Did you know? Knowing two percentiles of a normal distribution can reveal both its unknown mean and its unknown standard deviation. Suppose the 10th percentile is 40 and the 90th percentile is 60. These cutoffs are equally far below and above the mean, so μ=50. The standard normal 90th percentile is approximately 1.28155, giving σ≈10/1.28155≈7.803. A single percentile would usually not identify both unknowns, but two independent percentile conditions can.
This JC2 H2 Mathematics Tutor Bukit Timah guide explains how to recover μ and σ from two known normal-distribution probabilities and their cutoff values. It is deliberately distinct from the earlier guide about finding a top-10% cutoff when μ and σ are already known. Here the direction is reversed: two cutoffs and their percentiles are known, while the population parameters must be determined. A complete worked example is followed by symmetric and asymmetric cases, modelling checks and 36 original numerical clinics.
The SEAB H2 Mathematics 9758 syllabus for examination in 2027 includes normal distributions and finding relationships between a cutoff, μ and σ from a given probability. This article uses two such relationships simultaneously. It does not invoke the excluded normal approximation to a binomial distribution. All measurements and percentiles below are hypothetical mathematical examples, not actual pupils’ marks or published local datasets.
Quick answer: two z-score equations determine μ and σ
Let X~N(μ,σ²), with σ>0. Suppose you know that P(X≤x₁)=p₁ and P(X≤x₂)=p₂, where 0<p₁<p₂<1. Obtain the standard normal percentile z₁ and z₂ satisfying Φ(z₁)=p₁ and Φ(z₂)=p₂. Then (x₁−μ)/σ=z₁ and (x₂−μ)/σ=z₂. Subtracting gives σ=(x₂−x₁)/(z₂−z₁), and then μ=x₁−σz₁. These two equations are the entire method.
- Read probability direction: top 10% means the 90th percentile, not the 10th percentile.
- Convert to standard normal: look up the correct z-values for both cumulative probabilities.
- Write both equations: x₁=μ+σz₁ and x₂=μ+σz₂.
- Subtract: eliminate μ and solve for σ using the difference in cutoffs.
- Back-substitute: calculate μ from either equation.
- Check σ>0: a larger cumulative percentile must have a larger cutoff for a nondegenerate normal distribution.
- Verify both cutoffs: (x₁−μ)/σ and (x₂−μ)/σ should reproduce z₁ and z₂ within rounding.
Worked example one: symmetric 10th and 90th percentiles
Suppose X is normal, with the 10th percentile at x₁=40 and the 90th at x₂=60. The standard normal quantiles are approximately z₁=−1.2815516 and z₂=+1.2815516. Therefore the cutoff equations are 40=μ−1.2815516σ and 60=μ+1.2815516σ. Adding them gives 100=2μ, so μ=50.
Subtracting gives 20=2(1.2815516)σ, hence σ=20/2.5631032≈7.80304. The corresponding variance is σ²≈60.88746, because the second parameter in N(μ,σ²) is variance, not standard deviation. The cutoffs 40 and 60 are equally far from 50 only because the selected cumulative probabilities are symmetric about 50%.
Worked example two: symmetric 25th and 75th percentiles
A fictional normal variable has lower quartile 72 and upper quartile 88. The 25th and 75th standard normal quantiles are approximately −0.67448975 and +0.67448975. The mean is their midpoint, μ=(72+88)/2=80. The standard deviation is σ=(88−72)/(2×0.67448975)≈11.86082.
The width between the upper and lower quartile cutoffs is the interquartile range, IQR=16. In a normal distribution, IQR≈1.34898σ. This relationship is useful as a check, but it assumes a normal model. In a general skewed dataset, the mean need not equal the midpoint of its 25th and 75th percentiles.
Worked example three: asymmetric percentiles need algebra, not a midpoint
Now suppose the 10th percentile is 30 and the 75th percentile is 45. The z-values are z₁≈−1.2815516 and z₂≈0.67448975. Subtracting the two cutoff equations gives σ=15/(0.67448975−(−1.2815516))≈7.66855.
Substitute into the first equation: μ=30+1.2815516σ≈39.82764. This is not the midpoint (30+45)/2=37.5, because 10% and 75% are not symmetric positions under the normal curve. The correct mean is closer to the 75th cutoff than a naive midpoint calculation would predict.
Why the 50th percentile equals the normal mean
A normal curve is symmetric about μ, so the probability of falling below μ is 0.5. Consequently the 50th percentile, median and mean all coincide for an exact normal model. If one of the given conditions says P(X≤50)=0.5, the mean is immediately μ=50. A second distinct percentile can then determine σ. This is a useful shortcut but does not imply the same equality for every non-normal distribution.
For example, if P(X≤50)=0.5 and P(X≤62)=0.9 under a normal model, then σ=(62−50)/1.2815516≈9.3637. The mean required no algebra because the median was directly supplied. But the standard deviation still required the normal quantile corresponding to the second cutoff.
A top-tail phrase must be converted to a left cumulative probability
If a question says “the highest 10% exceed 82”, then the cutoff 82 is the 90th percentile: P(X≤82)=0.90 in the continuous normal model. If it says “the lowest 10% are below 82”, then 82 is the 10th percentile. These statements use opposite tails and therefore opposite signs of z for the same 10% fraction.
Changing the tail can change both inferred parameters. Before pressing inverse normal on a calculator, draw a quick bell curve, mark which side contains the specified 10%, and write the cumulative probability to the left of the threshold. The previous Top 10% and Inverse Normal Cutoffs guide explains finding the cutoff given μ and σ; here we infer both parameters from two labelled cutoffs.
Why one percentile is not enough for two unknowns
Suppose all you know is that the 90th percentile is 60. The equation is 60=μ+1.2815516σ. There are infinitely many positive σ values and corresponding μ values that satisfy it. For σ=5, the mean is about 53.592; for σ=10, the mean is about 47.184. Both normal models have the same 90th percentile 60. A second independent condition is needed to determine both unknown parameters.
Likewise, two statements referring to the same percentile do not add independent information. If both describe the 90th percentile with the same cutoff, they repeat one equation. If they give contradictory cutoffs for exactly the same percentile, no normal distribution can satisfy both. The issue is mathematical information, not calculator precision.
An impossible pair: percentile order and cutoff order disagree
A normal distribution with positive σ has a strictly increasing cumulative distribution function. Therefore if p₁<p₂, the corresponding cutoffs must satisfy x₁<x₂. Suppose a question claims P(X≤50)=0.20 and P(X≤40)=0.80. The 80th percentile is numerically below the 20th percentile, which is impossible for a proper normal model. The formula for σ would produce a negative result, warning that the data are inconsistent.
Do not take the absolute value of that negative σ and continue. A standard deviation cannot be negative, and changing the sign would hide the contradiction rather than resolve it. The correct answer is that the claimed exact percentiles cannot come from one normal distribution with σ>0. A question involving estimated quantiles from a small sample would require a separate uncertainty discussion, but exact theoretical percentile statements do not.
Mean and variance are separate parameters
The notation N(μ,σ²) uses σ² as its second parameter. If your calculation gives σ≈8, the distribution is N(μ,64), not N(μ,8), under the usual variance-parameter convention. A student can solve the two percentile equations correctly but lose marks by writing the standard deviation where the variance belongs in the final normal-model notation.
