Infer a normal mean from a probability condition
Translate a known tail into a signed quantile and solve the standardisation when standard deviation is supplied.
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
- Find signed standard normal quantiles.
- Rearrange a linear standardisation equation.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
A boundary leaves 10% above it. What sign does its standard normal quantile have?
What you will learn
- Infer μ using a known σ and a stated cumulative or upper-tail probability.
- Check the inferred centre against the required side of the boundary.
Write the event before the equation
Suppose X is normal with SD 5 and P(X > 74) = 0.10. The cumulative probability at 74 is 0.90, so (74 − μ)/5 ≈ 1.282 using the quoted critical value. Therefore μ ≈ 74 − 6.41 = 67.59. The mean lies below 74, as expected for a boundary leaving only a small upper tail.
A known σ gives the scale but not the centre. Keep the signed quantile in the equation before rearranging to μ = x − zσ. Substituting the resulting μ back should recover the same standardised boundary. The variance of this model is 25, not 5.
Large upper tails have negative z
If P(X < 30) = 0.10 and σ = 4, the boundary has z ≈ −1.282. Hence (30 − μ)/4 ≈ −1.282 and μ ≈ 35.128, approximately 35.13. A lower tenth-percentile value lies below the mean; a positive z would contradict that geometry.
If P(X > 50) = 0.90 and σ = 3, the cumulative probability at 50 is only 0.10. Use z ≈ −1.282 to get μ ≈ 53.846, approximately 53.85. The greater-than wording alone does not determine the quantile’s sign: convert the probability to the cumulative convention and then locate the boundary.
Pause and explain
With σ = 5 and P(X > 74) = 0.10, which equation uses the quoted critical value?
Worked example
X has a normal distribution with SD 5 and P(X > 74) = 0.10. Find its mean using z₀.₉₀ ≈ 1.282.
Show the worked solution
- Convert the upper tail to P(X ≤ 74) = 0.90.
- Show (74 − μ)/5 ≈ 1.282.
- Rearrange μ ≈ 74 − 5(1.282) = 67.59.
Answer μ ≈ 67.59.
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.
A normal variable has SD 5 and P(X > 74) = 0.10. Find μ using z₀.₉₀ ≈ 1.282.
Give me a hint
Convert to the cumulative ninetieth percentile, then rearrange.
Compare my reasoning
- P(X ≤ 74) = 0.90.
- (74 − μ)/5 ≈ 1.282.
- μ ≈ 74 − 6.41 = 67.59, below the boundary.
μ ≈ 67.59.
Look for these in your work
- Show the signed standardisation equation.
- Check the mean’s position relative to 74.
A normal variable has SD 4 and P(X < 30) = 0.10. Find μ using the critical value 1.282.
Give me a hint
The lower tenth-percentile quantile is negative.
Compare my reasoning
- Use z ≈ −1.282.
- (30 − μ)/4 ≈ −1.282.
- μ ≈ 30 + 5.128 = 35.128, approximately 35.13.
μ ≈ 35.13.
Look for these in your work
- Use the negative lower-tail quantile.
- Retain precision while solving the equation.
A normal variable has SD 3 and P(X > 50) = 0.90. Find μ and explain its position relative to 50.
Give me a hint
Ninety percent above means ten percent below.
Compare my reasoning
- The cumulative probability at 50 is 0.10, so z ≈ −1.282.
- (50 − μ)/3 ≈ −1.282 gives μ ≈ 53.846.
- Thus μ ≈ 53.85 lies above 50, making the required upper area large.
μ ≈ 53.85; the boundary 50 is below the mean.
Look for these in your work
- Convert the large upper tail correctly.
- Check the inferred centre against the sketch.
Common mistakes
- Convert a tail into the cumulative probability before choosing z.
- Retain a negative quantile for a boundary below the mean.
- Substitute back and check the location of the inferred centre.
What can you explain now?
Can a 90% upper-tail boundary lie below the mean?
Compare with the explanation
Yes; its cumulative probability is only 10%.
The upper region includes most of the symmetric curve.
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Make it stick
Convert the probability to a signed cumulative quantile and solve x − μ = zσ, checking the boundary’s side of the mean.
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.