical distribution for errors in disciplines in which fluctuations in nature were believed to behave randomly. Although he may not have been the first to discover the formula, the normal distribution was named the Gaussian distribution after Carl Friedrich Gauss. A well-known and respected German mathematician, physicist, and astronomer, Gauss applied the normal distribution while studying the motion of planets and stars. Gauss s prowess as a mathematician was exemplified by one of his most important discoveries: At the young age of 22, Gauss constructed a regular 17-gon by ruler and compasses a feat that was the most major advance in mathematics since the time of the ancient Greeks. In addition to publishing close to 200 scientific papers, Gauss invented the heliograph as well as a primitive telegraph. f(x) 4 0 3 x Figure 7 Several normal distributions with different means and standard deviations s. The formula for the normal probability distribution is shown in the next box. When plotted, this formula yields a curve like that shown in Figure 6. Probability Distribution for a Normal Random Variable x Probability density function: f1x2 = where 1 s12p e-11>2231x -m2>s42 m = Mean of the normal random variable x s = Standard deviation p = 3.1416 c e = 2.71828 c P(x 6 a) is obtained from a table of normal probabilities. Note that the mean m and standard deviation s appear in this formula, so that no separate formulas for m and s are necessary. To graph the normal curve, we have to know the numerical values of m and s. Computing the area over intervals under the normal probability distribution is a difficult task. * Consequently, we will use the computed areas listed in Table IV of Appendix: Tables. Although there are an infinitely large number of normal curves one for each pair of values of m and s we have formed a single table that will apply to any normal curve. Table IV is based on a normal distribution with mean m = 0 and standard deviation s = 1, called a standard normal distribution. A random variable with a standard normal distribution is typically denoted by the symbol. The formula for the probability distribution of is f12 = 1 12p e-11>222 Figure 8 shows the graph of a standard normal distribution. f() Figure 8 Standard normal distribution: m = 0, s = 1 3 2 1 0 1 2 3 240 * The student with a knowledge of calculus should note that there is no closed-form expression for P(a 6 x 6 b) = 1 b a f(x) dx for the normal probability distribution. The value of this definite integral can be obtained to any desired degree of accuracy by numerical approximation. For this reason, it is tabulated for the user.
The standard normal distribution is a normal distribution with m = 0 and s = 1. A random variable with a standard normal distribution, denoted by the symbol, is called a standard normal random variable. Since we will ultimately convert all normal random variables to standard normal variables in order to use Table IV to find probabilities, it is important that you learn to use Table IV well. A partial reproduction of that table is shown in Table 1. Note that the values of the standard normal random variable are listed in the left-hand column. The entries in the body of the table give the area (probability) between 0 and. Examples 3 6 illustrate the use of the table. Table 1 Reproduction of Part of Table IV in Appendix: Tables f() 0.00.01.02.03.04.05.06.07.08.09.0.0000.0040.0080.0120.0160.0199.0239.0279.0319.0359.1.0398.0438.0478.0517.0557.0596.0636.0675.0714.0753.2.0793.0832.0871.0910.0948.0987.1026.1064.1103.1141.3.1179.1217.1255.1293.1331.1368.1406.1443.1480.1517.4.1554.1591.1628.1664.1700.1736.1772.1808.1844.1879.5.1915.1950.1985.2019.2054.2088.2123.2157.2190.2224.6.2257.2291.2324.2357.2389.2422.2454.2486.2517.2549.7.2580.2611.2642.2673.2704.2734.2764.2794.2823.2852.8.2881.2910.2939.2967.2995.3023.3051.3078.3106.3133.9.3159.3186.3212.3238.3264.3289.3315.3340.3365.3389 1.0.3413.3438.3461.3485.3508.3531.3554.3577.3599.3621 