A clear final sentence can report both: “The mean is μ≈50, the standard deviation is σ≈7.803, so the variance is σ²≈60.89.” State whichever parameters the question asks for. The fact that a standard normal calculator input requests a standard deviation while the distribution notation prints a variance is an especially important interface difference.
How to check using the original probability statements
Once μ and σ have been recovered, restandardise the two original cutoffs. For the symmetric example with μ=50 and σ≈7.803, x₁=40 becomes (40−50)/7.803≈−1.28155, and x₂=60 becomes +1.28155. The corresponding normal cumulative probabilities are approximately 0.10 and 0.90. These checks directly test the information supplied in the question.
A calculator can evaluate those cumulative probabilities as an audit, but the solution should still show how two unknown parameters were determined. If a reported answer fails one of the original percentile conditions beyond expected rounding, a sign or probability-direction error is likely. Checking only one threshold is insufficient when the question gave two.
When input cutoffs are rounded, the recovered parameters are approximate
If the problem prints measured cutoffs rounded to two or three decimal places, the recovered μ and σ are also estimates based on the rounded data. The underlying normal model may have more precise quantiles. It is appropriate to use approximate equality and to avoid reporting many artificial decimal places that were not supported by the supplied cutoffs.
The 36 practice clinics below print cutoffs to three decimal places. Their stated μ and σ are constructed from exact model parameters before rounding, so reconstructing them from the printed cutoffs can differ by a few ten-thousandths. The worked solutions are explicitly approximate. A small mismatch caused by documented rounding is not evidence that the normal model is impossible.
A two-line solver for any pair of normal quantiles
Write x₂−x₁=σ(z₂−z₁). Compute σ first, then μ=x₁−σz₁. Because z₂>z₁ when p₂>p₁, the denominator is positive; a valid model with σ>0 therefore requires x₂>x₁. This is the first diagnostic before calculating any long decimal.
If the probabilities are symmetric, z₂=−z₁ and the formula simplifies to μ=(x₁+x₂)/2. If one is the median, z=0 and μ equals that cutoff. If neither is true, keep the general two-equation method. These are shortcuts derived from the same equations, not three unrelated formulas that must be memorised separately.
One-minute Bukit Timah diagnostic
Give the learner the 10th and 90th percentile cutoffs 40 and 60. They should identify the symmetric z-values, find μ=50, and calculate σ≈7.803. Change the 90th percentile to the 75th while keeping its cutoff 60: now the mean is no longer the midpoint, and the two z equations must be solved. Finally reverse the two cutoffs while keeping the probabilities in increasing order: the negative inferred σ signals incompatible information.
Thirty-six original two-percentile clinics
These 36 original problems are generated from hypothetical normal distributions with integer means and positive integer standard deviations. Each lists two cumulative probabilities and corresponding cutoffs rounded to three decimal places. The solution reconstructs μ and σ using the two z-score equations, then checks the recovered values against the generating model. The values are invented mathematical inputs, not official exam results or actual assessment distributions.
1. 10th and 90th percentile cutoffs
Original model question. A fictional continuous normal variable X has cumulative probability P(X≤28.592)=0.1 and P(X≤41.408)=0.9. The cutoffs have been printed to three decimal places. Determine the unknown mean μ and standard deviation σ. Before entering numbers into a calculator, state why the second cutoff must exceed the first and identify the two standard normal quantiles.
Complete two-equation calculation. The corresponding z values are z₁≈-1.2815516 and z₂≈1.2815516. Thus 28.592≈μ+(-1.2815516)σ and 41.408≈μ+(1.2815516)σ. Subtracting eliminates μ: σ≈(12.816)/(2.5631031)≈5.00019. Back-substitution gives μ≈35. The underlying constructed model used μ=35, σ=5; the small difference results from printing the percentiles to three decimals. Its variance is approximately σ²≈25.00189.
Interpretation and independent check. This pair is symmetric around the median, so the midpoint (28.592+41.408)/2≈35 independently checks the mean. When the percentiles are symmetric around 50%, the mean is their midpoint. Recalculate (x₁−μ)/σ and (x₂−μ)/σ with the recovered values to recover the two original standardised quantiles. If an examination question says “above” rather than “below”, convert the relevant tail probability before this calculation.
2. 25th and 75th percentile cutoffs
Original model question. A fictional continuous normal variable X has cumulative probability P(X≤35.255)=0.25 and P(X≤48.745)=0.75. The cutoffs have been printed to three decimal places. Determine the unknown mean μ and standard deviation σ. Before entering numbers into a calculator, state why the second cutoff must exceed the first and identify the two standard normal quantiles.
Complete two-equation calculation. The corresponding z values are z₁≈-0.6744898 and z₂≈0.6744898. Thus 35.255≈μ+(-0.6744898)σ and 48.745≈μ+(0.6744898)σ. Subtracting eliminates μ: σ≈(13.49)/(1.3489795)≈10.00015. Back-substitution gives μ≈42. The underlying constructed model used μ=42, σ=10; the small difference results from printing the percentiles to three decimals. Its variance is approximately σ²≈100.00304.
Interpretation and independent check. This pair is symmetric around the median, so the midpoint (35.255+48.745)/2≈42 independently checks the mean. For non-symmetric percentiles, do not use a midpoint rule; solve both linear z-score equations. Recalculate (x₁−μ)/σ and (x₂−μ)/σ with the recovered values to recover the two original standardised quantiles. If an examination question says “above” rather than “below”, convert the relevant tail probability before this calculation.
3. 20th and 80th percentile cutoffs
Original model question. A fictional continuous normal variable X has cumulative probability P(X≤47.317)=0.2 and P(X≤50.683)=0.8. The cutoffs have been printed to three decimal places. Determine the unknown mean μ and standard deviation σ. Before entering numbers into a calculator, state why the second cutoff must exceed the first and identify the two standard normal quantiles.
Complete two-equation calculation. The corresponding z values are z₁≈-0.8416212 and z₂≈0.8416212. Thus 47.317≈μ+(-0.8416212)σ and 50.683≈μ+(0.8416212)σ. Subtracting eliminates μ: σ≈(3.366)/(1.6832425)≈1.99971. Back-substitution gives μ≈49. The underlying constructed model used μ=49, σ=2; the small difference results from printing the percentiles to three decimals. Its variance is approximately σ²≈3.99885.
Interpretation and independent check. This pair is symmetric around the median, so the midpoint (47.317+50.683)/2≈49 independently checks the mean. Check that the higher cumulative probability corresponds to the larger cutoff. Recalculate (x₁−μ)/σ and (x₂−μ)/σ with the recovered values to recover the two original standardised quantiles. If an examination question says “above” rather than “below”, convert the relevant tail probability before this calculation.
4. 5th and 95th percentile cutoffs
Original model question. A fictional continuous normal variable X has cumulative probability P(X≤44.486)=0.05 and P(X≤67.514)=0.95. The cutoffs have been printed to three decimal places. Determine the unknown mean μ and standard deviation σ. Before entering numbers into a calculator, state why the second cutoff must exceed the first and identify the two standard normal quantiles.
Complete two-equation calculation. The corresponding z values are z₁≈-1.6448536 and z₂≈1.6448536. Thus 44.486≈μ+(-1.6448536)σ and 67.514≈μ+(1.6448536)σ. Subtracting eliminates μ: σ≈(23.028)/(3.2897073)≈7.00001. Back-substitution gives μ≈56. The underlying constructed model used μ=56, σ=7; the small difference results from printing the percentiles to three decimals. Its variance is approximately σ²≈49.00021.
