Infer normal spread and reject incompatible conditions
Solve for a positive standard deviation from a probability condition, then check whether the stated event is compatible with the known 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
- Translate normal tail probabilities into signed quantiles.
- Distinguish σ from σ².
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
If σ ≈ 4.6802, which quantity is the variance?
What you will learn
- Infer σ and variance from a known μ and a probability boundary.
- Reject conditions that would require non-positive spread or contradictory cumulative order.
Solve for the scale, then square if needed
A normal variable has mean 60 and P(X < 66) = 0.90. Using z ≈ 1.282, write (66 − 60)/σ ≈ 1.282, hence σ ≈ 6/1.282 = 4.6802. Its variance is approximately 21.9042, or 21.9 to three significant figures. The inferred SD is not the second parameter in N(μ,σ²).
A model with mean 20 and P(X > 32) = 0.10 has the same cumulative quantile. Thus σ ≈ 12/1.282 = 9.3604. A positive spread is consistent with a boundary above the mean and a small upper tail. Keep the quotient unrounded before squaring or calculating another probability.
Check whether a valid scale can exist
If μ = 60 and P(X < 54) = 0.90 is claimed, the boundary is below the mean but is assigned a cumulative probability above one half. Algebra gives σ ≈ (54 − 60)/1.282, which is negative. There is no valid positive-standard-deviation normal model satisfying that claim.
A negative inferred σ is not repaired by taking its absolute value: that changes the standardised boundary and the original event probability. Instead identify the incompatibility. Similarly, σ = 0 is outside the non-degenerate normal model in this section and cannot be used as a standardisation denominator.
Pause and explain
Mean 60 with P(X < 54) = 0.90 produces a negative σ. What should you conclude?
Worked example
A normal variable has mean 60 and P(X < 66) = 0.90. Find SD and variance using z₀.₉₀ ≈ 1.282.
Show the worked solution
- Show (66 − 60)/σ ≈ 1.282.
- Solve σ ≈ 6/1.282 = 4.6802.
- Square the unrounded quotient to obtain variance ≈ 21.9042.
Answer SD ≈ 4.68; variance ≈ 21.9.
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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.
A normal variable has mean 60 and P(X < 66) = 0.90. Find SD and variance using z ≈ 1.282.
Give me a hint
Rearrange the standardisation for σ and square its unrounded value.
Compare my reasoning
- 6/σ ≈ 1.282.
- σ ≈ 6/1.282 = 4.6802.
- Variance ≈ (6/1.282)² = 21.9042.
SD ≈ 4.68; variance ≈ 21.9.
Look for these in your work
- Show the full parameter equation.
- Check positive spread and square before final rounding.
A normal variable has mean 20 and P(X > 32) = 0.10. Find its SD using z₀.₉₀ ≈ 1.282.
Give me a hint
First convert the upper ten-percent tail to cumulative probability 0.90.
Compare my reasoning
- P(X ≤ 32) = 0.90.
- (32 − 20)/σ ≈ 1.282.
- σ ≈ 12/1.282 = 9.3604, approximately 9.36.
SD ≈ 9.36.
Look for these in your work
- Convert the tail to the cumulative convention.
- Verify the inferred SD is positive.
Can a normal variable with mean 60 and positive SD satisfy P(X < 54) = 0.90? Explain with a sketch and an equation.
Give me a hint
Compare a below-mean boundary with the positive ninetieth-percentile quantile.
Compare my reasoning
- 54 lies below the mean, so its left area must be less than one half.
- The claimed probability gives (54 − 60)/σ ≈ 1.282.
- This forces a negative σ, contradicting a valid normal model.
No valid positive SD; the below-mean boundary cannot have cumulative probability 0.90.
Look for these in your work
- State the geometric contradiction.
- Do not silently change the probability or parameter sign.
Common mistakes
- Solve for SD before writing variance.
- An admissible standard deviation must be positive.
- Do not take an absolute value to hide a contradictory probability condition.
What can you explain now?
Can a below-mean boundary have cumulative probability 0.90 in a non-degenerate normal model?
Compare with the explanation
No; below the mean the cumulative probability is below one half.
Changing spread cannot move a below-mean boundary to the positive side of Z.
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Make it stick
Infer a positive σ from the signed quantile, square it for variance, and reject conditions that contradict the mean and cumulative area.
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.