Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Chapter 4 Continuous Random Variables & Probability Distribution (Cont.) Normal approximation to the Binomial distribution • Example 4-17: • For many physical systems, the binomial model is appropriate with an extremely large values for n. • In these cases, it is difficult to calculate probabilities by using the binomial distribution. • Normal approximation is most effective in these cases. Normal approximation to the Binomial distribution • If X is a binomial random variable, Z X np np(1 p) is approximately a standard normal random variable. • The approximation is good for np > 5 and n(1-p) > 5 • The approximation is good when n is large relative to p. • Examples: 4-18, & 4-19 Normal approximation to the Poisson distributions • Example: 4-20 • If X is a Poisson random variable with E(X) = and V(X) = , Z X is approximately a standard normal random variable. The approximation is good for > 5. Exponential distribution • In Poisson distribution, X is the random variable representing number of (flaws along a length of wire). • The distance between flaws is another random variable of interest. • The random variable X that equals the distance between successive counts of a Poisson distribution with means > 0 is an exponential random variable with parameter . The probability density function of X is f(x) = e-x for 0 x < Probability distribution function of exponential random variable. Note how the distribution is skewed. Exponential distribution (cont.) • Cumulative distribution function: F(x) = P(X x) = 1 – e-x, x 0 • If a random variable X has an exponential distribution with parameter , = E(X) = 1/ 2 = V(X) = 1/ 2 Example 4-21 Exponential distribution (cont.) (1) Poisson process assumes that events occur uniformly throughout the interval of observation, i.e., there is no cluster of event. Thus, our starting point for observation does not matter. This due to the fact that the number of events in an interval of a Poisson process depends only on the length of the interval, not on the location. (2) In Poisson distribution, we assumed an interval could be partitioned into small intervals that are independent. THUS, knowledge of previous results does not affect the probabilities of events in future subintervals. This is called lack of memory property. P(X<t1+t2|X>t1)=P(X<t2) • The exponential distribution function is the only continuous distribution with this property. Example 4-22 P(X<t1+t2|X>t1)=P(X<t2) Exponential distribution (cont.) • Exponential distribution is often used in reliability studies as the model for the time until failure of a device (taking into consideration lack of memory property). Example: • Life time of a semiconductor chip might be modeled as an exponential random variable with a mean of 40,000 hours. Due to lack of memory property, regardless of how long the device has been operating, the probability of a failure in the next 1000 hours is the same as the probability of a failure in the first 1000 hours of operation (i.e., the device does not wear out). Erlang distribution • An exponential random variable describes the length until the first count is obtained in a Poisson distribution. • Generalization of the exponential distribution: The length until r counts occur in a Poisson distribution. The random variable that equals the interval length until r counts occur in a Poisson process has an Erlang random variable. Erlang distribution (Cont.) • The random variable X that equals the interval length until r counts occur in a Poisson process with mean > 0 has an Erlang random Variable with parameters and r . The probability density function of X is f ( x) r 1 x x e r (r 1)! for x > 0 and r = 1, 2 , … Example 4-24 Erlang distribution (Cont.) • Erlang random variable with r = 1 is an exponential random variable. • An Erlang random variable can be represented as the sum of r exponential random variables. • If X is an Erlang random variable with parameters and r, = E(X) = r/ 2 = V(X) = r/ 2 Weibul distribution • Weibul distribution is often used to model the time until failure of many different physical systems. • Distribution parameters provide a great deal of flexibility to model systems in which the number of failures: – Increases with time (e.g., bearing wear). – Decreases with time (some semiconductors). – Remains constant (failures caused by external shocks to the system). Weibul distribution (Cont.) • The random variable X with probability density function x f ( x) 1 x exp for x > 0 is a Weibul random variable with scale parameter > 0 and shape parameter > 0. Weibul distribution (Cont.) • By inspecting the probability density function, it is seen that when = 1, the Weibull distribution is identical to the exponential distribution. • If X has a Weibull distribution with parameters and , then the cumulative distribution function of X is: F ( x) 1 e x Weibul distribution (Cont.) • If X has a Weibul distribution with parameters and , 1 E (x) 1 2 2 1 2 2 2 V ( x) 1 1 (r) = (r-1)! Revision • A Poisson random variable describes the number of counts in an interval. • An exponential random variable describes the length until the first count is obtained in a Poisson distribution. • An Erlang random variable describes the length until r counts occur in a Poisson distribution. • Weibul distribution is used to model the time until failure of many different physical systems. ANNOUNCEMENTS • Homework III: 3, 19, 27, 38, 46, 50, 61, 82 • Due on: Monday, 16th of May