Interpretation and independent check. This pair is symmetric around the median, so the midpoint (44.486+67.514)/2≈56 independently checks the mean. A larger spread σ separates any pair of fixed normal percentiles further apart. Recalculate (x₁−μ)/σ and (x₂−μ)/σ with the recovered values to recover the two original standardised quantiles. If an examination question says “above” rather than “below”, convert the relevant tail probability before this calculation.
5. 10th and 75th percentile cutoffs
Original model question. A fictional continuous normal variable X has cumulative probability P(X≤47.621)=0.1 and P(X≤71.094)=0.75. The cutoffs have been printed to three decimal places. Determine the unknown mean μ and standard deviation σ. Before entering numbers into a calculator, state why the second cutoff must exceed the first and identify the two standard normal quantiles.
Complete two-equation calculation. The corresponding z values are z₁≈-1.2815516 and z₂≈0.6744898. Thus 47.621≈μ+(-1.2815516)σ and 71.094≈μ+(0.6744898)σ. Subtracting eliminates μ: σ≈(23.473)/(1.9560413)≈12.00026. Back-substitution gives μ≈62.99995. The underlying constructed model used μ=63, σ=12; the small difference results from printing the percentiles to three decimals. Its variance is approximately σ²≈144.00619.
Interpretation and independent check. The percentile positions are not symmetric, so the arithmetic midpoint of the two cutoffs is generally not μ; using both z-scores is essential. The second parameter of N(μ,σ²) is variance, not standard deviation. Recalculate (x₁−μ)/σ and (x₂−μ)/σ with the recovered values to recover the two original standardised quantiles. If an examination question says “above” rather than “below”, convert the relevant tail probability before this calculation.
6. 25th and 95th percentile cutoffs
Original model question. A fictional continuous normal variable X has cumulative probability P(X≤67.302)=0.25 and P(X≤76.579)=0.95. The cutoffs have been printed to three decimal places. Determine the unknown mean μ and standard deviation σ. Before entering numbers into a calculator, state why the second cutoff must exceed the first and identify the two standard normal quantiles.
Complete two-equation calculation. The corresponding z values are z₁≈-0.6744898 and z₂≈1.6448536. Thus 67.302≈μ+(-0.6744898)σ and 76.579≈μ+(1.6448536)σ. Subtracting eliminates μ: σ≈(9.277)/(2.3193434)≈3.99984. Back-substitution gives μ≈69.99985. The underlying constructed model used μ=70, σ=4; the small difference results from printing the percentiles to three decimals. Its variance is approximately σ²≈15.99871.
Interpretation and independent check. The percentile positions are not symmetric, so the arithmetic midpoint of the two cutoffs is generally not μ; using both z-scores is essential. Use approximate equality because the supplied cutoffs were rounded to three decimal places. Recalculate (x₁−μ)/σ and (x₂−μ)/σ with the recovered values to recover the two original standardised quantiles. If an examination question says “above” rather than “below”, convert the relevant tail probability before this calculation.
7. 2.5th and 97.5th percentile cutoffs
Original model question. A fictional continuous normal variable X has cumulative probability P(X≤59.36)=0.025 and P(X≤94.64)=0.975. The cutoffs have been printed to three decimal places. Determine the unknown mean μ and standard deviation σ. Before entering numbers into a calculator, state why the second cutoff must exceed the first and identify the two standard normal quantiles.
Complete two-equation calculation. The corresponding z values are z₁≈-1.959964 and z₂≈1.959964. Thus 59.36≈μ+(-1.959964)σ and 94.64≈μ+(1.959964)σ. Subtracting eliminates μ: σ≈(35.28)/(3.919928)≈9.00017. Back-substitution gives μ≈77. The underlying constructed model used μ=77, σ=9; the small difference results from printing the percentiles to three decimals. Its variance is approximately σ²≈81.00298.
Interpretation and independent check. This pair is symmetric around the median, so the midpoint (59.36+94.64)/2≈77 independently checks the mean. A single percentile cannot normally determine both μ and σ without another independent fact. Recalculate (x₁−μ)/σ and (x₂−μ)/σ with the recovered values to recover the two original standardised quantiles. If an examination question says “above” rather than “below”, convert the relevant tail probability before this calculation.
8. 15.87th (approx.) and 84.13th (approx.) percentile cutoffs
Original model question. A fictional continuous normal variable X has cumulative probability P(X≤70)=0.158655254 and P(X≤98)=0.841344746. The cutoffs have been printed to three decimal places. Determine the unknown mean μ and standard deviation σ. Before entering numbers into a calculator, state why the second cutoff must exceed the first and identify the two standard normal quantiles.
Complete two-equation calculation. The corresponding z values are z₁≈-1 and z₂≈1. Thus 70≈μ+(-1)σ and 98≈μ+(1)σ. Subtracting eliminates μ: σ≈(28)/(2)≈14. Back-substitution gives μ≈84. The underlying constructed model used μ=84, σ=14; the small difference results from printing the percentiles to three decimals. Its variance is approximately σ²≈196.
Interpretation and independent check. This pair is symmetric around the median, so the midpoint (70+98)/2≈84 independently checks the mean. The standardised distances from μ should reproduce both original z-quantile values. Recalculate (x₁−μ)/σ and (x₂−μ)/σ with the recovered values to recover the two original standardised quantiles. If an examination question says “above” rather than “below”, convert the relevant tail probability before this calculation.
9. 50th and 90th percentile cutoffs
Original model question. A fictional continuous normal variable X has cumulative probability P(X≤30)=0.5 and P(X≤37.689)=0.9. The cutoffs have been printed to three decimal places. Determine the unknown mean μ and standard deviation σ. Before entering numbers into a calculator, state why the second cutoff must exceed the first and identify the two standard normal quantiles.
Complete two-equation calculation. The corresponding z values are z₁≈0 and z₂≈1.2815516. Thus 30≈μ+(0)σ and 37.689≈μ+(1.2815516)σ. Subtracting eliminates μ: σ≈(7.689)/(1.2815516)≈5.99976. Back-substitution gives μ≈30. The underlying constructed model used μ=30, σ=6; the small difference results from printing the percentiles to three decimals. Its variance is approximately σ²≈35.9971.
Interpretation and independent check. The percentile positions are not symmetric, so the arithmetic midpoint of the two cutoffs is generally not μ; using both z-scores is essential. Top 10% above a cutoff corresponds to the 90th percentile, not the 10th. Recalculate (x₁−μ)/σ and (x₂−μ)/σ with the recovered values to recover the two original standardised quantiles. If an examination question says “above” rather than “below”, convert the relevant tail probability before this calculation.
10. 25th and 90th percentile cutoffs
Original model question. A fictional continuous normal variable X has cumulative probability P(X≤29.581)=0.25 and P(X≤51.097)=0.9. The cutoffs have been printed to three decimal places. Determine the unknown mean μ and standard deviation σ. Before entering numbers into a calculator, state why the second cutoff must exceed the first and identify the two standard normal quantiles.
Complete two-equation calculation. The corresponding z values are z₁≈-0.6744898 and z₂≈1.2815516. Thus 29.581≈μ+(-0.6744898)σ and 51.097≈μ+(1.2815516)σ. Subtracting eliminates μ: σ≈(21.516)/(1.9560413)≈10.99977. Back-substitution gives μ≈37.00023. The underlying constructed model used μ=37, σ=11; the small difference results from printing the percentiles to three decimals. Its variance is approximately σ²≈120.99489.
