OxfordAQA International AS and A-level Mathematics: The Poisson distribution (9660) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover section S2.1 (Poisson distribution) of the OxfordAQA International AS and A-level Mathematics specification (9660), Version 5.2. The specification places this content in International A-level (A2) Unit S2 (Statistics); it is not examined at International AS. Where a question needs a table value, the value is given in the question.

Related pages: the study guide, the revision notes, the course hub and the printable checklist.

Questions

1. State three conditions needed for the number of customers entering a shop in a 10-minute period to be modelled by a Poisson distribution, referring to the shop in each case. [3]

2. X ~ Po(2.5). Find

(a) P(X = 0) [2] (b) P(X = 2) [2]

3. The number of complaints received by a company in one day is modelled by X ~ Po(4.5). Find

(a) P(X = 3) [2] (b) P(X ≥ 2) [3]

4. X ~ Po(7). You are given these cumulative values: P(X ≤ 4) = 0.1730, P(X ≤ 5) = 0.3007, P(X ≤ 6) = 0.4497, P(X ≤ 9) = 0.8305. Find

(a) P(X = 6) [2] (b) P(5 ≤ X ≤ 9) [2] (c) P(X ≥ 10) [1]

5. Flaws occur at random, independently, at a mean rate of 0.6 per metre along a roll of tape. Find

(a) the distribution of the number of flaws in a 4 m length [1] (b) the probability of exactly 3 flaws in a 4 m length [2] (c) the probability of at most 1 flaw in a 4 m length [2]

6. The number of late trains at a station each evening is thought to follow a Poisson distribution.

(a) Over 60 evenings the sample mean is 4.1 and the sample variance is 9.3. Comment on whether the Poisson model is suitable. [3] (b) Y ~ Po(λ) has standard deviation 1.5. Find P(Y = 1). [3]

7. X ~ Po(λ) and P(X = 2) = 4 × P(X = 4).

(a) Show that λ = √3. [3] (b) Find P(X = 0). [2]

8. A machine makes components, and each is faulty with probability 0.02, independently. A batch contains 150 components. Let X be the number of faulty components in a batch.

(a) Explain why X can be approximated by a Poisson distribution, and state its parameter. [2] (b) Use the approximation to find P(X ≤ 1). [2] (c) The exact binomial value of P(X ≤ 1) is 0.1961. Comment on the accuracy of your answer to (b). [2]

9. Telephone calls to two independent departments arrive at random. In one hour, the number of calls to department A has distribution Po(2.4) and the number to department B has distribution Po(1.6). Let T be the total number of calls in one hour.

(a) State the distribution of T, giving a reason. [2] (b) Find P(T ≤ 2). [2] (c) Given that P(T ≤ 5) = 0.7851, find P(T ≥ 6). [2]

10. An online shop receives orders at random at a mean rate of 1.2 per day. Assume that days are independent.

(a) Find the probability that there are no orders on a given day. [2] (b) Find the probability that, in a 5-day week, exactly 2 days have no orders. [3] (c) Given that for Po(3.6), P(X ≤ 6) = 0.9267, find the probability that the shop receives at least 7 orders in a 3-day period. [3]

11. Packets of 250 items are checked. Each item is faulty with probability 0.012, independently. Let X be the number of faulty items in a packet.

(a) Show that a Poisson distribution is suitable for X, and state its mean and standard deviation. [3] (b) Use this distribution to find P(X = 4). [2] (c) A packet is rejected if it has more than k faulty items. You are given that for Po(3), P(X ≤ 5) = 0.9161 and P(X ≤ 6) = 0.9665. Find the smallest k for which the probability of rejecting a packet is less than 0.05. [3]

Answers

1. Any three of: customers arrive singly, so two customers do not enter at exactly the same instant [1]; customers arrive at random [1]; customers arrive independently of one another, so one customer entering does not make another more or less likely [1]; customers arrive at a constant average rate during the period, with no busy or quiet spells. Only three of the four are required. [3] Examiner insight: A condition stated in general words (“they are independent”) is weaker than one tied to the shop. Write what is independent of what, and name the situation each time.

2. (a) P(X = 0) = e^(−2.5) [1] = 0.0821 [1]. (b) P(X = 2) = e^(−2.5) × 2.5² / 2! [1] = 0.08208 × 3.125 = 0.2565 [1]. [4] Examiner insight: Show the substitution into the formula before the decimal. A bare decimal that is slightly out has nothing to support it, whereas a correct substitution with a slip in arithmetic still shows the method.

3. (a) P(X = 3) = e^(−4.5) × 4.5³ / 3! [1] = 0.011109 × 15.1875 = 0.1687 [1]. (b) P(X ≥ 2) = 1 − P(X = 0) − P(X = 1) [1] = 1 − e^(−4.5) (1 + 4.5) [1] = 1 − 0.06110 = 0.9389 [1]. [5] Examiner insight: “At least 2” means the complement is P(X ≤ 1). A common loss is removing only P(X = 0), or writing 1 − P(X ≤ 2).

4. (a) P(X = 6) = P(X ≤ 6) − P(X ≤ 5) [1] = 0.4497 − 0.3007 = 0.1490 [1]. (b) P(5 ≤ X ≤ 9) = P(X ≤ 9) − P(X ≤ 4) [1] = 0.8305 − 0.1730 = 0.6575 [1]. (c) P(X ≥ 10) = 1 − P(X ≤ 9) = 1 − 0.8305 = 0.1695 [1]. [5] Examiner insight: Part (b) is where the index slips. P(X ≤ 5) is the wrong subtraction because it removes the value 5, which the question includes. Write out the integers included.

