Questions on the Poisson distribution

Question 1

State the conditions under which a count can be modelled by a Poisson distribution, and name its one parameter.

Accepted Answer

Events must occur singly, at random, independently of one another, and at a constant average rate over the window in question. The only parameter is that mean rate λ, which is also the variance. B1 B1 for the four conditions, B1 for naming λ.

Question 2

X ~ Po(4). Find P(X = 3) and P(X ≤ 2), each to 4 decimal places.

Accepted Answer

P(X = 3) = e⁻⁴ × 4³/3! = 0.1954. P(X ≤ 2) = e⁻⁴(1 + 4 + 8) = 0.2381. M1 for the Poisson formula, A1 for 0.1954, A1 for 0.2381. The cumulative value comes fastest from a calculator's own Poisson function.

Question 3

X ~ B(1000, 0.003). State the Poisson distribution that approximates X, and the two conditions that make the approximation appropriate here.

Accepted Answer

Po(3), since λ = np = 1000 × 0.003 = 3. The approximation is appropriate because n = 1000 is large and p = 0.003 is small. B1 for Po(3), B1 for both conditions. The mean is preserved exactly; the variance np(1 − p) = 2.991 is replaced by 3, which is the small cost of the approximation.

Question 4

X ~ Po(1.8). Find P(X = 2) and P(X > 2), each to 4 decimal places.

Accepted Answer

P(X = 2) = e^−1.8 × 1.8²/2! = 0.1653 × 1.62 = 0.2678. P(X ≤ 2) = e^−1.8(1 + 1.8 + 1.62) = 0.7306, so P(X > 2) = 1 − 0.7306 = 0.2694. M1 for the Poisson formula, A1 for 0.2678, A1 for 0.2694. 'More than 2' is the complement of 'at most 2', so the cumulative probability to subtract includes the term for 2.

Question 5

Calls arrive at a switchboard at a mean rate of 2.5 per minute, at random. Find the probability of no calls in a two-minute period, and state the variance of the count over that period.

Accepted Answer

Scale λ with the window. Over two minutes the count is Po(2.5 × 2) = Po(5), so P(X = 0) = e⁻⁵ = 0.0067 and the variance is 5, equal to the mean. M1 for scaling λ, A1 for Po(5), A1 for 0.0067, B1 for the variance. Leaving λ at 2.5 gives 0.0821.

Question 6

X ~ Po(6). Find P(X ≥ 8) to 4 decimal places.

Accepted Answer

P(X ≥ 8) = 1 − P(X ≤ 7) = 1 − 0.7440 = 0.2560. M1 for the one minus form, A1 for 0.2560.

Question 7

A component fails with probability 0.01, independently. In a batch of 200, use a Poisson approximation to estimate the probability of at most one failure, and compare with the exact binomial value.

Accepted Answer

n is large and p small, so B(200, 0.01) ≈ Po(2) with λ = np = 2. Then P(X ≤ 1) = e⁻²(1 + 2) = 0.4060. The exact binomial gives 0.4046, so the approximation is out by 0.0014: close enough for any practical purpose. B1 for λ = 2, M1 for the Poisson probability, A1 for 0.4060, B1 for the binomial value, A1 for the comparison.

Question 8

Over 100 randomly chosen five-minute intervals, 248 vehicles in total passed a rural junction, and the variance of the 100 counts was 2.51. State, with a reason, whether these figures support a Poisson model for the number passing in a five-minute interval. Explain also why a Poisson model would be a poor choice for the number of people entering a cinema in a five-minute interval on a Saturday evening.

Accepted Answer

The sample mean is 248/100 = 2.48 and the sample variance is 2.51. A Poisson distribution has its variance equal to its mean, and two figures this close are consistent with that, so the junction data support the model. The cinema fails the model on two of its four conditions. People arrive in couples, families and parties, so the events do not occur singly, and one person arriving makes several more arrivals certain rather than leaving them independent. The rate is also not constant across the evening, because arrivals bunch into the few minutes before each showing. B1 for the mean 2.48, B1 for comparing it with the variance and drawing the conclusion, B1 for arrivals coming in groups rather than singly, B1 for the rate not being constant. A mean close to the variance is evidence for the model, not proof of it.