Interpretation and independent check. The percentile positions are not symmetric, so the arithmetic midpoint of the two cutoffs is generally not μ; using both z-scores is essential. The mean of a normal distribution is its median, but this is not generally true of a skew distribution. Recalculate (x₁−μ)/σ and (x₂−μ)/σ with the recovered values to recover the two original standardised quantiles. If an examination question says “above” rather than “below”, convert the relevant tail probability before this calculation.
11. 10th and 80th percentile cutoffs
Original model question. A fictional continuous normal variable X has cumulative probability P(X≤40.155)=0.1 and P(X≤46.525)=0.8. The cutoffs have been printed to three decimal places. Determine the unknown mean μ and standard deviation σ. Before entering numbers into a calculator, state why the second cutoff must exceed the first and identify the two standard normal quantiles.
Complete two-equation calculation. The corresponding z values are z₁≈-1.2815516 and z₂≈0.8416212. Thus 40.155≈μ+(-1.2815516)σ and 46.525≈μ+(0.8416212)σ. Subtracting eliminates μ: σ≈(6.37)/(2.1231728)≈3.00023. Back-substitution gives μ≈43.99995. The underlying constructed model used μ=44, σ=3; the small difference results from printing the percentiles to three decimals. Its variance is approximately σ²≈9.00136.
Interpretation and independent check. The percentile positions are not symmetric, so the arithmetic midpoint of the two cutoffs is generally not μ; using both z-scores is essential. A negative recovered standard deviation signals an incompatible percentile order. Recalculate (x₁−μ)/σ and (x₂−μ)/σ with the recovered values to recover the two original standardised quantiles. If an examination question says “above” rather than “below”, convert the relevant tail probability before this calculation.
12. 20th and 95th percentile cutoffs
Original model question. A fictional continuous normal variable X has cumulative probability P(X≤44.267)=0.2 and P(X≤64.159)=0.95. The cutoffs have been printed to three decimal places. Determine the unknown mean μ and standard deviation σ. Before entering numbers into a calculator, state why the second cutoff must exceed the first and identify the two standard normal quantiles.
Complete two-equation calculation. The corresponding z values are z₁≈-0.8416212 and z₂≈1.6448536. Thus 44.267≈μ+(-0.8416212)σ and 64.159≈μ+(1.6448536)σ. Subtracting eliminates μ: σ≈(19.892)/(2.4864749)≈8.00008. Back-substitution gives μ≈51.00004. The underlying constructed model used μ=51, σ=8; the small difference results from printing the percentiles to three decimals. Its variance is approximately σ²≈64.00129.
Interpretation and independent check. The percentile positions are not symmetric, so the arithmetic midpoint of the two cutoffs is generally not μ; using both z-scores is essential. The chosen percentiles and cutoffs describe a population model; no real sample data are being claimed. Recalculate (x₁−μ)/σ and (x₂−μ)/σ with the recovered values to recover the two original standardised quantiles. If an examination question says “above” rather than “below”, convert the relevant tail probability before this calculation.
13. 10th and 90th percentile cutoffs
Original model question. A fictional continuous normal variable X has cumulative probability P(X≤41.34)=0.1 and P(X≤74.66)=0.9. The cutoffs have been printed to three decimal places. Determine the unknown mean μ and standard deviation σ. Before entering numbers into a calculator, state why the second cutoff must exceed the first and identify the two standard normal quantiles.
Complete two-equation calculation. The corresponding z values are z₁≈-1.2815516 and z₂≈1.2815516. Thus 41.34≈μ+(-1.2815516)σ and 74.66≈μ+(1.2815516)σ. Subtracting eliminates μ: σ≈(33.32)/(2.5631031)≈12.99987. Back-substitution gives μ≈58. The underlying constructed model used μ=58, σ=13; the small difference results from printing the percentiles to three decimals. Its variance is approximately σ²≈168.99654.
Interpretation and independent check. This pair is symmetric around the median, so the midpoint (41.34+74.66)/2≈58 independently checks the mean. When the percentiles are symmetric around 50%, the mean is their midpoint. Recalculate (x₁−μ)/σ and (x₂−μ)/σ with the recovered values to recover the two original standardised quantiles. If an examination question says “above” rather than “below”, convert the relevant tail probability before this calculation.
14. 25th and 75th percentile cutoffs
Original model question. A fictional continuous normal variable X has cumulative probability P(X≤61.628)=0.25 and P(X≤68.372)=0.75. The cutoffs have been printed to three decimal places. Determine the unknown mean μ and standard deviation σ. Before entering numbers into a calculator, state why the second cutoff must exceed the first and identify the two standard normal quantiles.
Complete two-equation calculation. The corresponding z values are z₁≈-0.6744898 and z₂≈0.6744898. Thus 61.628≈μ+(-0.6744898)σ and 68.372≈μ+(0.6744898)σ. Subtracting eliminates μ: σ≈(6.744)/(1.3489795)≈4.99933. Back-substitution gives μ≈65. The underlying constructed model used μ=65, σ=5; the small difference results from printing the percentiles to three decimals. Its variance is approximately σ²≈24.99335.
Interpretation and independent check. This pair is symmetric around the median, so the midpoint (61.628+68.372)/2≈65 independently checks the mean. For non-symmetric percentiles, do not use a midpoint rule; solve both linear z-score equations. Recalculate (x₁−μ)/σ and (x₂−μ)/σ with the recovered values to recover the two original standardised quantiles. If an examination question says “above” rather than “below”, convert the relevant tail probability before this calculation.
15. 20th and 80th percentile cutoffs
Original model question. A fictional continuous normal variable X has cumulative probability P(X≤63.584)=0.2 and P(X≤80.416)=0.8. The cutoffs have been printed to three decimal places. Determine the unknown mean μ and standard deviation σ. Before entering numbers into a calculator, state why the second cutoff must exceed the first and identify the two standard normal quantiles.
Complete two-equation calculation. The corresponding z values are z₁≈-0.8416212 and z₂≈0.8416212. Thus 63.584≈μ+(-0.8416212)σ and 80.416≈μ+(0.8416212)σ. Subtracting eliminates μ: σ≈(16.832)/(1.6832425)≈9.99975. Back-substitution gives μ≈72. The underlying constructed model used μ=72, σ=10; the small difference results from printing the percentiles to three decimals. Its variance is approximately σ²≈99.99495.
Interpretation and independent check. This pair is symmetric around the median, so the midpoint (63.584+80.416)/2≈72 independently checks the mean. Check that the higher cumulative probability corresponds to the larger cutoff. Recalculate (x₁−μ)/σ and (x₂−μ)/σ with the recovered values to recover the two original standardised quantiles. If an examination question says “above” rather than “below”, convert the relevant tail probability before this calculation.
16. 5th and 95th percentile cutoffs
Original model question. A fictional continuous normal variable X has cumulative probability P(X≤75.71)=0.05 and P(X≤82.29)=0.95. The cutoffs have been printed to three decimal places. Determine the unknown mean μ and standard deviation σ. Before entering numbers into a calculator, state why the second cutoff must exceed the first and identify the two standard normal quantiles.
