Interpret a normal model and probability as area
Read the mean and variance correctly, distinguish continuous areas from exact-value probabilities, and recognise modelling limits.
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
- Distinguish variance from standard deviation.
- Identify continuous measurements and discrete counts.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is the standard deviation of N(40,25)?
What you will learn
- Interpret N(μ,σ²) and the role of a positive standard deviation.
- Use areas under a normal curve and distinguish a model from empirical evidence.
Read the parameters and the curve
X ~ N(40,25) means the normal model has mean 40 and variance 25, hence standard deviation 5. The second parameter is σ², not σ. The curve is symmetric about 40 and its total area is one. Probability over an interval is the area above that interval, not the height of the curve at one point.
For a continuous normal variable, P(X = 40) = 0 while P(X < 40) = 1/2 by symmetry. This does not say that a recorded rounded measurement of 40 has probability zero: rounding to the nearest whole unit represents an interval such as 39.5 ≤ X < 40.5. Distinguish an exact mathematical value from a displayed measurement.
Use a model with stated limits
The mean sets the centre and standard deviation sets spread. N(120,144) therefore has centre 120 and standard deviation 12, rather than 144. A larger standard deviation spreads probability over a wider range. A bell-shaped sample display can suggest a normal model, but does not prove the population is exactly normal.
A normal model is continuous and theoretically extends over every real value. For a non-negative physical quantity, check whether its negative tail is negligible before accepting that model. Counts remain discrete; a normal distribution can sometimes approximate their probabilities, with specific conditions and integer boundary corrections, but does not make the count itself continuous. No measurements or populations are assumed normal merely because their average is known.
Pause and explain
What is P(X = μ) for a continuous normal variable?
Worked example
X ~ N(40,25). State its mean, standard deviation, P(X < 40) and P(X = 40).
Show the worked solution
- The mean is 40 and the variance is 25, so σ = 5.
- Symmetry places half the total area below the mean.
- One exact value has no interval width and therefore zero probability.
Answer Mean 40; SD 5; P(X < 40) = 1/2; P(X = 40) = 0.
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.
X ~ N(40,25). State its mean, variance, standard deviation and probability below 40.
Give me a hint
Read the variance, take its positive square root and use symmetry.
Compare my reasoning
- Mean = 40 and variance = 25.
- Standard deviation = √25 = 5.
- The mean divides the symmetric area equally, so P(X < 40) = 1/2.
Mean 40; variance 25; SD 5; P(X < 40) = 1/2.
Look for these in your work
- Read the second parameter as variance.
- Explain the area argument for one half.
X ~ N(120,144). Find its standard deviation and P(X = 120). Explain why a reading rounded to 120 is a different event.
Give me a hint
Separate the exact value from the interval represented by a rounded reading.
Compare my reasoning
- σ = √144 = 12.
- The exact-value probability is zero for this continuous model.
- Rounding to the nearest whole unit records 120 for values in [119.5,120.5), an interval of positive width.
SD 12; P(X = 120) = 0; rounded 120 represents [119.5,120.5).
Look for these in your work
- Use the positive square root of variance.
- Distinguish exact and rounded events.
A count of correct answers and a continuous manufactured length are each proposed as exactly normal. What must be checked before using those claims?
Give me a hint
Classify the variable and ask what evidence supports its distribution.
Compare my reasoning
- The answer count is discrete, so it is not an exactly continuous normal variable.
- A length can be modelled continuously, but normal shape and relevant parameter values still need evidence.
- Check physical bounds; any normal approximation for a count needs separate suitability and continuity-correction checks.
The count is discrete; the length needs modelling evidence and physical-bound checks.
Look for these in your work
- Separate a variable’s type from its proposed distribution.
- State the limitations rather than assuming a normal model.
Common mistakes
- The second normal parameter is variance.
- Density height is not interval probability.
- A rounded reading represents an interval, not one exact real value.
What can you explain now?
Does P(X = 40) = 0 imply a rounded display can never read 40?
Compare with the explanation
No; rounding groups a whole interval of underlying values.
The exact-value event and the rounded-recording event are different.
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Make it stick
Read mean and variance, use areas for probabilities, and keep exact-value, rounded-measurement and model claims distinct.
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.