Check exact Poisson probabilities before a threshold decision
Verify close integer decisions using exact cumulative sums and compare them with corrected normal estimates.
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
- Sum count masses through an integer.
- Apply normal continuity correction.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which is exact Po(20) P(X ≤ 25)?
What you will learn
- Compare exact Poisson sums with normal estimates.
- Check neighbouring integers before claiming an exact target.
An estimate differs from the exact model expression
For Po(20), exact P(X ≤ 25) is e⁻²⁰∑ᵣ₌₀²⁵ 20ʳ/r! ≈ 0.887815. Its corrected normal estimate is approximately 0.890621, higher by about 0.002806. Correction does not remove every approximation error.
A useful recurrence starts at p₀ = e⁻²⁰ and uses pᵣ₊₁ = pᵣ × 20/(r + 1). Sum through the requested integer. Exact means the Poisson model expression; displayed decimals are rounded evaluations, not guarantees that the real process follows this model.
Check the original decision
For exact P(X ≤ k) ≥ 0.90, CDF at 25 is approximately 0.887815 and at 26 is 0.922113. Thus the minimum integer is 26; increasing cumulative probability makes the preceding failed candidate prove minimality.
For target 0.89, normal CDF at 25 is approximately 0.890621 and passes, but exact CDF 0.887815 fails. At 24 both are below target; the approximate minimum is 25 and exact minimum 26. These are probability targets, not hypothesis-test critical regions; section 6.5 covers testing.
Pause and explain
Exact CDF(25) = 0.887815 and normal estimate = 0.890621. Does 25 meet an exact target 0.89?
Worked example
Find the minimum integer k with exact Po(20) P(X ≤ k) ≥ 0.90.
Show the worked solution
- Exact CDF(25) ≈ 0.887815 < 0.90.
- Exact CDF(26) ≈ 0.922113 ≥ 0.90.
- Monotonicity proves minimum integer k = 26.
Answer k = 26 for the exact model target.
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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.
Compare exact Po(20) P(X ≤ 25) and its corrected normal estimate.
Give me a hint
Calculate the exact sum separately from the matched normal area.
Compare my reasoning
- Exact sum is ≈ 0.887815.
- Normal Φ(5.5/√20) is ≈ 0.890621.
- The estimate is higher by ≈ 0.002806.
Exact ≈ 0.887815; normal ≈ 0.890621; difference ≈ 0.002806.
Look for these in your work
- Use separate formulae.
- Label rounded numerical values.
Find minimum k for exact Po(20) P(X ≤ k) ≥ 0.90 given CDF(25) ≈ 0.887815 and CDF(26) ≈ 0.922113.
Give me a hint
Compare adjacent values with the target.
Compare my reasoning
- Twenty-five fails.
- Twenty-six passes.
- Increasing cumulative probability rules out every smaller integer.
Minimum k = 26.
Look for these in your work
- Check both sides of the target.
- Explain minimality.
For target 0.89, compare exact and normal minimum integers. At 24 CDFs are 0.843227 and 0.842848; at 25 they are 0.887815 and 0.890621; exact CDF(26) is 0.922113.
Give me a hint
Keep the exact and approximate sequences separate.
Compare my reasoning
- Both values at twenty-four fail.
- Normal estimate at twenty-five passes, giving approximate minimum 25; exact value fails.
- Exact CDF at twenty-six passes, giving exact minimum 26.
Approximate minimum 25; exact minimum 26.
Look for these in your work
- Do not substitute an estimate for exact requirements.
- Verify neighbouring cumulative values.
Common mistakes
- Corrected normal probabilities remain estimates.
- Check the preceding candidate for minimality.
- Model probabilities are conditional on model assumptions.
What can you explain now?
Can a normal estimate pass a target that the exact count probability fails?
Compare with the explanation
Yes; at k = 25 and target 0.89 here.
Close decisions may require exact verification.
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Make it stick
Label exact and approximate probabilities, then verify close integer targets with the exact model and adjacent count.
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.