Complete two-equation calculation. The corresponding z values are z₁≈-1.6448536 and z₂≈1.6448536. Thus 75.71≈μ+(-1.6448536)σ and 82.29≈μ+(1.6448536)σ. Subtracting eliminates μ: σ≈(6.58)/(3.2897073)≈2.00018. Back-substitution gives μ≈79. The underlying constructed model used μ=79, σ=2; the small difference results from printing the percentiles to three decimals. Its variance is approximately σ²≈4.00071.
Interpretation and independent check. This pair is symmetric around the median, so the midpoint (75.71+82.29)/2≈79 independently checks the mean. A larger spread σ separates any pair of fixed normal percentiles further apart. Recalculate (x₁−μ)/σ and (x₂−μ)/σ with the recovered values to recover the two original standardised quantiles. If an examination question says “above” rather than “below”, convert the relevant tail probability before this calculation.
17. 10th and 75th percentile cutoffs
Original model question. A fictional continuous normal variable X has cumulative probability P(X≤77.029)=0.1 and P(X≤90.721)=0.75. The cutoffs have been printed to three decimal places. Determine the unknown mean μ and standard deviation σ. Before entering numbers into a calculator, state why the second cutoff must exceed the first and identify the two standard normal quantiles.
Complete two-equation calculation. The corresponding z values are z₁≈-1.2815516 and z₂≈0.6744898. Thus 77.029≈μ+(-1.2815516)σ and 90.721≈μ+(0.6744898)σ. Subtracting eliminates μ: σ≈(13.692)/(1.9560413)≈6.99985. Back-substitution gives μ≈85.99967. The underlying constructed model used μ=86, σ=7; the small difference results from printing the percentiles to three decimals. Its variance is approximately σ²≈48.99793.
Interpretation and independent check. The percentile positions are not symmetric, so the arithmetic midpoint of the two cutoffs is generally not μ; using both z-scores is essential. The second parameter of N(μ,σ²) is variance, not standard deviation. Recalculate (x₁−μ)/σ and (x₂−μ)/σ with the recovered values to recover the two original standardised quantiles. If an examination question says “above” rather than “below”, convert the relevant tail probability before this calculation.
18. 25th and 95th percentile cutoffs
Original model question. A fictional continuous normal variable X has cumulative probability P(X≤23.906)=0.25 and P(X≤51.738)=0.95. The cutoffs have been printed to three decimal places. Determine the unknown mean μ and standard deviation σ. Before entering numbers into a calculator, state why the second cutoff must exceed the first and identify the two standard normal quantiles.
Complete two-equation calculation. The corresponding z values are z₁≈-0.6744898 and z₂≈1.6448536. Thus 23.906≈μ+(-0.6744898)σ and 51.738≈μ+(1.6448536)σ. Subtracting eliminates μ: σ≈(27.832)/(2.3193434)≈11.99995. Back-substitution gives μ≈31.99984. The underlying constructed model used μ=32, σ=12; the small difference results from printing the percentiles to three decimals. Its variance is approximately σ²≈143.99875.
Interpretation and independent check. The percentile positions are not symmetric, so the arithmetic midpoint of the two cutoffs is generally not μ; using both z-scores is essential. Use approximate equality because the supplied cutoffs were rounded to three decimal places. Recalculate (x₁−μ)/σ and (x₂−μ)/σ with the recovered values to recover the two original standardised quantiles. If an examination question says “above” rather than “below”, convert the relevant tail probability before this calculation.
19. 2.5th and 97.5th percentile cutoffs
Original model question. A fictional continuous normal variable X has cumulative probability P(X≤31.16)=0.025 and P(X≤46.84)=0.975. The cutoffs have been printed to three decimal places. Determine the unknown mean μ and standard deviation σ. Before entering numbers into a calculator, state why the second cutoff must exceed the first and identify the two standard normal quantiles.
Complete two-equation calculation. The corresponding z values are z₁≈-1.959964 and z₂≈1.959964. Thus 31.16≈μ+(-1.959964)σ and 46.84≈μ+(1.959964)σ. Subtracting eliminates μ: σ≈(15.68)/(3.919928)≈4.00007. Back-substitution gives μ≈39. The underlying constructed model used μ=39, σ=4; the small difference results from printing the percentiles to three decimals. Its variance is approximately σ²≈16.00059.
Interpretation and independent check. This pair is symmetric around the median, so the midpoint (31.16+46.84)/2≈39 independently checks the mean. A single percentile cannot normally determine both μ and σ without another independent fact. Recalculate (x₁−μ)/σ and (x₂−μ)/σ with the recovered values to recover the two original standardised quantiles. If an examination question says “above” rather than “below”, convert the relevant tail probability before this calculation.
20. 15.87th (approx.) and 84.13th (approx.) percentile cutoffs
Original model question. A fictional continuous normal variable X has cumulative probability P(X≤37)=0.158655254 and P(X≤55)=0.841344746. The cutoffs have been printed to three decimal places. Determine the unknown mean μ and standard deviation σ. Before entering numbers into a calculator, state why the second cutoff must exceed the first and identify the two standard normal quantiles.
Complete two-equation calculation. The corresponding z values are z₁≈-1 and z₂≈1. Thus 37≈μ+(-1)σ and 55≈μ+(1)σ. Subtracting eliminates μ: σ≈(18)/(2)≈9. Back-substitution gives μ≈46. The underlying constructed model used μ=46, σ=9; the small difference results from printing the percentiles to three decimals. Its variance is approximately σ²≈81.
Interpretation and independent check. This pair is symmetric around the median, so the midpoint (37+55)/2≈46 independently checks the mean. The standardised distances from μ should reproduce both original z-quantile values. Recalculate (x₁−μ)/σ and (x₂−μ)/σ with the recovered values to recover the two original standardised quantiles. If an examination question says “above” rather than “below”, convert the relevant tail probability before this calculation.
21. 50th and 90th percentile cutoffs
Original model question. A fictional continuous normal variable X has cumulative probability P(X≤53)=0.5 and P(X≤70.942)=0.9. The cutoffs have been printed to three decimal places. Determine the unknown mean μ and standard deviation σ. Before entering numbers into a calculator, state why the second cutoff must exceed the first and identify the two standard normal quantiles.
Complete two-equation calculation. The corresponding z values are z₁≈0 and z₂≈1.2815516. Thus 53≈μ+(0)σ and 70.942≈μ+(1.2815516)σ. Subtracting eliminates μ: σ≈(17.942)/(1.2815516)≈14.00022. Back-substitution gives μ≈53. The underlying constructed model used μ=53, σ=14; the small difference results from printing the percentiles to three decimals. Its variance is approximately σ²≈196.00608.
Interpretation and independent check. The percentile positions are not symmetric, so the arithmetic midpoint of the two cutoffs is generally not μ; using both z-scores is essential. Top 10% above a cutoff corresponds to the 90th percentile, not the 10th. Recalculate (x₁−μ)/σ and (x₂−μ)/σ with the recovered values to recover the two original standardised quantiles. If an examination question says “above” rather than “below”, convert the relevant tail probability before this calculation.
22. 25th and 90th percentile cutoffs
Original model question. A fictional continuous normal variable X has cumulative probability P(X≤55.953)=0.25 and P(X≤67.689)=0.9. The cutoffs have been printed to three decimal places. Determine the unknown mean μ and standard deviation σ. Before entering numbers into a calculator, state why the second cutoff must exceed the first and identify the two standard normal quantiles.