5. (a) λ = 0.6 × 4 = 2.4, so X ~ Po(2.4) [1]. (b) P(X = 3) = e^(−2.4) × 2.4³ / 3! [1] = 0.090718 × 13.824 / 6 = 0.2090 [1]. (c) P(X ≤ 1) = e^(−2.4) (1 + 2.4) [1] = 3.4e^(−2.4) = 0.3084 [1]. [5] Examiner insight: Stating the new distribution in part (a) is the step that protects the rest of the question. If the mean for one metre, 0.6, is used instead, every later value is wrong for the length asked about.

6. (a) For a Poisson distribution the mean and the variance are equal [1]. Here the variance 9.3 is more than twice the mean 4.1, so the model looks unsuitable [1]. Late trains may occur in clusters, for example when one delay causes another, so they are not independent [1]. (b) Variance = 1.5² = 2.25, so λ = 2.25 [1]. P(Y = 1) = e^(−2.25) × 2.25 [1] = 0.10540 × 2.25 = 0.2371 [1]. [6] Examiner insight: A comment that stops at “the numbers are different” gains little. Say which property is compared, state the conclusion, and link it to the context.

7. (a) P(X = 2) = e^(−λ) λ² / 2 and P(X = 4) = e^(−λ) λ⁴ / 24 [1]. Substituting: e^(−λ) λ² / 2 = 4 × e^(−λ) λ⁴ / 24 [1]. Cancel e^(−λ) and λ²: 1/2 = λ²/6, so λ² = 3 and λ = √3 (λ > 0) [1]. (b) P(X = 0) = e^(−√3) [1] = 0.1769 [1]. [5] Examiner insight: In a “show that” question every line of working must be on the page, including the cancelling of e^(−λ). Finish on the stated result and explain why only the positive root is taken.

8. (a) n = 150 is large and p = 0.02 is small [1], so X is approximately Po(np) = Po(3) [1]. (b) P(X ≤ 1) = e^(−3) (1 + 3) [1] = 4e^(−3) = 0.1991 [1]. (c) The difference is 0.1991 − 0.1961 = 0.0030 [1]. This is small, so the approximation is accurate, which is expected because p is small [1]. [6] Examiner insight: A comment on accuracy needs a figure for the difference and a conclusion. Naming “n large, p small” as the reason for the good agreement shows you know why the approximation works.

9. (a) The two counts are independent, so the sum is Poisson with the rates added [1]: T ~ Po(4) [1]. (b) P(T ≤ 2) = e^(−4) (1 + 4 + 8) [1] = 0.018316 × 13 = 0.2381 [1]. (c) P(T ≥ 6) = 1 − P(T ≤ 5) [1] = 1 − 0.7851 = 0.2149 [1]. [6] Examiner insight: The reason in part (a) must mention independence. The sum rule does not apply to dependent counts, so naming it is part of a complete answer.

10. (a) Let X be the number of orders in a day, so X ~ Po(1.2) [1]. P(X = 0) = e^(−1.2) = 0.3012 [1]. (b) Let Y be the number of days out of 5 with no orders. Y ~ B(5, 0.3012) [1]. P(Y = 2) = 10 × 0.3012² × 0.6988³ [1] = 0.3096 [1]. (c) The total over 3 days has distribution Po(3.6) [1]. P(at least 7) = 1 − P(≤ 6) [1] = 1 − 0.9267 = 0.0733 [1]. [8] Examiner insight: Part (b) links two models: the Poisson gives the probability of a quiet day, and the binomial counts the quiet days. Define Y and name its distribution before calculating, so the structure is clear.

11. (a) n = 250 is large and p = 0.012 is small [1], so X is approximately Po(250 × 0.012) = Po(3), with mean 3 [1] and standard deviation √3 = 1.73 [1]. (b) P(X = 4) = e^(−3) × 3⁴ / 4! [1] = 0.049787 × 81 / 24 = 0.1680 [1]. (c) Rejection probability is P(X > k) = 1 − P(X ≤ k), and this needs to be less than 0.05, so P(X ≤ k) > 0.95 [1]. P(X ≤ 5) = 0.9161 is too small [1]. P(X ≤ 6) = 0.9665 > 0.95, so the smallest value is k = 6 [1]. [8] Examiner insight: In part (c), show the comparison with 0.95 for the value just below your answer as well as for your answer. Stating only that k = 6 works does not show that it is the smallest.

Where marks are usually lost

• Giving a condition without referring to the context.
• Using λ for the wrong interval after a change of length or time.
• Writing P(X ≥ r) = 1 − P(X ≤ r) instead of 1 − P(X ≤ r − 1).
• Removing only P(X = 0) when “at least 2” needs P(X = 0) and P(X = 1) removed.
• Approximating a binomial by a Poisson when p is not small.
• Commenting that mean and variance “differ” without saying what that means for the model.
• Leaving out the statement of independence when adding Poisson variables.
• Missing out working in a “show that” question.
• Rounding e^(−λ) to two or three figures before multiplying.

Next steps

• Read the study guide for full explanations.
• Recap with the revision notes.
• Return to the course hub and tick off the checklist.
• Try all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

These questions follow the OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2, section 3.4 (International A-level Unit S2: Statistics), S2.1 Poisson distribution, published by OxfordAQA.