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You may be more familiar with what are called discrete random variables, for example the number of heads obtained in ten tosses of a coin, which can only take a finite number of discrete values. In the case of a discrete random variable, the probability of a single outcome can be positive. For example, the probability that a single flip of a coin produces tails is 50%. The situation is very different when we consider a random variable like the number of miles a tire can be driven before failure, which can take any value from zero to something over 100,000 miles. Since there are an infinite number of possible outcomes, the probability that the tire fails at exactly some number of miles, for example 50,000 miles, is zero. However, we would expect that the probability that the tire would fail between 40,000 miles and 100,000 miles would not be zero, but would be a positive number.
A random variable that can take on a continuous range of values is called a continuous random variable. There turn out to be lots of applications of continuous random variables in science, engineering, and business, so a lot of effort has gone into devising mathematical models. These mathematical models are all based on the following definition.
We say that a random variable X is continuous if there is a function f(x), called the probability density function, such that
For example, consider the following function.
This function is non-negative, and also satisfies the second condition, since
which is pretty easy to show. So this could be a probability density function for a continuous random variable X.
A lot of the effort involved in modeling a random process, that is, a process whose outcome is a random variable, is in finding a suitable probability density function. Over the years, lots of different functions have been proposed and used. One thing that they all have in common, though, is that they depend on parameters. For example, the general exponential probability density function is defined as
where is a parameter that can be adjusted to get the best fit to any particular situation. In the exercises, you will be asked to show that only positive values of make sense.
The process of deciding what probability density function to use and how to determine the parameters is very complicated and can involve very sophisticated mathematics. However, in the simple approach we are taking here, the problem of determining the parameter value(s) often depends on quantities that can be determined experimentally, for example by collecting data on tire failure. For our purposes, the two most important quantities are the mean, and the standard deviation . The mean is defined by
and the standard deviation is the square root of the variance, V, which is defined by
In practice, the variance V is often computed as follows,
which can be easily be obtained by expanding and writing V as the sum of three integrals.
Probably the most important distribution is the normal distribution, widely referred to as the bell-shaped curve. The probability density function for a normal distribution with mean and standard deviation is given by the following equation.
This distribution has a tremendous number of applications in science, engineering, and business. The exercises provide a few simple ones.
In applications, one generally has to know in advance that the random variable you want to model has, approximately, a certain kind of distribution. How one would determine this is way beyond the scope of this course, so we won't really discuss it. On the other hand, once you know, for example, that your random variable has a normal distribution you only need the values of the mean and the standard deviation to be able to model it. The exponential distribution is even simpler, since it only has one parameter, and you only need to know the mean of your random variable to use this distribution to model it.
One thing to keep in mind when you are using the normal distribution as a model is that calculations can involve values of your random variable that don't make physical sense. For example, suppose that a machining operation produces steel shafts whose diameters have a normal distribution, with a mean of 1.005 inches and a standard deviation of 0.01 inch. If you were asked to compute the percentage of the shafts in a certain production run had diameters less than 0.9 inches you would use the following integral
even though negative values for the shaft diameters don't make sense.
satisfies the condition
as long as is a positive number. What would happen if was negative?
First, compute the value of c. Then find the mean .
William W. Farr