Complete two-equation calculation. The corresponding z values are z₁≈-0.6744898 and z₂≈1.2815516. Thus 55.953≈μ+(-0.6744898)σ and 67.689≈μ+(1.2815516)σ. Subtracting eliminates μ: σ≈(11.736)/(1.9560413)≈5.99987. Back-substitution gives μ≈59.99985. The underlying constructed model used μ=60, σ=6; the small difference results from printing the percentiles to three decimals. Its variance is approximately σ²≈35.99848.
Interpretation and independent check. The percentile positions are not symmetric, so the arithmetic midpoint of the two cutoffs is generally not μ; using both z-scores is essential. The mean of a normal distribution is its median, but this is not generally true of a skew distribution. Recalculate (x₁−μ)/σ and (x₂−μ)/σ with the recovered values to recover the two original standardised quantiles. If an examination question says “above” rather than “below”, convert the relevant tail probability before this calculation.
23. 10th and 80th percentile cutoffs
Original model question. A fictional continuous normal variable X has cumulative probability P(X≤52.903)=0.1 and P(X≤76.258)=0.8. The cutoffs have been printed to three decimal places. Determine the unknown mean μ and standard deviation σ. Before entering numbers into a calculator, state why the second cutoff must exceed the first and identify the two standard normal quantiles.
Complete two-equation calculation. The corresponding z values are z₁≈-1.2815516 and z₂≈0.8416212. Thus 52.903≈μ+(-1.2815516)σ and 76.258≈μ+(0.8416212)σ. Subtracting eliminates μ: σ≈(23.355)/(2.1231728)≈11.00005. Back-substitution gives μ≈67.00013. The underlying constructed model used μ=67, σ=11; the small difference results from printing the percentiles to three decimals. Its variance is approximately σ²≈121.00103.
Interpretation and independent check. The percentile positions are not symmetric, so the arithmetic midpoint of the two cutoffs is generally not μ; using both z-scores is essential. A negative recovered standard deviation signals an incompatible percentile order. Recalculate (x₁−μ)/σ and (x₂−μ)/σ with the recovered values to recover the two original standardised quantiles. If an examination question says “above” rather than “below”, convert the relevant tail probability before this calculation.
24. 20th and 95th percentile cutoffs
Original model question. A fictional continuous normal variable X has cumulative probability P(X≤71.475)=0.2 and P(X≤78.935)=0.95. The cutoffs have been printed to three decimal places. Determine the unknown mean μ and standard deviation σ. Before entering numbers into a calculator, state why the second cutoff must exceed the first and identify the two standard normal quantiles.
Complete two-equation calculation. The corresponding z values are z₁≈-0.8416212 and z₂≈1.6448536. Thus 71.475≈μ+(-0.8416212)σ and 78.935≈μ+(1.6448536)σ. Subtracting eliminates μ: σ≈(7.46)/(2.4864749)≈3.00023. Back-substitution gives μ≈74.00006. The underlying constructed model used μ=74, σ=3; the small difference results from printing the percentiles to three decimals. Its variance is approximately σ²≈9.00139.
Interpretation and independent check. The percentile positions are not symmetric, so the arithmetic midpoint of the two cutoffs is generally not μ; using both z-scores is essential. The chosen percentiles and cutoffs describe a population model; no real sample data are being claimed. Recalculate (x₁−μ)/σ and (x₂−μ)/σ with the recovered values to recover the two original standardised quantiles. If an examination question says “above” rather than “below”, convert the relevant tail probability before this calculation.
25. 10th and 90th percentile cutoffs
Original model question. A fictional continuous normal variable X has cumulative probability P(X≤70.748)=0.1 and P(X≤91.252)=0.9. The cutoffs have been printed to three decimal places. Determine the unknown mean μ and standard deviation σ. Before entering numbers into a calculator, state why the second cutoff must exceed the first and identify the two standard normal quantiles.
Complete two-equation calculation. The corresponding z values are z₁≈-1.2815516 and z₂≈1.2815516. Thus 70.748≈μ+(-1.2815516)σ and 91.252≈μ+(1.2815516)σ. Subtracting eliminates μ: σ≈(20.504)/(2.5631031)≈7.99968. Back-substitution gives μ≈81. The underlying constructed model used μ=81, σ=8; the small difference results from printing the percentiles to three decimals. Its variance is approximately σ²≈63.99485.
Interpretation and independent check. This pair is symmetric around the median, so the midpoint (70.748+91.252)/2≈81 independently checks the mean. When the percentiles are symmetric around 50%, the mean is their midpoint. Recalculate (x₁−μ)/σ and (x₂−μ)/σ with the recovered values to recover the two original standardised quantiles. If an examination question says “above” rather than “below”, convert the relevant tail probability before this calculation.
26. 25th and 75th percentile cutoffs
Original model question. A fictional continuous normal variable X has cumulative probability P(X≤79.232)=0.25 and P(X≤96.768)=0.75. The cutoffs have been printed to three decimal places. Determine the unknown mean μ and standard deviation σ. Before entering numbers into a calculator, state why the second cutoff must exceed the first and identify the two standard normal quantiles.
Complete two-equation calculation. The corresponding z values are z₁≈-0.6744898 and z₂≈0.6744898. Thus 79.232≈μ+(-0.6744898)σ and 96.768≈μ+(0.6744898)σ. Subtracting eliminates μ: σ≈(17.536)/(1.3489795)≈12.99946. Back-substitution gives μ≈88. The underlying constructed model used μ=88, σ=13; the small difference results from printing the percentiles to three decimals. Its variance is approximately σ²≈168.98586.
Interpretation and independent check. This pair is symmetric around the median, so the midpoint (79.232+96.768)/2≈88 independently checks the mean. For non-symmetric percentiles, do not use a midpoint rule; solve both linear z-score equations. Recalculate (x₁−μ)/σ and (x₂−μ)/σ with the recovered values to recover the two original standardised quantiles. If an examination question says “above” rather than “below”, convert the relevant tail probability before this calculation.
27. 20th and 80th percentile cutoffs
Original model question. A fictional continuous normal variable X has cumulative probability P(X≤29.792)=0.2 and P(X≤38.208)=0.8. The cutoffs have been printed to three decimal places. Determine the unknown mean μ and standard deviation σ. Before entering numbers into a calculator, state why the second cutoff must exceed the first and identify the two standard normal quantiles.
Complete two-equation calculation. The corresponding z values are z₁≈-0.8416212 and z₂≈0.8416212. Thus 29.792≈μ+(-0.8416212)σ and 38.208≈μ+(0.8416212)σ. Subtracting eliminates μ: σ≈(8.416)/(1.6832425)≈4.99987. Back-substitution gives μ≈34. The underlying constructed model used μ=34, σ=5; the small difference results from printing the percentiles to three decimals. Its variance is approximately σ²≈24.99874.
Interpretation and independent check. This pair is symmetric around the median, so the midpoint (29.792+38.208)/2≈34 independently checks the mean. Check that the higher cumulative probability corresponds to the larger cutoff. Recalculate (x₁−μ)/σ and (x₂−μ)/σ with the recovered values to recover the two original standardised quantiles. If an examination question says “above” rather than “below”, convert the relevant tail probability before this calculation.
