Discrete Probability Distributions: Random Variables, Probability Distributions, and Their Properties

Discrete Probability Distributions

Random Variables

A random variable is a numerical value associated with the outcome of a probability experiment. The value is determined by chance, and random variables are fundamental in probability and statistics.

• Discrete Random Variable: Has a finite or countable number of possible outcomes that can be listed. Typically represents counted data (e.g., number of calls made in a day).

• Continuous Random Variable: Has an uncountable number of possible outcomes, usually represented by an interval on the number line. Typically represents measured data (e.g., time spent making calls).

Example:

• Let x be the number of Fortune 500 companies that lost money last year. Since this can be counted, x is discrete.

• Let x be the volume of gasoline in a 21-gallon tank. Since this can take any value between 0 and 21 (including fractions), x is continuous.

Study Tip: Even if measured data (like age or weight) are rounded to whole numbers, they are still considered continuous random variables.

Discrete Probability Distributions

A discrete probability distribution lists each possible value of a discrete random variable along with its probability. Such a distribution must satisfy:

• Each probability is between 0 and 1, inclusive: $0 \leq P(x) \leq 1$

• The sum of all probabilities is 1: $\sum P(x) = 1$

These distributions can be represented graphically using a relative frequency histogram.

Constructing a Discrete Probability Distribution

1. Make a frequency distribution for the possible outcomes.

2. Find the sum of the frequencies.

3. Calculate the probability of each outcome: $P(x) = \frac{\text{frequency of } x}{\text{total frequency}}$

4. Check that each probability is between 0 and 1, and that the sum is 1.

Example: Constructing a Probability Distribution

An industrial psychologist gives a personality test to 150 employees, scoring from 1 (extremely passive) to 5 (extremely aggressive). The frequency distribution is:

Score, x	Frequency, f
1	24
2	33
3	42
4	30
5	21

To find the probability distribution, divide each frequency by 150:

x	P(x)
1	0.16
2	0.22
3	0.28
4	0.20
5	0.14

The sum of probabilities is 1, so this is a valid probability distribution.

Verifying a Probability Distribution

To verify a distribution:

• Check that all probabilities are between 0 and 1.

• Check that the sum of all probabilities is 1.

Example:

Days of rain, x	P(x)
0	0.216
1	0.432
2	0.288
3	0.064

All probabilities are between 0 and 1, and their sum is 1. Thus, this is a valid probability distribution.

Common Errors in Probability Distributions

• If the sum of probabilities is not 1, it is not a valid distribution.

• If any probability is negative or greater than 1, it is not valid.

Mean, Variance, and Standard Deviation of a Discrete Probability Distribution

The mean (expected value), variance, and standard deviation describe the center and spread of a probability distribution.

• Mean (μ): $\mu = \sum [x \cdot P(x)]$

• Variance (σ²): $\sigma^2 = \sum [(x - \mu)^2 \cdot P(x)]$

• Standard Deviation (σ): $\sigma = \sqrt{\sigma^2}$

Example: Finding the Mean

x	P(x)	x·P(x)
1	0.16	0.16
2	0.22	0.44
3	0.28	0.84
4	0.20	0.80
5	0.14	0.70

Sum of x·P(x): $0.16 + 0.44 + 0.84 + 0.80 + 0.70 = 2.94$

So, $\mu = 2.94$

Example: Finding the Variance and Standard Deviation

x	P(x)	x - μ	(x - μ)2	(x - μ)2P(x)
1	0.16	-1.94	3.7636	0.6022
2	0.22	-0.94	0.8836	0.1944
3	0.28	0.06	0.0036	0.0010
4	0.20	1.06	1.1236	0.2247
5	0.14	2.06	4.2436	0.5941

Sum of (x - μ)2P(x): $0.6022 + 0.1944 + 0.0010 + 0.2247 + 0.5941 = 1.6164$

So, $\sigma^2 = 1.62$ (rounded), $\sigma = 1.27$ (rounded)

Study Tip: Round the mean, variance, and standard deviation to one more decimal place than the random variable values.

Expected Value

The expected value of a discrete random variable is the same as its mean. It represents the long-run average outcome of a probability experiment.

• Formula: $E(x) = \sum [x \cdot P(x)]$

Example: Raffle Expected Value

Suppose 1500 tickets are sold at $2 each for four prizes: $500, $250, $150, and $75. You buy one ticket. The possible gains (prize minus ticket cost) and their probabilities are:

Gain, x	P(x)
$498	1/1500
$248	1/1500
$148	1/1500
$73	1/1500
-$2	1496/1500

Expected value:

$E(x) = 498 \cdot \frac{1}{1500} + 248 \cdot \frac{1}{1500} + 148 \cdot \frac{1}{1500} + 73 \cdot \frac{1}{1500} + (-2) \cdot \frac{1496}{1500}$

Interpretation: The expected value is negative, so on average, you lose money per ticket. If $E(x) = 0$, the game is fair.

Summary Table: Properties of Discrete Probability Distributions

Property	Description
Probability Range	Each $P(x)$ must satisfy $0 \leq P(x) \leq 1$
Sum of Probabilities	$\sum P(x) = 1$
Mean (μ)	$\mu = \sum [x \cdot P(x)]$
Variance (σ²)	$\sigma^2 = \sum [(x - \mu)^2 \cdot P(x)]$
Standard Deviation (σ)	$\sigma = \sqrt{\sigma^2}$
Expected Value	Same as mean; $E(x) = \mu$

Additional info: Technology such as calculators or software (e.g., TI-84 Plus, Excel) can be used to compute mean and standard deviation for discrete random variables by entering values and their probabilities.