Question 9

The number of vehicles passing a checkpoint in a one-hour period is modelled by X ~ Po(λ), and P(X = 2) = P(X = 3). Find λ, and hence find P(X ≥ 2) and the probability that exactly one vehicle passes in a 20-minute period.

Accepted Answer

Write both probabilities out from the formula: e^−λλ²/2! = e^−λλ³/3!. The factor e^−λ is never zero, so it cancels, and λ > 0 allows division by λ² as well. That leaves 1/2 = λ/6, so λ = 3.
P(X ≥ 2) = 1 − P(X ≤ 1) = 1 − e⁻³(1 + 3) = 1 − 0.1991 = 0.8009.
A 20-minute period is a third of an hour, so the count over it is Po(3 × 1/3) = Po(1) and P(X = 1) = e⁻¹ = 0.3679.
M1 for equating the two expressions, A1 for λ = 3, M1 for the one minus P(X ≤ 1) form, A1 for 0.8009, M1 for rescaling λ to 1, A1 for 0.3679.
Cancelling the exponential but not the λ² leaves the cubic λ³/6 = λ²/2, whose other root, λ = 0, describes a checkpoint no vehicle ever passes.

Question 10

X ~ Po(2) and Y ~ Po(3) are independent. Find P(X + Y = 4), and the probability that X = 1 given that X + Y = 4.

Accepted Answer

Independent Poisson variables add, so X + Y ~ Po(5) and P(X + Y = 4) = e⁻⁵ × 5⁴/4! = e⁻⁵ × 625/24 = 0.1755.
The event 'X = 1 and X + Y = 4' is 'X = 1 and Y = 3', which by independence has probability P(X = 1)P(Y = 3) = (2e⁻²)(4.5e⁻³) = 9e⁻⁵ = 0.06064. So P(X = 1 | X + Y = 4) = 0.06064/0.17547 = 0.3456. M1 for Po(5), A1 for 0.1755, M1 for P(X = 1)P(Y = 3) over P(X + Y = 4), A1 for 0.3456. Cancelling e⁻⁵ gives 9 × 24/625 = 0.3456 exactly, which is the binomial probability 4 × 0.4 × 0.6³: given the total, each event belongs to X with probability 2/5.

Question 11

The number of misprints on a page follows a Poisson distribution, and the probability that a page has no misprints is 0.2. Find the mean number of misprints per page, and the probability that a page has at least 3 misprints.

Accepted Answer

P(X = 0) = e^−λ = 0.2, so λ = ln 5 = 1.609. P(X ≥ 3) = 1 − P(X ≤ 2) = 1 − e^−λ(1 + λ + λ²/2) = 1 − 0.2(1 + 1.6094 + 1.2951) = 1 − 0.7809 = 0.2191. M1 for e^−λ = 0.2, A1 for λ = ln 5 = 1.609, M1 for 1 − P(X ≤ 2) with e^−λ replaced by 0.2, A1 for 0.2191. Keep λ as ln 5 in the working: e^−λ is then exactly 0.2 and no rounding enters until the end.

Question 12

Two independent sources of faults occur at rates following Po(1.5) and Po(2) per hour. State the distribution of the total number of faults in a three-hour shift, explain why its mean and variance are equal, and find the probability that the shift produces at most 8 faults.

Accepted Answer

Independent Poisson variables add, so the two sources pool to Po(1.5 + 2) = Po(3.5) per hour. Scaling the window to three hours multiplies the rate by 3, giving Po(10.5) for the shift.
Both the mean and the variance are 10.5. A Poisson distribution has only one parameter, and it serves as both, so a sample whose variance is far from its mean is evidence against the model.
P(X ≤ 8) = 0.279 to three decimal places, from the cumulative Poisson function with λ = 10.5.
M1 for pooling the rates, M1 for scaling to three hours, A1 for Po(10.5), B1 for the mean and variance, B1 for why they are equal, M1 for the cumulative probability, A1 for 0.279.
Both rates must be scaled to the shift. Pool the rates first and then scale, or scale each and then pool, but never scale only one of them. The additive property needs the two sources to be independent, and that should be stated.

Question 13

A helpdesk receives routine emails at a mean rate of 3 per hour and urgent emails at a mean rate of 1 per hour, the two streams being independent of each other and of the time of day. State the distribution of the total number of emails received in a 15-minute period. Given that at least one email arrives in a particular 15-minute period, find the probability that exactly two arrive. Find also, to the nearest minute, the length of period over which the probability of receiving no email at all is 0.05.

