Continuous Probability Distributions Types: 1. Uniform Distribution 2. Exponential Distribution 3. Normal Distribution 4. Gamma Distribution 5. Lognormal Distribution 6. Beta Distribution 7. Weibull Distribution Many random variables seen in practice have more than a countable collection of possible values. For example: Diameter of machine rods from a certain industrial process may be anywhere from 1.2 to 1.5 centimeters Proportions of impurities in ore samples may run from 0.10 to 0.80 The height of students in the College of Engineering, 4.8 to 6.5 feet Water depth along the coast of Manila Bay, 0.5 feet to 20.5 feet Etc A random variable X is said to be continuous if it can take on the infinite number of possible values associated with intervals of real numbers. A function f(x) of a random variable X is called the probability density function if 1. f(x) 0 for all x 2. 3. Note that for a continuous random variable X for any specific value of a. If random variable X is continuous with the probability density function f(x), then the distribution function X is Expected Values of Continuous Random Variables For a continuous random variable X having probability density function f(x), The expected value, also known as the mean, of X is given by The variance of X is given by For a continuous random variable and constants a and b, E(aX + b) = aE(X) + b V(aX + b) = a2V(X) Example: For a lathe in a machine shop, let X denote the percentage of time out of a 40-hour workweek that the lathe is actually in use. Suppose X has a probability density function given by Find the mean and variance of X. Solution: From the definition of expected value we have, Thus, on the average, the lathe is in use 75% of the time. To compute V(X), we first find E(X2) as follows Then, . The Uniform Distribution Example: The failure of a circuit board causes a computing system to shut down until a new board is delivered. The delivery time X is uniformly distributed over the interval of 1 to 5 days. The cost C of this failure and shut down consists of a fixed cost, c0, for the new part plus lost time and a cost (c1) that increases proportional to X2, so that a. Find the probability that the delivery time is two or more days b. Find the expected cost of a single component failure. Solution: The delivery time X is uniformly distributed form 1 to 5 days, which gives a. The probability that the delivery time is two or more days (See Figure 6.7) is Figure 6.7 0.25 0 b. We know that get Thus, Because 1 2 3 4 5 , we The Exponential Distribution E(X) = and V(X) = 2 F(x) = where is a constant that determines the rate at which the curve decreases. Example: A sugar refinery has three processing plants, all receiving raw sugar in bulk. The amount of raw sugar (in tons) that one plant can process in one day can be modelled using an exponential distribution with mean of 4 tons for each of three plants. If each plant operates independently, a. Find the probability that any given plant processes more than 5 tons of raw sugar on a given day. b. Find the probability that exactly two of the three plants process more than 5 tons of raw sugar on a given day. c. How much raw sugar should be stocked for the plant each day so that the chance of running out of the raw sugar is only 0.05? Solution: Let X = amount of sugar processed in a day. a. Knowledge of the distribution function could allow us to evaluate this probability easily as b. Assuming that three of the plants operate independently, the problem is to find the probability of two successes out of three tries, where 0.37 denotes the probability of success. This is a binomial problem, and the solution is P(exactly 2 use more than 5 tons) = c. Let a = the amount of raw sugar to be stocked. The chance of running out of the raw sugar means there is a probability that more than the stocked amount is needed. We want to choose a so that Solving this equation yields a = 11.98 tons. The Normal Distribution The most widely used of all continuous probability distributions is the normal distribution. The normal probability density function is bell-shaped and centered at the mean value . Its spread is measured by the standard deviation . The functional form is given by Some useful properties of the normal distribution The mean determines the location of the distribution The standard deviation determines the spread of the distribution. The normal distribution with larger is shorter and more spread (Figure 6.4) Any linear function of the normally distributed random variable is also normally distributed. That is, if X has a normal distribution with mean and variance 2 and Y = aX + b for constants a and b, then Y is also normally distributed with mean a and b and variance a22. Standard Normal Distribution A normal distribution with = 0 and = 1 is known as a standard normal distribution. The letter Z is used to indicate the standard normal variable. For any normally distributed random variable X, with parameters and 2, will have a standard normal distribution. Because any normally distributed random variable can be transformed to the standard normal, probabilities can be evaluated for any normal distribution by having a table of standard normal integrals available and provided in Table 16 of the Appendix. Table 16 gives numerical values for the cumulative distribution function using Example: Find P(Z 1.53) for a standard normal variable Z. Solution: The P(Z 1.53) is equal to the shaded area in Figure 6.15. From the table, this is found in the cross-section of the row corresponding to 1.5 and the column corresponding to 0.03. Hence, P(Z 1.53) = 0.9370 Example: If Z denotes a standard normal variable, find a. P(Z 1) b. P(Z < -1.5) c. P(Z > 1) d. P(-1.5 Z 0.5) e. Find the value of Z, say z0, such that P(Z z0) = 0.99. Solution: a. P(Z 1) = 0.8413 as shown in Figure 6.17 b. P(Z < -1.5) = 0.0668, as shown in Figure 6.18 Figure 6.17 P(Z 1) Figure 6.18 P(Z < -1.5) c. P(Z > 1) = 1 – P(Z 1) = 1 – 0.8413 = 0.1587, shown in Figure 6.19 d. P(-1.5 Z 0.5) = P(Z 0.5) – P(Z < -1.5) = 0.6915 – 0.0668 = 0.6247 as shown in Figure 6.20. Figure 6.19 P(Z > 1) Figure 6.20 P(-1.5 Z 0.5) e. Find the value of Z, say z0, such that P(Z z0) = 0.99. We must look for the given probability 0.99 on the area side of Appendix Table 4. The closest we can come is 0.9901, which corresponds to the z-value of 2.33. Hence, z0 = 2.33. Figure 6.21 (Z z0) = 0.99 Example: A firm that manufactures and bottles apple juice has a machine that automatically fills 16-ounce bottles. There is, however, some variation in the amount of liquid dispensed (in ounces) into each bottle by the machine. Over a long period of time, the average amount dispensed into the bottles was 16 ounces, but there is a standard deviation of 1 ounce in these measurements. If the amount filled per bottle can be assumed to be normally distributed, find the probability that the machine will dispense more than 17 ounces of liquid into any one bottle. Solution: Let Z denote the ounces of liquid dispensed into one bottle by the filling machine. Then X is assumed to be normally distributed with mean 16 and standard deviation 1 (Figure 6.23). Hence, See Figure 6.23. Figure 6.23 Example: Suppose that another machine, similar to the one in the previous example that operate so that ounces of fill have a mean equal to the dial setting for “amount of liquid” but have a standard deviation of 1.2 ounces. Find the proper setting for the dial so that 17-ounce bottles will overflow only 5% of the time. Assume that the amounts dispensed have a normal distribution. Solution: Letting X denote the amount of liquid dispensed, we are now looking for a value of such that P(X > 17) = 0.05, as depicted in Figure 6.24. Hence From Appendix Table 4, we know that if P(Z > z0) = 0.05, then z0 = 1.645. Thus, it must be that and See Figure 6.24 Figure 6.24 P(X > 17) = 0.05 The Relationship between the Poisson and the Exponential Distribution The Gamma Distribution Some useful results: The Lognormal Distribution If Y is a normal random variable, then X = eY follows a lognormal distribution. The lognormal distribution is very commonly used to describe lifetime distribution, such as lifetime of products, components, and systems. It is a right-skewed distribution and takes values in the range (0, ) The lognormal distribution has the following form, and the graph of the density function is shown in Figure 6.38. Because the probabilities of a lognormal random variable are obtained by transforming a normal random variable, we can obtain the cumulative distribution function of lognormal using the cumulative distribution function of a standard normal variable, Z, as follows: Figure 6.38 The lognormal density function The Lognormal Distribution If Y = ln (X) has a normal distribution with mean and variance 2, then then X has a lognormal distribution with probability density function Example: The Department of Energy (DOE) has a Chronic Beryllium Disease Prevention Program. The focus of this program is to monitor equipment and buildings for beryllium contamination and exposure of workers to beryllium in the workplace. Data given in Table 6.3 were collected from a smelter facility used to recycle metal, where beryllium contaminants tend to deposit on all surface. Wipe samples were taken randomly from the surface of equipment in the workplace and tested for possible beryllium contaminations (measured in gm/100 cm2). The distribution of data (Figure 6.39) shows the lognormal distribution (given by the curve) is a reasonable choice to describe beryllium contamination at this smelter. From the samples, the estimated is 2.291 and to be 1.276. Find the following: a. If a wipe sample is taken randomly form this smelter, what is the probability that it will have beryllium contamination exceeding 0.50 gm/100 cm2? b. The DOE has established a safe limit for beryllium contamination to be L0.95 = 0.20 gm/100 cm2, where L0.95 indicate the 95th percentile. Use this data to determine the safety of workers’ health at this smelter c. The expected amount of beryllium and the variance. Solution: a. Let X the beryllium contamination then ln X follows a normal distribution with mean – 2.291 and standard deviation 1.276. Therefore, using the above stated result and standard normal cumulative distribution function, we get b. The DOE has established a safe limit for beryllium contamination to be L0.95 = 0.20 gm/100cm2, where L0.95 indicates 95th percentile. Use this data to determine the safety of workers’ health at this smelter. We need to determine L0.95 value for this smelter such that P(X L0.95) = 0.95. Therefore, from the above result relating normal and lognormal distributions we get Using the standard normal cumulative distribution function, Solving this equation, we get L0.95 = exp[-2.291] + 1.645(1.276)]=0.8253 > 0.20. Because this estimate of 95 th percentile exceeds the DOE safety limit of 0.20, we can conclude that the beryllium contamination at this smelter is at an unhealthy level for workers. c. The expected amount of beryllium and the variance can be calculated as follows: and Even the average amount of beryllium exceeds the safety limit. The Beta Distribution For , > 0 The following relation is used by the beta distribution. For , > 0, Example: A gasoline wholesale distributor has bulk storage tanks holding a fixed supply. The tanks are filled every Monday. Of interest to the wholesaler is the proportion of this supply that is sold during the week. Over many weeks, this proportion has been observed to be modelled fairly well by a beta distribution with = 4 and = 2. a. Find the expected value of this proportion. b. Is it highly likely that the wholesaler will sell at least 90% of the stock in a given week? Solution: Let x denote the proportion of the total supply sold in a given week. a. The expected proportion of supply sold in a given week is b. We are interested in It is not very likely that 90% of the stock will be sold in a given week. The Weibull Distribution Example: The length of service time during which a certain type of thermistor produces resistance within its specifications has been observed to follow a Weibull distribution with = 50 and = 2. The measurements are in thousands of hours. a. Find the probability that one of these randomly selected thermistors, to be installed in a system today, will function properly for over 10,000 hours. b. Find the expected lifetimes for thermistors of this type. Solution: Let X represent the lifetime of the thermistor in question. a. As seen earlier, the Weibull distribution has a closed-form expression for F(x). Thus, P(X > 10) = 1 – F(10) = b. The expected lifetime for thermistors of this type is Thus, the average service time for these thermistors is 6,270 hours.