Loav · VCE Mathematical Methods · Units 3 and 4
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The normal distribution is a continuous model with a symmetric bell-shaped density. Two parameters specify its location and spread, allowing many different measurements to be handled through one standard distribution. This guide explains the parameters, standardisation, interval and conditional probabilities, unknown cutoffs and unknown parameters. You need the area interpretation of continuous probability and basic algebra.
A normal model is a stated assumption about a random variable. A familiar-looking histogram can suggest the model, but does not guarantee it. Normal densities extend across the whole real line, so a model for a physically nonnegative quantity may assign a tiny probability to negative values. Whether that approximation is reasonable depends on the location and spread in the context.
1. Read the notation carefully
X ∼ N(μ, σ2), where σ > 0.
E(X) = μ, Var(X) = σ2, sd(X) = σ.
f(x) = [1/(σ√(2π))] exp[−(x − μ)2/(2σ2)].
The second parameter in N(μ,σ2) is the variance. If X ∼ N(40,9), the mean is 40 and the standard deviation is 3. Many calculator commands ask for standard deviation, so entering nine in that input would create a distribution three times as spread out. Label both parameters before using technology.
The density is symmetric about μ, has total area one, and approaches the horizontal axis in both tails. Changing μ shifts its location. Increasing σ spreads the same total area over a wider region, lowering the central height. It does not increase the total probability. These parameter effects should be visible in any sketch.
The probability of one exact value is zero. For this continuous model, P(X < a) = P(X ≤ a). Intervals with different included endpoints therefore have the same probability. This useful property does not carry over to an exact binomial count.
2. Standardisation measures distance in standard deviations
Subtracting μ measures how far a value lies from the centre. Dividing by σ expresses that distance in standard-deviation units. The resulting variable has the standard normal distribution.
Z = (X − μ)/σ ∼ N(0,1).
P(a < X < b) = P((a − μ)/σ < Z < (b − μ)/σ).
A positive z-value lies above the mean; a negative z-value lies below it. A value with z = 2 lies two standard deviations above its mean regardless of the original units. Because σ is positive, dividing by it does not reverse the inequality. Standardisation changes the coordinate scale while preserving the relevant area.
Write Φ(z) for the probability P(Z ≤ z) when this shorthand helps. Then an interval probability is Φ(zupper) − Φ(zlower), and an upper tail is 1 − Φ(z). This is only a convenient label for an accumulated area; you can write the full probability statements throughout instead.
3. Sketch the requested area before calculating
Mark the mean and the requested boundaries, then shade the event. Decide whether the required area is a left tail, right tail, central interval or two tails. A sketch often detects using the wrong complement immediately. For example, a threshold well above the mean should have a small upper-tail probability and a large lower-tail probability.
Approximately 68% of a normal distribution lies within one standard deviation of its mean, 95% within two, and 99.7% within three. These rounded rules are useful reasonableness checks. They do not replace a more precise calculation when the requested accuracy is four decimal places. In particular, the central 95% cutoff is approximately 1.96 standard deviations rather than exactly two.
Worked examples
Worked example 1: an interval and an upper-tail cutoff
An original timing model has X ∼ N(72,82) seconds. Find P(60 < X < 80), then find c such that only 10% of times exceed c.
- At 60 seconds, z = (60 − 72)/8 = −1.5. At 80 seconds, z = 1.
- P(60 < X < 80) = P(−1.5 < Z < 1) ≈ 0.7745.
- For the cutoff, P(X > c) = 0.10 means P(X ≤ c) = 0.90. An inverse-normal calculation gives z ≈ 1.2815516.
- Convert back: c = 72 + 8(1.2815516) ≈ 82.2524 seconds.
The cutoff lies above the mean because 90% of the distribution is to its left. Using lower-tail area 0.10 instead would produce a value below the mean and answer the opposite question.
4. Inverse probability is an equation about area
Ordinary probability calculations start with boundaries and find an area. Inverse calculations start with an area and find a boundary. Before using an inverse-normal function, translate the question into a lower-tail area, since many devices use that convention. “Exceeded by 5%” corresponds to a lower-tail area of 0.95.
If μ and σ are known, obtain z and use x = μ + σz. If a parameter is unknown, standardisation produces an equation involving that parameter. Two independent area conditions can determine two unknown parameters. Solve with unrounded z-values and verify that the resulting σ is positive.
Worked example 2: recover two model parameters
A normal model for a production measurement has P(X < 80) = 0.10 and P(X < 120) = 0.90. Find its mean and standard deviation.
- The standard-normal boundaries are approximately −1.2815516 and +1.2815516. Thus 80 = μ − 1.2815516σ and 120 = μ + 1.2815516σ.
- Add the equations: 200 = 2μ, so μ = 100.
- Subtract them: 40 = 2(1.2815516)σ, giving σ ≈ 15.6061.
- Check that 80 and 120 are equally far below and above 100, matching the complementary tail probabilities.
The standard deviation is not 20: that is the distance from the mean to either supplied boundary. The boundary is 1.2815516 standard deviations away, so the distance must be divided by that factor.
