The concept of the probability density function was introduced in Chapter 5. There are many specific probability density functions. Each represents a different population that is characteristic of some physical process. In the following, a few of the more common ones will be examined.
Some of the probability density functions are for discrete processes (those having only discrete outcomes), such as the binomial probability density
FIGURE 5.11
Graphical approach to determine p(x)dx.
function. This describes the probability of the number of successful out- comes, n, in N repeated trials, given that only either success (with proba- bility P ) or failure (with probability Q = 1 − P ) is possible. The binomial probability density function, for example, describes the probability of ob- taining a certain sum of the numbers on a pair of dice when tossed or the probability of getting a particular number of heads and tails for a series of coin tosses. The Poisson probability density function models the probability of rarely occurring events. It can be derived from the binomial probability density function. Two examples of processes that can be modeled by the Poisson probability density function are number of disintegrative emissions from an isotope and the number of micrometeoroid impacts on a spacecraft. Although the outcomes of these processes are discrete whole numbers, the process is considered continuous because of the very large number of events considered. This essentially amounts to possible outcomes that span a large, continuous range of whole numbers.
Other probability density functions are for continuous processes. The most common one is the normal (Gaussian) probability density function. Many situations closely follow a normal distribution, such as the times of runners finishing a marathon, the scores on an exam for a very large class, and the IQs of everyone without a college degree (or with one). The Weibull probability density function is used to determine the probability of fatigue- induced failure times for components. The lognormal probability density
function is similar to the normal probability density function but considers its variable to be related to the logarithm of another variable. The diameters of raindrops are lognormally distributed, as are the populations of various biological systems. Most recently, scientists have suggested a new probability density function that can be used quite successfully to model the occurrence of clear-air turbulence and earthquakes. This probability density function is similar to the normal probability density function but is skewed to the left and has a larger tail to the right to account for the observed higher frequency of more rarely occurring events.
5.6.1
Binomial Distribution
Consider first the binomial distribution. In a repeated trials experiment con- sisting of N independent trials with a probability of success, P , for an in- dividual trial, the probability of getting exactly n successes (for n ≤ N) is given by the binomial probability density function
p(n) = N ! (N − n)!n! Pn(1 − P )N −n. (5.14)
The mean, ¯n, and the variance, σ2, are N P and N P Q, respectively, where
Q is the probability of failure, which equals 1 − P . The higher-order central moments of the skewness and kurtosis are (Q − P )/(NP Q)0.5and 3 + [(1 −
6P Q)/N P Q], respectively.
As shown in Figure 5.12, for a fixed N , as P becomes larger, the proba- bility density function becomes skewed more to the right. For a fixed P , as N becomes larger, the probability density function becomes more symmet- ric. Tending to the limit of large N and small but finite P , the probability density function approaches a normal one. The MATLAB M-file bipdfs.m was used to generate this figure based upon the binopdf(n,N,P) command.
Example Problem 5.4
Statement: Suppose five students are taking a probability course. Typically, only 75 % of the students pass this course. Determine the probabilities that exactly 0, 1, 2, 3, 4, or 5 students will pass the course.
Solution: These probabilities are calculated using Equation 5.14, where N = 5, P = 0.75, and n = 0, 1, 2, 3, 4, and 5. They are displayed immediately below and plotted in Figure 5.13. p(0) = 1 × 0.750 × 0.255 = 0.0010 p(1) = 5 × 0.751 × 0.254 = 0.0146 p(2) = 10 × 0.752 × 0.253 = 0.0879 p(3) = 10 × 0.753 × 0.252 = 0.2637 p(4) = 5 × 0.754 × 0.251 = 0.3955 p(5) = 1 × 0.755 × 0.250 = 0.2373 sum = 1.0000
FIGURE 5.12
Binomial probability density functions for various N and P .
5.6.2
Poisson Distribution
Next consider the Poisson distribution. In the limit when N becomes very large and P becomes very small (close to zero, which implies a rare event) in such a way that the mean (= N P ) remains finite, the binomial probability density function very closely approximates the Poisson probability density function.
For these conditions, the Poisson probability density function allows us to determine the probability of n rare event successes (occurrences) out of a large number of N repeated trial experiments (during a series of N time intervals) with the probability P of an event success (during a time interval) as given by the probability density function
p(n) = (N P )
n
n! e
−N P. (5.15)
The MATLAB command poisspdf(n,N*P) can be used to calculate the probabilities given by Equation 5.15. The mean and variance both equal N P , noting (1 − P ) ≈ 1. The skewness and kurtosis are (NP )−0.5 and
3 + 1/N P , respectively. As N P is increased, the Poisson probability density function approaches a normal probability density function.
FIGURE 5.13
Probabilities for various numbers of students passed.
Example Problem 5.5
Statement: There are 2 × 10−20α particles per second emitted from the nucleus
of an isotope. This implies that the probability for an emission from a nucleus to occur in one second is 2 × 10−20. Assume that the total material to be observed
is comprised of 1020 atoms. Emissions from the material are observed at one-second
intervals. Determine the resulting probabilities that a total of 0, 1, 2, ..., 8 emissions occur in the interval.
Solution: The probabilities are calculated using Equation 5.15, where N = 1020, P
= 2 × 10−20and n = 0 through 8. The results are
p(0) = 0.135 p(1) = 0.271 p(2) = 0.271 p(3) = 0.180 p(4) = 0.090 p(5) = 0.036 p(6) = 0.012 p(7) = 0.003 p(8) = 0.001 sum = 0.999
FIGURE 5.14
Poisson example of isotope emissions.