Correct normal approximations to Poisson events
Use half-unit edges to preserve integer cells, then standardise with √λ and choose the required normal area.
Exam scope and assessment notes
Probability & Statistics 2 (Paper 6, A Level) · Mathematics 9709, 2026–2027, section 6.1. Paper 6 builds on Paper 5 and assumes Paper 3 calculus. Po(λ) counts non-negative integers over a stated exposure, with finite λ ≥ 0. Mean and variance are each λ; proofs are not required. The homogeneous process examples stipulate constant rate, independent counts in disjoint exposures, and negligible multiple events in sufficiently small exposures; real applications need evidence. A binomial-to-Poisson approximation uses large n and small p, with n > 50 and np < 5 approximately as the syllabus guide. A Poisson-to-normal approximation uses λ > 15 approximately, matching N(λ,λ) and applying continuity correction. These guides are not universal error guarantees. Exact and approximate calculations are labelled separately; numerical normal probabilities use unrounded z and can differ slightly from rounded-table working. Independent Poisson sums and linear combinations, sample means and hypothesis testing belong to later Statistics 2 sections. Written practice is self-checked; cumulative assessment and human subject-teacher review remain pending.
Before you begin
- Match Po(λ) to N(λ,λ).
- Use cumulative normal areas and complements.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What SD is used to standardise N(20,20)?
What you will learn
- Correct tails, intervals and exact-count events.
- Calculate labelled approximate areas using SD √λ.
Retain the complete boundary cell
For Po(20), use Y ~ N(20,20). X ≤ 25 includes the count-25 cell through its upper edge 25.5. Thus P(X ≤ 25) is approximately Φ((25.5 − 20)/√20) ≈ 0.8906.
X ≥ 25 instead starts at the lower edge 24.5, giving approximately 1 − Φ(4.5/√20) = 0.1572. X > 25 excludes that cell and uses Y > 25.5. Discrete strict and inclusive events differ, even though continuous endpoints of zero area do not.
Correct both ends and single counts
15 ≤ X ≤ 25 becomes 14.5 < Y < 25.5, with standardised limits ±5.5/√20 and area approximately 0.7812. Correct both endpoints before standardising.
X = 20 becomes 19.5 < Y < 20.5, area approximately 0.0890. The zero continuous point probability P(Y = 20) does not estimate a whole discrete mass. Numerical values here use unrounded z; rounded table working may differ slightly. The corrected result remains approximate.
Pause and explain
Which corrected normal event represents X ≥ 25?
Worked example
Approximate P(X ≥ 25) for Po(20).
Show the worked solution
- Use N(20,20), SD √20.
- Preserve count twenty-five with Y > 24.5.
- Calculate 1 − Φ(4.5/√20) ≈ 0.1572.
Answer Approximately 0.1572; corrected boundary 24.5.
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Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Approximate P(X ≤ 25) for Po(20).
Give me a hint
Use the upper edge of the included count-twenty-five cell.
Compare my reasoning
- N(20,20) meets the guide.
- Use Y < 25.5 with z = 5.5/√20.
- Left normal area is approximately 0.8906.
Approximately 0.8906; Y < 25.5.
Look for these in your work
- Correct the upper endpoint.
- Use SD √20 and label approximate.
Approximate P(X ≥ 25) for Po(20) and explain why 25.5 is wrong.
Give me a hint
Keep the count-twenty-five cell in the upper event.
Compare my reasoning
- Use Y > 24.5.
- Right area at z = 4.5/√20 is approximately 0.1572.
- A boundary of 25.5 excludes count twenty-five and estimates X > 25.
Approximately 0.1572; use 24.5.
Look for these in your work
- Preserve the event.
- Take the right normal tail.
For Po(20), approximate P(15 ≤ X ≤ 25) and P(X = 20), explaining the positive latter area.
Give me a hint
Use outer half-unit edges for each included cell.
Compare my reasoning
- Interval becomes 14.5 < Y < 25.5, area ≈ 0.7812.
- Count twenty becomes 19.5 < Y < 20.5, area ≈ 0.0890.
- That whole cell differs from a zero-width continuous point.
Interval ≈ 0.7812; count-twenty estimate ≈ 0.0890.
Look for these in your work
- Correct both interval ends.
- Represent the single discrete mass by a cell.
Common mistakes
- Correct before standardising.
- X ≥ k begins at k − 0.5.
- A discrete mass is a whole cell, not a continuous point.
What can you explain now?
Which event represents X ≥ 25?
Compare with the explanation
Y > 24.5.
The lower edge includes all of the count-25 cell.
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Make it stick
Draw cells, choose their outer half-unit edges, standardise with √λ and calculate a labelled approximate area.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.