5. Probability requirements can constrain variation
A model can be used to determine how small a standard deviation must be to keep enough outcomes within a tolerance interval. A symmetric tolerance around a known mean is particularly convenient because the remaining probability is split equally between two tails.
Worked example 3: determine a maximum standard deviation
A dispenser’s output is modelled as normal with mean 500 mL. At least 95% of outputs must lie between 496 and 504 mL. Find the largest standard deviation allowed by this model.
- The central requirement is P(496 < X < 504) ≥ 0.95. The bounds lie four millilitres from the mean.
- At the limiting case, P(−4/σ < Z < 4/σ) = 0.95. Each outside tail has probability 0.025.
- Therefore 4/σ = z where P(Z ≤ z) = 0.975. Numerically z ≈ 1.959964.
- σ = 4/1.959964 ≈ 2.0409 mL. Smaller positive σ values concentrate more area within the tolerance, so σ ≤ approximately 2.0409 mL.
If the mean shifts away from 500 mL, this symmetric calculation no longer applies unchanged. Location and spread both affect the probability of meeting fixed limits.
6. Conditional normal probabilities still use the event rule
For a conditional question, find the area of the intersection and divide by the area of the conditioning event. If b > a, then P(X > b | X > a) = P(X > b)/P(X > a). If intervals overlap only partly, use the actual intersection. Normality does not justify ignoring the condition or treating two events involving the same measurement as independent.
Keep the original normal model for both areas. Do not move the original mean to the conditioning threshold and call the remaining tail a new normal distribution. The tail-only conditional distribution has a different shape. The ratio formula gives the required answer without needing a separate named model.
When communicating a result, include the model, the event, the standardized boundaries if used, and the final probability or cutoff with units. Calculator syntax alone does not explain the mathematics. For inverse problems, substitute the proposed cutoff back into the original probability statement as a final check.
7. Use numerical tools without hiding the model
The normal density does not have an elementary antiderivative in the functions used in this course, so normal probabilities are obtained numerically. This does not change their meaning as areas. A numerical result should still be accompanied by the distribution parameters and the probability statement. For a two-tail event, add the left-tail and right-tail probabilities, or subtract the central interval from one. Do not assume the tails are equal unless their boundaries are equally far from the mean.
In an inverse question, distinguish a calculated boundary from a practical rounded decision. A cutoff rounded to the nearest unit may no longer meet a strict probability requirement exactly. If the task asks for the smallest whole-unit cutoff meeting a condition, test the adjacent integers in the original model. Keep the unrounded inverse result during this decision and report the final direction of rounding with a reason.
Practice and reasoned answers
1. X ∼ N(40,9). Find P(37 < X < 43).
The standard deviation is √9 = 3. The bounds standardise to −1 and 1, so the probability is approximately 0.6827. Treating nine as the standard deviation would produce an incorrectly narrow standardized interval.
2. X ∼ N(100,10²). Find P(X > 115).
The boundary has z = 1.5. The required upper tail is 1 − P(Z ≤ 1.5) ≈ 0.0668. The answer should be below 0.5 because the threshold lies above the mean.
3. X ∼ N(160,12²). Find c such that P(X > c) = 0.05.
Use lower-tail area 0.95, giving z ≈ 1.644854. Then c = 160 + 12z ≈ 179.7382. The requested object is a value of X, not the standardized value 1.644854.
4. X has normal mean 30 and P(20 < X < 40) = 0.95. Find σ.
The interval is symmetric with half-width ten. Its standardized upper boundary is approximately 1.959964, so σ = 10/1.959964 ≈ 5.1021. The normal notation would be N(30,5.1021²), with the unrounded value retained during checks.
5. Z ∼ N(0,1). Find P(Z > 1 | Z > 0).
The event Z > 1 lies inside Z > 0. Divide its area, approximately 0.1586553, by P(Z > 0) = 0.5. The result is approximately 0.3173. It is larger than the unconditional upper-tail probability because the reference group is restricted.
6. Two normal models have the same mean and standard deviations 2 and 5. Which has more probability within one unit of the mean?
The model with standard deviation 2 does. Its interval becomes −0.5 < Z < 0.5, while the wider model gives −0.2 < Z < 0.2. The first standardized interval contains the second, so its area is larger.
7. X ∼ N(50,4²). Find P(42 < X < 52), and explain why the interval is not symmetric.
The boundaries standardise to −2 and 0.5. Therefore the probability is approximately 0.6687. The lower boundary is eight units below the mean, while the upper boundary is only two above it; a symmetric central-area shortcut is inappropriate.
Learning check
Distinguish variance from standard deviation, shade an event, and standardise both boundaries. Reverse the process for an unknown cutoff and solve a parameter question from area conditions. Explain why a conditional tail requires division. Next, use the normal distribution as an approximation to the distribution of sample proportions, while preserving the difference between a population parameter and a sample statistic.
Curriculum and study route
Loav Mathematical Methods Units 3 and 4. These module divisions are a teaching sequence. Use the current VCAA Mathematics study design, Mathematical Methods Units 3 and 4, Area of Study 4, and the VCAA Mathematical Methods FAQ for course requirements. Scope checked 15 October 2026. All worked scenarios, questions and explanations in this guide are original.
