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Probability Distributions

Random Variables

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Definition

A random variable is a variable (typically represented by x) that has a single numerical value that is determined by chance.

A probability distribution is a graph, table, or formula that gives the probability for each value of the random variable.

If x is a random variable then denote by P(x) to be the probability that x occurs. It must be the case that $0\le P(x) \le 1$ for each value of x and $\sum P(x) = 1$ (the sum of all the probabilities is 1.)

Example of a probability distribution

Below is a table that gives the probabilities of obtaining exactly x heads in 4 throws.

x	P(x)
0	.0625
1	.2500
2	.3750
3	.2500
4	.0625

Is this a probability distribution? Solution: For each value of x the probability is between 0 and 1. The sum of the probabilities is 1. So the answer is YES, it is a probability distribution.

Many times it is useful to determine the mean and standard deviation for the data. The TI-83 is a very useful tool to do the calculations for you. Here is a step by step way to do this.

Computing the TI-83 to compute mean and standard deviation

1. Enter the values of the random variable x in the list L1.
2. Enter the corresponding probabilities in the list L2.
3. Press STAT, select CALC option and choose 1-Var Stats
4. Enter "L1,L2" and Press the ENTER key.

Example (cont.)

For the example above determine the mean and standard deviation. Solution: Here is what you should see.

There are two items that you are interested in. The first is x, the mean, and the other is $\sigma x$, the standard deviation.

The example above is a special type of probability distribution, which will be discussed in the next section.

Binomial Probability Distributions

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A binomial probability distribution is useful when dealing with two outcomes. Here is the definition of a binomial distribution.

Definition

A binomial probability distribution occurs when the following requirements are met.

1. The procedure has a fixed number of trials.
2. The trials must be independent.
3. Each trial must have all outcomes that fall into two categories.
4. The probabilities must remain constant for each trial.

There are many ways to compute P(x) when dealing with a binomial probability distribution. The TI-83 has this capability built in. In order to use the TI-83, some notation will be needed.

Notation for binomial probability distribution

n	denotes the number of fixed trials
x	denotes the number of successes in the n trials
p	denotes the probability of success
q	denotes the probability of failure (1-p)

How to use the TI-83 to get the probabilities for a binomial probability distribution

1. Press 2nd VARS.
2. Select the option binompdf(.
3. Complete the entry to obtain binompdf(n, p, x), with the appropriate values substituted in.

Flipping coins

What is the probability of getting exactly 2 heads when 4 tosses are made? Solution: Using the TI-83 with binompdf(4, .5, 2), it follows that the probability for getting 2 heads on 4 throws is .375. (see example)

Flipping coins (cont.)

What is the probability of getting at least 2 heads in 4 throws? Solution: To satisfy the condition, 2, 3 or 4 heads must be thrown. These events are independent, so the probabilities can be added together to find the total probability. Using the binomial probability distribution for x = 2, 3, 4, the probability of at least 2 heads is .3750 + .2500 + .0625 = .6875.

Mean and Standard Deviation for Binomial Probability Distribution

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When using binomial probability distribution the computation of the mean and standard deviation is VERY simple. Here are the formulas for computing the mean and standard deviation of a binomial probability distribution.

$$\mu =$$ (3.1)

$$\sigma = \sqrt{n\cdot p\cdot q}$$ (3.2)

Flipping coins (cont.)

What is the mean number of heads obtained in 4 throws? What is the standard deviation? Solution: Since there are only two outcomes and the probabilities are constant, the conditions for the binomial distribution are met. Using formulas (3.1) and (3.2) with n = 4, p = .5, q = .5, then mean is 2.0 and the standard deviation is 1.0.

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