Calculate normal intervals, outside regions and conditional areas
Subtract cumulative endpoints for a middle region, complement it for outside values and restrict the denominator when conditioning.
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
- Standardise both ends of an interval.
- Use complementary and conditional probability.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which expression gives P(a < Z < b) for a < b?
What you will learn
- Calculate interval and two-tail normal probabilities.
- Use a positive given-event area as a conditional denominator.
Standardise both interval endpoints
For X ~ N(50,16), the interval 46 < X < 58 becomes −1 < Z < 2. Its probability is Φ(2) − Φ(−1) = 0.9772 − 0.1587 = 0.8185 using the quoted table values. Keep the endpoints in increasing order and subtract lower cumulative area from upper.
The outside event X < 46 or X > 58 is complementary, with probability 1 − 0.8185 = 0.1815. Its two tails are disjoint, so they may also be added: 0.1587 + 0.0228. Doubling one tail is valid only for symmetric boundaries, not this asymmetric interval.
Condition on the area that remains
The symmetric interval 46 < X < 54 corresponds to −1 < Z < 1 and has probability 0.8413 − 0.1587 = 0.6826 using the table. Given X > 46, only the area 0.8413 remains available, so P(46 < X < 54 | X > 46) is the interval area divided by the given-event area.
Using unrounded numerical normal values gives approximately 0.8114 for this conditional probability; the quoted four-decimal table ratio 0.6826/0.8413 gives approximately 0.8114 as well. Do not keep the original denominator one after restricting the sample space. The interval event is contained in the given event here, so their intersection is just that interval.
Pause and explain
For X ~ N(50,16), what is P(46 < X < 58) using the stated table values?
Worked example
X ~ N(50,16). Find P(46 < X < 58) using Φ(1) = 0.8413 and Φ(2) = 0.9772.
Show the worked solution
- The endpoints standardise to −1 and 2.
- Reflect Φ(−1) = 1 − 0.8413 = 0.1587.
- Subtract: 0.9772 − 0.1587 = 0.8185.
Answer 0.8185 using the quoted table values.
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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 ~ N(50,16). Find P(46 < X < 54) using Φ(1) = 0.8413.
Give me a hint
The bounds lie one SD either side of the mean.
Compare my reasoning
- The standardised bounds are −1 and 1.
- Φ(−1) = 0.1587 by symmetry.
- Subtract 0.8413 − 0.1587 = 0.6826.
0.6826 using the quoted table value.
Look for these in your work
- Standardise both endpoints.
- Subtract the complete lower cumulative area.
X ~ N(50,16). Find P(X < 46 or X > 58) using Φ(1) = 0.8413 and Φ(2) = 0.9772.
Give me a hint
Use the complement of the interval or add its two disjoint tails.
Compare my reasoning
- The lower tail is Φ(−1) = 0.1587.
- The upper tail is 1 − Φ(2) = 0.0228.
- Their sum is 0.1815, also one minus the interval area 0.8185.
0.1815 using the quoted table values.
Look for these in your work
- Retain the asymmetric endpoints.
- Combine disjoint tails or the exact complementary region.
X ~ N(50,16). Given X > 46, find P(46 < X < 54 | X > 46).
Give me a hint
The numerator is the middle area and the denominator is the given upper region.
Compare my reasoning
- The interval probability is 0.6826 using the table.
- The given probability is P(X > 46) = 0.8413.
- Divide 0.6826/0.8413 ≈ 0.8114; the denominator is positive.
Approximately 0.8114.
Look for these in your work
- Identify the contained intersection event.
- Use the restricted positive denominator.
Common mistakes
- Standardise both boundaries using the same μ and σ.
- Do not double a tail for asymmetric boundaries.
- Conditioning changes the denominator to the given-event probability.
What can you explain now?
Can P(X < 46 or X > 58) be twice P(X > 58) for N(50,16)?
Compare with the explanation
No; the boundaries are not equally far from the mean.
Two-tail doubling requires symmetry of the two selected tails.
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Make it stick
Sketch the required region, standardise both boundaries and subtract, complement or condition according to the actual event.
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.