28. 5th and 95th percentile cutoffs
Original model question. A fictional continuous normal variable X has cumulative probability P(X≤24.551)=0.05 and P(X≤57.449)=0.95. The cutoffs have been printed to three decimal places. Determine the unknown mean μ and standard deviation σ. Before entering numbers into a calculator, state why the second cutoff must exceed the first and identify the two standard normal quantiles.
Complete two-equation calculation. The corresponding z values are z₁≈-1.6448536 and z₂≈1.6448536. Thus 24.551≈μ+(-1.6448536)σ and 57.449≈μ+(1.6448536)σ. Subtracting eliminates μ: σ≈(32.898)/(3.2897073)≈10.00028. Back-substitution gives μ≈41. The underlying constructed model used μ=41, σ=10; the small difference results from printing the percentiles to three decimals. Its variance is approximately σ²≈100.00564.
Interpretation and independent check. This pair is symmetric around the median, so the midpoint (24.551+57.449)/2≈41 independently checks the mean. A larger spread σ separates any pair of fixed normal percentiles further apart. Recalculate (x₁−μ)/σ and (x₂−μ)/σ with the recovered values to recover the two original standardised quantiles. If an examination question says “above” rather than “below”, convert the relevant tail probability before this calculation.
29. 10th and 75th percentile cutoffs
Original model question. A fictional continuous normal variable X has cumulative probability P(X≤45.437)=0.1 and P(X≤49.349)=0.75. The cutoffs have been printed to three decimal places. Determine the unknown mean μ and standard deviation σ. Before entering numbers into a calculator, state why the second cutoff must exceed the first and identify the two standard normal quantiles.
Complete two-equation calculation. The corresponding z values are z₁≈-1.2815516 and z₂≈0.6744898. Thus 45.437≈μ+(-1.2815516)σ and 49.349≈μ+(0.6744898)σ. Subtracting eliminates μ: σ≈(3.912)/(1.9560413)≈1.99996. Back-substitution gives μ≈48.00005. The underlying constructed model used μ=48, σ=2; the small difference results from printing the percentiles to three decimals. Its variance is approximately σ²≈3.99983.
Interpretation and independent check. The percentile positions are not symmetric, so the arithmetic midpoint of the two cutoffs is generally not μ; using both z-scores is essential. The second parameter of N(μ,σ²) is variance, not standard deviation. Recalculate (x₁−μ)/σ and (x₂−μ)/σ with the recovered values to recover the two original standardised quantiles. If an examination question says “above” rather than “below”, convert the relevant tail probability before this calculation.
30. 25th and 95th percentile cutoffs
Original model question. A fictional continuous normal variable X has cumulative probability P(X≤50.279)=0.25 and P(X≤66.514)=0.95. The cutoffs have been printed to three decimal places. Determine the unknown mean μ and standard deviation σ. Before entering numbers into a calculator, state why the second cutoff must exceed the first and identify the two standard normal quantiles.
Complete two-equation calculation. The corresponding z values are z₁≈-0.6744898 and z₂≈1.6448536. Thus 50.279≈μ+(-0.6744898)σ and 66.514≈μ+(1.6448536)σ. Subtracting eliminates μ: σ≈(16.235)/(2.3193434)≈6.99983. Back-substitution gives μ≈55.00031. The underlying constructed model used μ=55, σ=7; the small difference results from printing the percentiles to three decimals. Its variance is approximately σ²≈48.99756.
Interpretation and independent check. The percentile positions are not symmetric, so the arithmetic midpoint of the two cutoffs is generally not μ; using both z-scores is essential. Use approximate equality because the supplied cutoffs were rounded to three decimal places. Recalculate (x₁−μ)/σ and (x₂−μ)/σ with the recovered values to recover the two original standardised quantiles. If an examination question says “above” rather than “below”, convert the relevant tail probability before this calculation.
31. 2.5th and 97.5th percentile cutoffs
Original model question. A fictional continuous normal variable X has cumulative probability P(X≤38.48)=0.025 and P(X≤85.52)=0.975. The cutoffs have been printed to three decimal places. Determine the unknown mean μ and standard deviation σ. Before entering numbers into a calculator, state why the second cutoff must exceed the first and identify the two standard normal quantiles.
Complete two-equation calculation. The corresponding z values are z₁≈-1.959964 and z₂≈1.959964. Thus 38.48≈μ+(-1.959964)σ and 85.52≈μ+(1.959964)σ. Subtracting eliminates μ: σ≈(47.04)/(3.919928)≈12.00022. Back-substitution gives μ≈62. The underlying constructed model used μ=62, σ=12; the small difference results from printing the percentiles to three decimals. Its variance is approximately σ²≈144.00529.
Interpretation and independent check. This pair is symmetric around the median, so the midpoint (38.48+85.52)/2≈62 independently checks the mean. A single percentile cannot normally determine both μ and σ without another independent fact. Recalculate (x₁−μ)/σ and (x₂−μ)/σ with the recovered values to recover the two original standardised quantiles. If an examination question says “above” rather than “below”, convert the relevant tail probability before this calculation.
32. 15.87th (approx.) and 84.13th (approx.) percentile cutoffs
Original model question. A fictional continuous normal variable X has cumulative probability P(X≤65)=0.158655254 and P(X≤73)=0.841344746. The cutoffs have been printed to three decimal places. Determine the unknown mean μ and standard deviation σ. Before entering numbers into a calculator, state why the second cutoff must exceed the first and identify the two standard normal quantiles.
Complete two-equation calculation. The corresponding z values are z₁≈-1 and z₂≈1. Thus 65≈μ+(-1)σ and 73≈μ+(1)σ. Subtracting eliminates μ: σ≈(8)/(2)≈4. Back-substitution gives μ≈69. The underlying constructed model used μ=69, σ=4; the small difference results from printing the percentiles to three decimals. Its variance is approximately σ²≈16.
Interpretation and independent check. This pair is symmetric around the median, so the midpoint (65+73)/2≈69 independently checks the mean. The standardised distances from μ should reproduce both original z-quantile values. Recalculate (x₁−μ)/σ and (x₂−μ)/σ with the recovered values to recover the two original standardised quantiles. If an examination question says “above” rather than “below”, convert the relevant tail probability before this calculation.
33. 50th and 90th percentile cutoffs
Original model question. A fictional continuous normal variable X has cumulative probability P(X≤76)=0.5 and P(X≤87.534)=0.9. The cutoffs have been printed to three decimal places. Determine the unknown mean μ and standard deviation σ. Before entering numbers into a calculator, state why the second cutoff must exceed the first and identify the two standard normal quantiles.
Complete two-equation calculation. The corresponding z values are z₁≈0 and z₂≈1.2815516. Thus 76≈μ+(0)σ and 87.534≈μ+(1.2815516)σ. Subtracting eliminates μ: σ≈(11.534)/(1.2815516)≈9.00003. Back-substitution gives μ≈76. The underlying constructed model used μ=76, σ=9; the small difference results from printing the percentiles to three decimals. Its variance is approximately σ²≈81.0005.
Interpretation and independent check. The percentile positions are not symmetric, so the arithmetic midpoint of the two cutoffs is generally not μ; using both z-scores is essential. Top 10% above a cutoff corresponds to the 90th percentile, not the 10th. Recalculate (x₁−μ)/σ and (x₂−μ)/σ with the recovered values to recover the two original standardised quantiles. If an examination question says “above” rather than “below”, convert the relevant tail probability before this calculation.
