Poisson Tables Quiz: Constructing and Reading Poisson Tables

P(X = 2) = e^{−3} × 3² / 2! = e^{−3} × 9/2 = 4.5e^{−3} ≈ 0.2240. This is the probability of exactly two events when the mean rate is 3. In a Poisson table, this value appears at the intersection of the k = 2 row and the λ = 3 column.

Larger λ increases both the mean and variance (both equal λ), shifting the distribution's mass toward higher k values while also spreading it out. In a Poisson table this is visible as the peak probability moving to larger k, with more rows carrying substantial probability. Option A is incorrect — while P(X = 0) = e^{−λ} shrinks, probabilities at higher k values grow.

Using the complement rule: P(X > 4) = 1 − P(X ≤ 4) = 1 − 0.8153 = 0.1847. Confirming P(X ≤ 4) for λ = 3: P(0) ≈ 0.0498, P(1) ≈ 0.1494, P(2) ≈ 0.2240, P(3) ≈ 0.2240, P(4) ≈ 0.1680. Sum ≈ 0.8153. Poisson tables make such complement calculations straightforward.

The answer is False. The Poisson distribution is defined only for non-negative integers k = 0, 1, 2, … Negative counts are impossible — you cannot observe a negative number of events. All Poisson tables therefore begin at k = 0 and extend upward, with no rows for negative k values.

The mode of a Poisson distribution is ⌊λ⌋ when λ is not an integer. For λ = 2.5, the mode is ⌊2.5⌋ = 2. Confirming: P(2) = e^{−2.5} × 2.5² / 2! ≈ 0.2565 and P(3) = e^{−2.5} × 2.5³ / 3! ≈ 0.2138. Since P(2) > P(3), k = 2 is the most probable outcome in the table.

The answer is False. For λ > 1, P(X = k) first increases from k = 0, reaches its peak at the mode (approximately ⌊λ⌋), and then decreases. For example for λ = 3: P(0) ≈ 0.0498, P(1) ≈ 0.1494, P(2) ≈ 0.2240, P(3) ≈ 0.2240. The values rise then fall. Only for λ ≤ 1 does P(X = k) decrease monotonically from k = 0.

P(X ≤ 3) = P(0) + P(1) + P(2) + P(3). For λ = 4: P(0) = e^{−4} ≈ 0.0183, P(1) = 4e^{−4} ≈ 0.0733, P(2) = 8e^{−4} ≈ 0.1465, P(3) = (32/6)e^{−4} ≈ 0.1954. Sum = 0.0183 + 0.0733 + 0.1465 + 0.1954 = 0.4335. Option C (0.5665) is a common incorrect answer that results from using wrong individual probabilities.

The answer is True. At small λ the distribution is strongly right-skewed with most probability near k = 0. As λ grows, the mass shifts toward higher k values and the shape becomes more symmetric, converging toward a normal distribution with mean and variance both equal to λ. This progression is clearly visible in Poisson tables as the peak probability moves rightward.

Options A, B, and C are all correct. Choosing the k range defines the table's scope. Computing each P(X = k) using the formula fills each cell. Verifying the sum confirms the table is valid. Option D is incorrect: large k values should be included until their probabilities become negligibly small. Stopping too early makes the table incomplete and causes the probabilities to sum to less than 1.

Using the complement rule: P(X ≥ 2) = 1 − P(X ≤ 1) = 1 − [P(0) + P(1)]. P(0) = e^{−2} ≈ 0.1353 and P(1) = 2e^{−2} ≈ 0.2707. So P(X ≥ 2) = 1 − 0.1353 − 0.2707 = 0.5940. This is easily computed from a table by reading the first two rows and subtracting their sum from 1.

A Poisson probability table lists P(X = k) = e^{−λ} λ^k / k! for multiple values of k at fixed λ values. This allows easy lookup of exact probabilities without manual computation. The table is organised by rows of k and columns of λ, and all entries for a given λ sum to 1.

P(X ≤ 2) = P(0) + P(1) + P(2). For λ = 5 this equals e^{−5}(1 + 5 + 12.5) = 18.5e^{−5} ≈ 0.1247. The complement 1 − 0.1247 = 0.8753 gives P(X ≥ 3). Cumulative entries are useful for finding tail probabilities quickly without summing individual rows.

P(X ≤ 3) = P(0) + P(1) + P(2) + P(3). For λ = 2: P(0) = e^{−2} ≈ 0.1353, P(1) = 2e^{−2} ≈ 0.2707, P(2) = 2e^{−2} ≈ 0.2707, P(3) = (4/3)e^{−2} ≈ 0.1804. Sum = 0.1353 + 0.2707 + 0.2707 + 0.1804 = 0.8571.

P(X ≥ 3) = 1 − P(X ≤ 2) = 1 − [P(0) + P(1) + P(2)]. Option A is correct: the right λ column must be identified first. Option B is correct: the three rows for k = 0, 1, 2 are needed. Option C is correct: their sum is subtracted from 1. Option D is incorrect: reading P(X = 3) alone gives only P(X = 3), not P(X ≥ 3).

Standard Poisson tables use rows to index event counts k = 0, 1, 2, … and columns to index different λ values. Each cell contains P(X = k; λ). This layout allows a reader to locate any (k, λ) pair quickly without recomputing the formula.

P(X = 3) = 0.1804 means the probability of observing exactly three events is approximately 18%, given a mean rate of λ = 2. This is a single-row probability, not a cumulative one. Confirming: e^{−2} × 2³ / 3! = e^{−2} × 8/6 ≈ 0.1353 × 1.3333 ≈ 0.1804.

P(X = 0) = e^{−λ} × λ^0 / 0! = e^{−4} ≈ 0.0183. This represents the probability of zero events when the average rate is 4 per interval. In a Poisson table, the row k = 0 always contains e^{−λ}, which decreases as λ increases.

The answer is True. The Poisson PMF satisfies Σ_{k=0}^{∞} e^{−λ} λ^k / k! = e^{−λ} × e^{λ} = 1. This is the normalisation condition confirming that the total probability across all possible outcomes is exactly 1. Every valid Poisson table must satisfy this property.

A standard Poisson table includes the mean rate λ, the event count k, and the corresponding probabilities P(X = k). Standard deviation is not listed because it can be derived from λ since SD = √λ. Including it would add a redundant column that does not serve the table's core lookup purpose.

Each row in a Poisson table represents a specific count k. For k = 2, the entry gives P(X = 2) = e^{−3} × 3² / 2! = e^{−3} × 4.5 ≈ 0.2240, meaning the probability of observing exactly two events when the mean rate is 3.