Choose normal tails and check the boundary sign
Use a sketch and a half-area check to avoid reversing lower and upper probabilities, especially below the mean.
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 an original measurement.
- Reflect negative normal values.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Where is the boundary X = 65 for N(70,25)?
What you will learn
- Calculate left and right normal tails above or below the mean.
- Use symmetry to check probability direction and magnitude.
A right tail can be larger than one half
For X ~ N(70,25), standard deviation is 5. At 65 the z value is −1, so P(X > 65) = 1 − Φ(−1) = Φ(1) = 0.8413. The boundary is below the mean, leaving most of the curve to its right. ‘Greater than’ does not automatically mean a small tail.
The complementary P(X ≤ 65) is 0.1587. Sketch the boundary to the left of μ and shade the requested side before selecting a table expression. For a genuinely continuous normal variable, strict and inclusive comparisons at the same threshold give the same probability because one exact point has zero probability.
Use a sign and symmetry check
For Y ~ N(80,100), σ = 10 and P(Y > 95) = P(Z > 1.5) = 0.0668. This boundary is above the mean, so a right tail below one half is sensible. The lower-tail probability at 95 is 0.9332; the same z value does not make the two requested events identical.
For W ~ N(40,9), thresholds 37 and 43 are one standard deviation either side of μ. P(W > 37) = P(W < 43) = 0.8413 by symmetry. These two events overlap, so equal probabilities do not make them complements. The complementary pairs use the same boundary with opposite sides.
Pause and explain
For N(70,25), should P(X > 65) exceed one half?
Worked example
X ~ N(70,25). Calculate P(X > 65).
Show the worked solution
- Standardise the boundary: (65 − 70)/5 = −1.
- The right tail is 1 − Φ(−1), which equals Φ(1).
- Using Φ(1) = 0.8413 gives a probability greater than one half.
Answer 0.8413.
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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(70,25). Find P(X > 65).
Give me a hint
Standardise to a negative boundary, then shade its right side.
Compare my reasoning
- z = (65 − 70)/5 = −1.
- P(X > 65) = 1 − Φ(−1) = Φ(1).
- The table gives 0.8413, agreeing with the more-than-half sketch.
0.8413.
Look for these in your work
- Keep the right-tail event after standardising.
- Check its size against one half.
Y ~ N(80,100). Calculate P(Y > 95).
Give me a hint
Find σ and use the complement of the positive cumulative boundary.
Compare my reasoning
- σ = 10 and z = (95 − 80)/10 = 1.5.
- P(Y > 95) = 1 − Φ(1.5).
- Use 1 − 0.9332 = 0.0668.
0.0668.
Look for these in your work
- Show the complete standardisation.
- Check the above-mean right tail is small.
W ~ N(40,9). Compare P(W > 37) with P(W < 43). Are the events complementary?
Give me a hint
The boundaries are symmetric about the mean, but the events overlap.
Compare my reasoning
- The z boundaries are −1 and 1 respectively.
- Both probabilities equal Φ(1) = 0.8413.
- They overlap on 37 < W < 43, so they are not complementary; their probabilities need not sum to one.
Both 0.8413; not complementary because they overlap.
Look for these in your work
- Use symmetry without changing the event definitions.
- Distinguish equal probabilities from complementary regions.
Common mistakes
- Do not assume every right tail is small.
- A negative z locates the boundary, not a negative probability.
- Equal symmetric probabilities need not be complementary events.
What can you explain now?
Should P(X > a) be above one half when a < μ in a normal model?
Compare with the explanation
Yes; the right side contains the mean and more than half the area.
A boundary-side check catches many complement and sign errors.
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Make it stick
Sketch the mean, boundary and requested side; a sign check and half-area check should agree with the table calculation.
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.