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The concept of random variables and their classification as discrete or continuous. Discrete random variables have a finite or countable number of possible outcomes, while continuous random variables have an uncountable number of possible outcomes. The document also covers probability distributions for discrete random variables, which list each possible value and its probability, and provides examples of probability distributions and their expected value, variance, and standard deviation calculations.
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A random variable x represents a numerical value associated with each outcome of a probability experiment. (Note: Many other books make a distinction between an unknown random variable X and the value that it takes x.)
There are two kind of random variables: discrete and continuous. The kind of random variable determines how we will analyze it.
A random variable is discrete if it has a finite number of possible outcomes, or a countable number (i.e. the integers are infinite, but are able to be counted). For example, the number of heads you get when flip a coin 100 times is discrete, since it can only be a whole number between 0 and 100.
A random variable is continuous if it has an uncountable number of possible outcomes, such as an interval on a number line (the interval does not need to be bounded). For example, the amount of time you spend studying for this class is a continuous random variable, since it can be any number that is at least zero hours, and you can spend any percentage of an hour or minute studying.
Examples Determine whether each of the following variables x is discrete or continuous:
We’ll be spending the rest of this chapter analyzing discrete random variables. In Chapter 5 we will be working with a very important class of continuous random variables called Normal random variables.
A discrete probability distribution lists each possible value that a random variable can take, along with its probability. It has the following properties:
P (x) = 1.
Examples Determine if each of the following tables represents a probability distribution:
x − 5 6 9 P (x) 0.5 0.25 0. Yes, this is a probability distribution, since all of the probabilities are between 0 and 1, and they add to 1.
x 1 2 3 4 P (x) 0.4 0.4 0.4 0. This is not a probability distribution, since the probabilities add to 1.4, not 1.
x 1 2 3 4 P (x) 0.4 0.4 0.4 -0. This is not a probability distribution, since P (4) = − 0 .2, and probabilities cannot be negative.
Using a frequency distribution, you can make a probability distribution by using the relative fre- quencies for the probabilities.
Example Make a probability distribution from the following frequency distribution represent the number of fish caught in a 6-hour period: Number of fish caught 0 1 2 3 4 Frequency 88 72 30 8 2
To find the probabilities, we need to divide the frequencies by the total number of observations, 200, so we get:
Number of fish caught 0 1 2 3 4 Probability 0.44 0.36 0.15 0.04 0.
The mean, or expected value of a discrete random variable, μ, is given by
μ = E(x) =
xP (x).
The variance of a discrete random variable, σ^2 is given by
σ^2 =
(x − μ)^2 P (x),
and its standard deviation, σ is given
σ =
σ^2 =
(x − μ)^2 P (x).
Examples Calculate the expected value, variance, and standard deviation for each of the following probability distributions:
x − 5 6 9 P (x) 0.5 0.25 0. First, we need to calculate the expected value (as it is used in the rest of the calculations), so
μ = − 5 · 0 .5 + 6 · 0 .25 + 9 · 0 .25 = 1. 25.
Then the variance is
σ^2 = (− 5 − 1 .25)^2 · 0 .5 + (6 − 1 .25)^2 · 0 .25 + (9 − 1 .25)^2 · 0 .25 = 40. 1875 ,
so the standard deviation is σ =
Number of fish caught 0 1 2 3 4 Probability 0.44 0.36 0.15 0.04 0. The expected value is
μ = 0 · 0 .44 + 1 · 0 .36 + 2 · 0 .15 + 3 · 0 .04 + 4 · 0 .01 = 0. 82.
The variance is
σ^2 = (0 − 0 .82)^2 · 0 .44 + (1 − 0 .82)^2 · 0 .36 + (2 − 0 .82)^2 · 0 .15 + (3 − 0 .82)^2 · 0. 04
and the standard deviation is σ =