Accepted Answer

Independent Poisson variables add, so the two streams pool to Po(3 + 1) = Po(4) per hour. A quarter of an hour scales that rate by 1/4, giving Po(1) for the 15-minute count.
The condition changes the denominator, not the numerator. The event 'X = 2 and X ≥ 1' is the event X = 2, so P(X = 2 | X ≥ 1) = P(X = 2)/P(X ≥ 1) = (e⁻¹ × 1²/2!)/(1 − e⁻¹) = 0.1839/0.6321 = 0.2910. That is larger than P(X = 2) = 0.1839 on its own, because conditioning on X ≥ 1 removes the outcome X = 0, which carries probability e⁻¹ = 0.3679.
Over t hours the count is Po(4t), so P(no email) = e^−4t = 0.05. Taking logarithms, 4t = ln 20 = 2.9957, so t = 0.7489 hours, which is 44.9 minutes, or 45 minutes to the nearest minute.
M1 for pooling the rates, A1 for Po(1), M1 for the conditional probability form, A1 for 0.2910, M1 for e^−4t = 0.05, A1 for t = 0.749 hours, A1 for 45 minutes.
In λ = 4t the time t is measured in hours, so work in hours throughout and convert to minutes only at the end.

Question 14

The number of insects on a leaf follows Po(0.4), independently from leaf to leaf. A gardener inspects 10 leaves. Find the probability that at least one of the 10 leaves carries 3 or more insects. Find also the probability that the 10 leaves carry 3 or more insects in total, and explain why the two probabilities are so different.

Accepted Answer

For one leaf, P(X ≥ 3) = 1 − P(X ≤ 2) = 1 − e^−0.4(1 + 0.4 + 0.08) = 1 − 0.9921 = 0.0079.
The leaves are independent, so the probability that none of the 10 has 3 or more is (1 − 0.0079)¹⁰, and P(at least one does) = 1 − 0.9921¹⁰ = 0.0765.
The total over 10 independent leaves is Po(10 × 0.4) = Po(4), so P(total ≥ 3) = 1 − e⁻⁴(1 + 4 + 8) = 1 − 0.2381 = 0.7619.
The first event needs three insects crowded on to a single leaf, which is rare when a leaf averages 0.4; the second only needs three insects spread anywhere across ten leaves that average 4 between them, which is likely. The additive property answers the second question and says nothing about the first. M1 for 1 − P(X ≤ 2) on one leaf, A1 for 0.0079, M1 for 1 − (1 − 0.0079)¹⁰, A1 for 0.0765, M1 for Po(4) with 1 − P(X ≤ 2), A1 for 0.7619 with the explanation. The count of leaves with 3 or more insects is binomial B(10, 0.0079), not Poisson, because it counts successes in a fixed number of independent trials.

Question 15

X ~ Po(λ). Show that P(X = k)/P(X = k − 1) = λ/k for k = 1, 2, 3, …, and use this to explain how the mode of X is found. Find the mode when λ = 3.6, and show that when λ = 4 there are two modes.

Accepted Answer

P(X = k)/P(X = k − 1) = [e^−λλ^k/k!] × [(k − 1)!/(e^−λλ^k−1)] = λ × (k − 1)!/k! = λ/k.
The ratio exceeds 1 while k < λ and is below 1 once k > λ, so the probabilities rise until k passes λ and fall after it: the mode is the largest integer k with k ≤ λ, the integer part of λ.
For λ = 3.6 the ratios are 3.6/3 = 1.2 > 1 and 3.6/4 = 0.9 < 1, so P(X = 3) exceeds both neighbours and the mode is 3 (the probabilities at 2, 3, 4 are 0.1771, 0.2125, 0.1912).
For λ = 4 the ratio at k = 4 is 4/4 = 1 exactly, so P(X = 4) = P(X = 3) = e⁻⁴ × 4³/3! = 0.1954, and the two are both modes, 3 and 4: an integer λ always gives a pair of equal largest probabilities at λ − 1 and λ. M1 for writing the two probabilities and cancelling, A1 for λ/k, M1 for comparing the ratio with 1 on either side of λ, A1 for the mode 3 at λ = 3.6, A1 for the equal probabilities at 3 and 4 when λ = 4. The mode is not the mean: the mean 3.6 is not even a possible value.