Source: Quiz 3 Solutions, Principles of Statistics I (Texas A&M University)
Tags: discrete probability distribution, PMF, probability mass function, random variable, expected value, mean, variance, standard deviation, STAT 211, probability table
A discrete probability distribution assigns probabilities to each possible value of a random variable, and those probabilities must sum to 1. From the distribution table you can compute the mean (expected value) and standard deviation, and you can find the probability of any event by summing the relevant entries.
Probability distribution (discrete)
A function f_X(x) that assigns a probability to each value x that a discrete random variable X can take. All probabilities must be non-negative and sum to exactly 1.
Probability mass function (PMF)
Another name for the probability distribution of a discrete random variable. Written f_X(x) = P(X = x).
Mean (expected value), μ_X
The long-run average value of a random variable. Computed as the sum of each value multiplied by its probability:
μ_X = Σ x · f_X(x), summed over all x
Variance, σ²_X
A measure of how spread out the distribution is around the mean. Computed as:
σ²_X = Σ (x − μ_X)² · f_X(x), summed over all x
Standard deviation, σ_X
The square root of the variance. Same units as the random variable itself, which makes it more interpretable than variance for describing spread.
Domain (support)
The set of x values where f_X(x) > 0. Any value outside the domain has probability zero.
For a table of f_X(x) values to qualify as a probability distribution, two conditions must hold:
Every probability must be ≥ 0
The sum of all probabilities must equal 1
If the table is written in terms of an unknown (like a), set the sum equal to 1 and solve.
Worked example from quiz:
The PMF was given as f_X(0) = 2a, f_X(1) = 2a, f_X(2) = a, f_X(3) = 4a, f_X(4) = a.
Setting the sum to 1:
2a + 2a + a + 4a + a = 10a = 1, so a = 1/10 = 0.1
This gives the completed table:
x | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
f_X(x) | 0.2 | 0.2 | 0.1 | 0.4 | 0.1 |
Multiply each x value by its probability, then sum. Using the table above:
μ_X = (0)(0.2) + (1)(0.2) + (2)(0.1) + (3)(0.4) + (4)(0.1)
μ_X = 0 + 0.2 + 0.2 + 1.2 + 0.4 = 2.0
The third column of the table (x · f_X(x)) gives these products, and their sum is the mean.
First compute the variance using the (x − μ_X)² · f_X(x) column, then take the square root.
x | f_X(x) | x · f_X(x) | (x − μ_X)² | (x − μ_X)² · f_X(x) |
|---|---|---|---|---|
0 | 0.2 | 0 | 4 | 0.8 |
1 | 0.2 | 0.2 | 1 | 0.2 |
2 | 0.1 | 0.2 | 0 | 0 |
3 | 0.4 | 1.2 | 1 | 0.4 |
4 | 0.1 | 0.4 | 4 | 0.4 |
Sum | 1.0 | 2.0 | 1.8 |
σ²_X = 1.8, so σ_X = √1.8 ≈ 1.34
P(X = specific value): read it directly from the table. If the value is outside the domain, the probability is 0.
P(X = 2016) = 0, because 2016 is not in the domain {0, 1, 2, 3, 4}
P(X < some value): sum the probabilities for all x values strictly less than that value.
P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.2 + 0.2 + 0.1 + 0.4 = 0.9
Complement shortcut: P(X < 4) = 1 − P(X = 4) = 1 − 0.1 = 0.9. This is often faster when summing the complement involves fewer terms.
Mean: μ_X = Σ x · f_X(x)
Variance: σ²_X = Σ (x − μ_X)² · f_X(x)
Standard deviation: σ_X = √(σ²_X)
Valid PMF conditions: f_X(x) ≥ 0 for all x, and Σ f_X(x) = 1
Complement rule: P(A) = 1 − P(A^c)
⚠️ Always check that probabilities sum to 1 before doing anything else. If you're asked to find an unknown, this is the equation to set up.
⚠️ The mean does not have to be one of the possible x values. Here μ_X = 2, which happens to be in the domain, but that is coincidence.
⚠️ When asked P(X < 4), do not include P(X = 4). The inequality is strict. Read inequality signs carefully on the exam.
⚠️ The complement rule (1 minus the opposite probability) is a common shortcut. Recognise when it saves work, especially for "at least" or "less than" questions.
⚠️ If asked for a probability at a value outside the domain, the answer is always 0.
Q: A discrete random variable Y has PMF: f_Y(1) = c, f_Y(2) = 2c, f_Y(3) = 3c. Find c.
A: c + 2c + 3c = 6c = 1, so c = 1/6.
Q: Using the distribution from the quiz (a = 0.1), what is P(1 ≤ X ≤ 3)?
A: P(X = 1) + P(X = 2) + P(X = 3) = 0.2 + 0.1 + 0.4 = 0.7.
Q: Why can't you just read the standard deviation from the table?
A: The standard deviation requires a weighted sum of squared deviations from the mean. You need to compute the mean first, then build the (x − μ)² column, weight each by f_X(x), sum, and take the square root.
Q: If P(X ≥ 2) = 0.6, what is P(X < 2)?
A: By the complement rule, P(X < 2) = 1 − 0.6 = 0.4.
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