Correct binomial intervals, point events and probability thresholds
Include both endpoint cells, approximate a single count by a finite-width interval and check neighbouring integers for a probability target.
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
- Use binomial normal suitability and continuity correction.
- Calculate normal interval probabilities and percentiles.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which corrected interval represents 35 ≤ X ≤ 45?
What you will learn
- Correct interval and exact-count binomial events.
- Verify an integer threshold under the approximation and distinguish it from an exact guarantee.
Two endpoints and one-count cells
For X ~ B(100,0.40), use Y ~ N(40,24). The event 35 ≤ X ≤ 45 includes all endpoint cells, giving 34.5 < Y < 45.5. Standardised limits are approximately −1.1227 and 1.1227; the normal interval probability is approximately 0.7384.
The exact-count event X = 40 is approximated by 39.5 < Y < 40.5, not by P(Y = 40). Its standardised limits are ±0.1021 and the interval probability is approximately 0.0813. A continuous single point has probability zero, but the discrete count has a whole cell of positive area in the approximation.
Find the first valid integer under the model
To find the smallest integer k with approximate P(X ≤ k) ≥ 0.90, require Φ((k + 0.5 − 40)/√24) ≥ 0.90. Using an inverse normal calculation gives a real threshold near k = 45.78, so test k = 46. The quoted critical value 1.282 gives the same candidate.
Check both neighbours with unrounded numerical probabilities: at k = 45 the corrected cumulative probability is approximately 0.8692, below the target; at k = 46 it is approximately 0.9077, above. Thus 46 is the first integer under the normal approximation. An exact binomial target requires checking exact binomial probabilities; this approximate decision alone does not prove an exact guarantee, particularly near a target.
Pause and explain
Which event approximates X = 40 using continuity correction?
Worked example
X ~ B(100,0.40). Approximate P(X = 40).
Show the worked solution
- Check the criterion and use Y ~ N(40,24).
- The count-40 cell is 39.5 < Y < 40.5, with z limits ±0.5/√24.
- Subtract the two cumulative areas to obtain approximately 0.0813.
Answer Approximately 0.0813.
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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.
X ~ B(100,0.40). Approximate P(35 ≤ X ≤ 45).
Give me a hint
Move the lower boundary down half a unit and the upper boundary up half a unit.
Compare my reasoning
- Use matching Y ~ N(40,24), with both expected counts above five.
- Correct to 34.5 < Y < 45.5, giving z limits ±5.5/√24.
- The interval probability is approximately 0.7384.
Approximately 0.7384; corrected interval 34.5 to 45.5.
Look for these in your work
- Include both endpoint cells.
- Show both standardised limits and the approximation label.
Each of 80 independent trials has success probability 0.30. Let X count the successes. Use a normal approximation to find the probability of exactly 27 successes.
Give me a hint
Represent the whole count-27 cell before finding the normal area.
Compare my reasoning
- Model X ~ B(80,0.30). Both np = 24 and n(1 − p) = 56 exceed five, so use Y ~ N(24,16.8), with SD √16.8.
- The event X = 27 corresponds to 26.5 < Y < 27.5. Standardise both edges: (26.5 − 24)/√16.8 ≈ 0.6099 and (27.5 − 24)/√16.8 ≈ 0.8539.
- Using unrounded standardised values, subtract the cumulative areas: Φ(0.8539126…) − Φ(0.6099375…) ≈ 0.0744. This is an approximate binomial probability, not a continuous point probability.
Approximately 0.0744; use 26.5 < Y < 27.5 with Y ~ N(24,16.8).
Look for these in your work
- Identify the binomial model and check both expected counts.
- Include the whole count-27 cell, which is above the mean of 24.
- Standardise both boundaries using the standard deviation, not the variance.
- Subtract cumulative areas and label the result approximate.
For X ~ B(100,0.40), find the minimum integer k giving approximate P(X ≤ k) ≥ 0.90. State the limitation for an exact target.
Give me a hint
Use k + 0.5 in the corrected boundary and check both neighbouring integers.
Compare my reasoning
- Require (k + 0.5 − 40)/√24 at least the ninetieth-percentile z value, giving k near 45.78.
- At k = 45 the corrected normal probability is approximately 0.8692; at k = 46 it is approximately 0.9077.
- Choose k = 46 under the approximation. An exact target still requires exact binomial verification.
Minimum k = 46 under the approximation; an exact target needs exact binomial verification.
Look for these in your work
- Correct the integer threshold before inverting.
- Check both neighbours and preserve the approximation limitation.
Common mistakes
- Correct both ends of an inclusive interval outward.
- Do not replace an exact discrete count by a continuous point probability.
- A threshold found by approximation is not an exact probability guarantee.
What can you explain now?
Why is P(Y = 40) not a valid approximation to P(X = 40)?
Compare with the explanation
The discrete count is represented by a unit-width cell, not a single continuous point.
Using zero-width area would erase the positive mass of that count.
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Make it stick
Map integer cells to continuous intervals, verify neighbouring threshold integers, and keep an approximate result separate from an exact binomial claim.
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.