Apply continuity correction to one-sided binomial events
Translate inclusive or strict integer counts into the correct half-unit boundary before standardising the normal approximation.
Exam scope and assessment notes
Probability & Statistics 1 (Paper 5, AS & A Level) · Mathematics 9709, 2026–2027, section 5.5. N(μ,σ²) uses variance as its second parameter, with σ > 0; probabilities are areas, not density heights. Show the full standardisation Z = (X − μ)/σ. The MF19 table gives Φ(z) = P(Z ≤ z), with negative values found by symmetry. Model sketches and shaded probabilities may be required. Probability answers use quoted four-decimal table values where supplied; calculations with non-tabulated z use unrounded standardisation and numerical normal probabilities, so table interpolation can differ slightly in the last digits. Inverse examples use the quoted three-decimal MF19 critical values and are approximate. A binomial normal approximation requires both np > 5 and n(1 − p) > 5, and a continuity correction. It remains approximate, including for integer probability thresholds. A continuous normal variable has zero probability at a single exact value; a discrete binomial count does not. Sample means, the Central Limit Theorem and hypothesis testing belong to Statistics 2 and are not covered here. Model assumptions need evidence in real applications. Written practice is self-checked; human subject-teacher review remains pending.
Before you begin
- Check normal-approximation suitability.
- Standardise a continuous normal variable.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is the integer equivalent of X > 45?
What you will learn
- Translate ≤, ≥, < and > integer count events into normal boundaries.
- Show full corrected standardisation and label approximate probabilities.
The integer cell determines the boundary
For X ~ B(100,0.40), use Y ~ N(40,24) after checking np = 40 and nq = 60. The integer count 45 is represented by a cell from 44.5 to 45.5. Therefore X ≤ 45 corresponds to Y < 45.5; standardise (45.5 − 40)/√24 ≈ 1.1227, giving probability approximately 0.8692.
For X ≥ 45, retain the entire cell at 45 and all those above it: use Y > 44.5. Standardisation gives (44.5 − 40)/√24 ≈ 0.9186 and a right-tail probability approximately 0.1792. The sign of the half-unit shift follows the included cells, not a rule to always add or always subtract 0.5.
Translate strict integer events first
X > 45 means X ≥ 46, so its lower boundary is 45.5 and its probability is approximately 0.1308. X < 45 means X ≤ 44, so its upper boundary is 44.5 and its probability is approximately 0.8208. The distinction between > and ≥ matters for X because it is discrete, even though strict and inclusive endpoints agree for Y.
Calculate with the unrounded standardised value when possible; the shown answers are four-decimal numerical normal approximations. A table rounded too early can change the last digits. A correctly corrected probability still remains an approximation to the original binomial event, rather than an exact binomial probability.
Pause and explain
Which normal event approximates X ≥ 45 with continuity correction?
Worked example
X ~ B(100,0.40). Use a normal approximation to find P(X ≤ 45).
Show the worked solution
- Check expected counts 40 and 60, then use Y ~ N(40,24).
- Include the whole cell at 45: P(X ≤ 45) ≈ P(Y < 45.5).
- Standardise z = (45.5 − 40)/√24 ≈ 1.1227, giving Φ(z) ≈ 0.8692.
Answer Approximately 0.8692.
Try a topic test
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Test this topic →From guided practice to a new situation
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.
X ~ B(100,0.40). Approximate P(X ≤ 45) using a corrected normal event.
Give me a hint
Include the upper edge of count forty-five before standardising.
Compare my reasoning
- The criterion holds and Y ~ N(40,24).
- Use P(Y < 45.5) with z = 5.5/√24 ≈ 1.1227.
- The normal cumulative probability is approximately 0.8692.
Approximately 0.8692; corrected boundary 45.5.
Look for these in your work
- Check suitability and show the corrected event.
- Standardise with SD √24 and retain the approximation label.
X ~ B(100,0.40). Approximate P(X ≥ 45).
Give me a hint
Start at the lower edge of the included count-forty-five cell.
Compare my reasoning
- Use the suitable matching Y ~ N(40,24).
- P(X ≥ 45) ≈ P(Y > 44.5), where z = 4.5/√24 ≈ 0.9186.
- The right-tail probability is approximately 0.1792.
Approximately 0.1792; corrected boundary 44.5.
Look for these in your work
- Preserve the inclusive lower count.
- Use the complementary normal tail.
For X ~ B(100,0.40), approximate P(X > 45) and P(X < 45).
Give me a hint
Translate strict inequalities to included integer values before correcting.
Compare my reasoning
- X > 45 means X ≥ 46, giving Y > 45.5.
- X < 45 means X ≤ 44, giving Y < 44.5.
- The corresponding probabilities are approximately 0.1308 and 0.8208; the excluded count-45 cell lies between them.
Approximately 0.1308 and 0.8208.
Look for these in your work
- Translate both strict integer boundaries.
- Explain why these events exclude the central count and are not complements.
Common mistakes
- Include the full boundary cell for an inclusive integer event.
- Translate a strict count event before applying the half-unit shift.
- Divide by √(npq), not the variance npq.
What can you explain now?
Why does X ≥ 45 use the normal boundary 44.5?
Compare with the explanation
The cell representing count 45 starts at 44.5 and must be included.
A boundary at 45.5 would omit the count-45 cell and approximate X > 45 instead.
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Make it stick
Write the integer event, include its entire boundary cell, then standardise the half-unit boundary under the matching normal model.
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.