1.1.3643.3665.3686.3708.3729.3749.3770.3790.3810.3830 1.2.3849.3869.3888.3907.3925.3944.3962.3980.3997.4015 1.3.4032.4049.4066.4082.4099.4115.4131.4147.4162.4177 1.4.4192.4207.4222.4236.4251.4265.4279.4292.4306.4319 1.5.4332.4345.4357.4370.4382.4394.4406.4418.4429.4441 Example 3 Using the Standard Normal Table to find P(- 0 6 6 0 ) Problem Find the probability that the standard normal random variable falls between -1.33 and +1.33. Solution The standard normal distribution is shown again in Figure 9. Since all probabilities associated with standard normal random variables can be depicted as areas under the standard normal curve, you should always draw the curve and then equate the desired probability to an area. In this example, we want to find the probability that falls between -1.33 and +1.33, which is equivalent to the area between -1.33 and +1.33, shown highlighted in Figure 9. Table IV gives the area between = 0 and any positive value of, so that if A 2 A 1 Figure 9 Areas under the standard normal curve for Example 3 3 2 1 0 1 2 3 1.33 1.33 241
.0.1.2.3.4.5.6.7.8..9 1.0 1.1 1.2 1.3....00.01.02.03....4082 Figure 10 Finding = 1.33 in the standard normal table, Example 3 we look up = 1.33 (the value in the 1.3 row and.03 column, as shown in Figure 10 ), we find that the area between = 0 and = 1.33 is.4082. This is the area labeled A 1 in Figure 9. To find the area A 2 located between = 0 and = -1.33, we note that the symmetry of the normal distribution implies that the area between = 0 and any point to the left is equal to the area between = 0 and the point equidistant to the right. Thus, in this example the area between = 0 and = -1.33 is equal to the area between = 0 and = +1.33. That is, A 1 = A 2 =.4082 The probability that falls between -1.33 and +1.33 is the sum of the areas of A 1 and A 2. We summarie in probabilistic notation: P1-1.33 6 6 +1.332 = P1-1.33 6 6 02 + P10 6 1.332 = A 1 + A 2 =.4082 +.4082 =.8164 Look Back Remember that 6 and are equivalent in events involving, because the inclusion (or exclusion) of a single point does not alter the probability of an event involving a continuous random variable. Now Work Exercise 25 e f Example 4 Using the Standard Normal Table to find P( 7 0 ) Problem Find the probability that a standard normal random variable exceeds 1.64; that is, find P( 7 1.64). Solution The area under the standard normal distribution to the right of 1.64 is the highlighted area labeled A 1 in Figure 11. This area represents the probability that exceeds 1.64. However, when we look up = 1.64 in Table IV, we must remember that the probability given in the table corresponds to the area between = 0 and = 1.64 (the area labeled A 2 in Figure 11 ). From Table IV, we find that A 2 =.4495. To find the area A 1 to the right of 1.64, we make use of two facts: 1. The standard normal distribution is symmetric about its mean, = 0. 2. The total area under the standard normal probability distribution equals 1. Taken together, these two facts imply that the areas on either side of the mean, = 0, equal.5; thus, the area to the right of = 0 in Figure 11 is A 1 + A 2 =.5. Then P1 7 1.642 = A 1 =.5 - A 2 =.5 -.4495 =.0505 P( > 1.64) = A 1 A 2 Figure 11 Areas under the standard normal curve for Example 4 3 2 1 0 1 2 3 1.64 Look Back To attach some practical significance to this probability, note that the implication is that the chance of a standard normal random variable exceeding 1.64 is only about.05. Now Work Exercise 26a Example 5 Using the Standard Normal Table to find P( 6 0 ) 242 Problem Find the probability that a standard normal random variable lies to the left of.67. Solution The event sought is shown as the highlighted area in Figure 12. We want to find P( 6.67). We divide the highlighted area into two parts: the area A 1 between = 0 and =.67, and the area A 2 to the left of = 0. We must always make such a division when the desired area lies on both sides of the mean ( = 0) because Table IV contains areas