34. 25th and 90th percentile cutoffs
Original model question. A fictional continuous normal variable X has cumulative probability P(X≤73.557)=0.25 and P(X≤100.942)=0.9. The cutoffs have been printed to three decimal places. Determine the unknown mean μ and standard deviation σ. Before entering numbers into a calculator, state why the second cutoff must exceed the first and identify the two standard normal quantiles.
Complete two-equation calculation. The corresponding z values are z₁≈-0.6744898 and z₂≈1.2815516. Thus 73.557≈μ+(-0.6744898)σ and 100.942≈μ+(1.2815516)σ. Subtracting eliminates μ: σ≈(27.385)/(1.9560413)≈14.00022. Back-substitution gives μ≈83. The underlying constructed model used μ=83, σ=14; the small difference results from printing the percentiles to three decimals. Its variance is approximately σ²≈196.00603.
Interpretation and independent check. The percentile positions are not symmetric, so the arithmetic midpoint of the two cutoffs is generally not μ; using both z-scores is essential. The mean of a normal distribution is its median, but this is not generally true of a skew distribution. Recalculate (x₁−μ)/σ and (x₂−μ)/σ with the recovered values to recover the two original standardised quantiles. If an examination question says “above” rather than “below”, convert the relevant tail probability before this calculation.
35. 10th and 80th percentile cutoffs
Original model question. A fictional continuous normal variable X has cumulative probability P(X≤82.311)=0.1 and P(X≤95.05)=0.8. The cutoffs have been printed to three decimal places. Determine the unknown mean μ and standard deviation σ. Before entering numbers into a calculator, state why the second cutoff must exceed the first and identify the two standard normal quantiles.
Complete two-equation calculation. The corresponding z values are z₁≈-1.2815516 and z₂≈0.8416212. Thus 82.311≈μ+(-1.2815516)σ and 95.05≈μ+(0.8416212)σ. Subtracting eliminates μ: σ≈(12.739)/(2.1231728)≈5.99998. Back-substitution gives μ≈90.00029. The underlying constructed model used μ=90, σ=6; the small difference results from printing the percentiles to three decimals. Its variance is approximately σ²≈35.99979.
Interpretation and independent check. The percentile positions are not symmetric, so the arithmetic midpoint of the two cutoffs is generally not μ; using both z-scores is essential. A negative recovered standard deviation signals an incompatible percentile order. Recalculate (x₁−μ)/σ and (x₂−μ)/σ with the recovered values to recover the two original standardised quantiles. If an examination question says “above” rather than “below”, convert the relevant tail probability before this calculation.
36. 20th and 95th percentile cutoffs
Original model question. A fictional continuous normal variable X has cumulative probability P(X≤26.742)=0.2 and P(X≤54.093)=0.95. The cutoffs have been printed to three decimal places. Determine the unknown mean μ and standard deviation σ. Before entering numbers into a calculator, state why the second cutoff must exceed the first and identify the two standard normal quantiles.
Complete two-equation calculation. The corresponding z values are z₁≈-0.8416212 and z₂≈1.6448536. Thus 26.742≈μ+(-0.8416212)σ and 54.093≈μ+(1.6448536)σ. Subtracting eliminates μ: σ≈(27.351)/(2.4864749)≈10.99991. Back-substitution gives μ≈35.99976. The underlying constructed model used μ=36, σ=11; the small difference results from printing the percentiles to three decimals. Its variance is approximately σ²≈120.99802.
Interpretation and independent check. The percentile positions are not symmetric, so the arithmetic midpoint of the two cutoffs is generally not μ; using both z-scores is essential. The chosen percentiles and cutoffs describe a population model; no real sample data are being claimed. Recalculate (x₁−μ)/σ and (x₂−μ)/σ with the recovered values to recover the two original standardised quantiles. If an examination question says “above” rather than “below”, convert the relevant tail probability before this calculation.
Ten quick answers and model checks
- The 10th and 90th percentile cutoffs 40 and 60 imply μ=50.
- Their σ is approximately 20/(2×1.2815516)≈7.803.
- The 25th and 75th cutoffs 72 and 88 imply μ=80.
- Their σ is approximately 16/(2×0.67448975)≈11.861.
- Top 10% above a cutoff means the cutoff is the 90th percentile.
- For P(X≤c)=0.5 in a normal distribution, c=μ.
- Use σ=(x₂−x₁)/(z₂−z₁) for any two distinct percentile conditions.
- Then μ=x₁−σz₁, with σ positive.
- Increasing cumulative percentile requires increasing cutoff in a nondegenerate normal model.
- The normal notation N(μ,σ²) uses variance as its second parameter.
Five-day independent learning plan
Day one: convert percentile phrases and upper-tail statements into left cumulative probabilities. Day two: learn the standard normal quantile positions for common symmetric percentile pairs, then identify a mean by midpoint only when justified. Day three: solve asymmetric two-cutoff systems using two z-score equations. Day four: check positive σ, compute variance and restandardise both original cutoffs as independent verification. Day five: mix rounded cutoffs, reversed probability phrases and incompatible data to test model validity before arithmetic.
In a three-student tutorial, one learner can translate the probability wording, another write and solve the two linear equations and a third verify both recovered cutoffs and parameter meanings. Rotate the roles and replace one percentile while keeping its cutoff fixed. A learner who can explain why the midpoint shortcut then fails has understood the normal model more deeply than one who memorises only a 10th/90th example.
Frequently asked questions
Can two percentiles determine both μ and σ?
Yes for a nondegenerate normal model when the two distinct cumulative probabilities and corresponding cutoffs are known consistently.
Why is the mean a midpoint for the 10th and 90th percentiles?
A normal curve is symmetric, so those two cutoffs have equal and opposite z-values and are equally far from μ.
Does the midpoint work for the 10th and 75th percentiles?
Not generally. Their z-values are not opposite, so the mean must be obtained from the two equations.
How do I convert the highest 5% statement?
The cutoff for the highest 5% is the 95th percentile, so its left cumulative probability is 0.95.
What if two percentile cutoffs are reversed?
If the higher cumulative percentile has a smaller cutoff, the claimed exact normal model is inconsistent with positive σ.
Why does the calculator request σ when N(μ,σ²) prints σ²?
Normal probability functions often take the standard deviation σ as input, while mathematical distribution notation uses variance σ² in its second parameter.
Why are recovered parameters approximate?
The printed percentiles and cutoffs may be rounded. Their equations then recover approximations rather than infinitely precise generating parameters.
Is this method in the 2027 H2 syllabus?
Yes. Section 6.3 includes finding relationships among normal probabilities, cutoffs, μ and σ, which can be combined to determine both unknown parameters.
Continue learning and official syllabus
- Hyperbola Graphs: Which Axis Holds the Branches?
- Complex Modulus: Distance Between Two Argand Points
- Skew Lines: Nonparallel Does Not Mean Intersecting
- Normal Distribution: Find μ and σ From Two Percentiles: current guide
For the earlier, reverse-direction problem, read Top 10% and Inverse Normal Cutoffs. For the broader topic, see Bukit Timah H2 Normal Distribution and Sampling and the JC2 Bukit Timah tuition hub. The immutable Clementi small-group Mathematics reference illustrates the teaching format at another level.
Official syllabus: SEAB 2027 H2 Mathematics 9758, §6.3 Normal Distribution. These numerical percentiles are original hypothetical model values and not actual student results, official paper questions or measured population distributions.