SCSA WACE 2021-style practice: Calculator-free. The masses of apples are normally distributed with mean μ=150 g and standard deviation σ=10 g. Using the empirical rule, find (a) P(140 X 160), (b) P(X>170) and (c) the proportion between 130 g and 150 g.

A calculator-free empirical-rule question. (a) 140=μ-σ and 160=μ+σ, so this is the central 1σ band: P≈ 0.68. (b) 170=μ+2σ. Beyond 2σ is 5\% split into two tails, so the upper tail is P(X>170)≈ 0.025. (c) 130=μ-2σ and 150=μ. From -2σ to the mean is half of the central 95\% band, so P≈ 0.475. Markers reward expressing each boundary as a multiple of σ and using symmetry.

SCSA WACE 2023-style practice: Calculator-free. A normal distribution has P(X 34) using the empirical rule and (b) the value 3 standard deviations above the mean.

A short symmetry-and-parameters question. (a) 34=μ+σ=30+4. Within 1σ is 68\%, leaving 32\% in the two tails, so the upper tail is P(X>34)≈ 0.16. (b) μ+3σ=30+3(4)=42. Markers reward identifying 34 as 1σ above the mean, halving the remaining 32\%, and the 3σ value.

§-Syllabus dot point

WAMath MethodsSyllabus dot point

What are the defining properties of the normal distribution and the empirical 68-95-99.7 rule?

Describe the normal distribution, its symmetry and parameters, and apply the empirical 68-95-99.7 rule

WACE Year 12 Mathematics Methods Unit 4 the normal distribution: the bell shape, symmetry about the mean, the role of mu and sigma, and the empirical 68-95-99.7 rule, with worked SCSA-style examples.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
Defining properties
The empirical rule
Building probabilities from the rule
Comparing distributions

What this dot point is asking

SCSA Unit 4 makes the normal distribution the central continuous model. This dot point asks you to describe its shape and parameters and to use the empirical 68-95-99.7 rule for quick probability estimates, examined in both sections, with the rule especially useful in the calculator-free section.

Defining properties

The mean, median and mode of a normal distribution all coincide at $\mu$ because of the symmetry. The notation $N(\mu,\sigma^{2})$ uses the variance $\sigma^{2}$ as the second parameter, not the standard deviation.

The empirical rule

For any normal distribution the proportion of values within a fixed number of standard deviations is the same, giving the 68-95-99.7 rule.

By symmetry, each tail beyond $2\sigma$ holds about $2.5\%$ , and beyond $3\sigma$ about $0.15\%$ . These let you answer many questions without a calculator.

Using the empirical rule

Seedling heights are normal with $\mu=20$ cm and $\sigma=3$ cm. Estimate the proportion taller than $26$ cm.

Step 1: Express the boundary as a multiple of $\sigma$

The empirical rule is stated in terms of standard deviations from the mean, so the first task is to check how many $\sigma$ the boundary lies from $\mu$ .

$26 = 20 + 2\times 3$ , so $26$ is exactly $2\sigma$ above the mean.

Step 2: Apply the empirical rule and use symmetry

By the 68-95-99.7 rule, $95\%$ of values lie within $2\sigma$ of the mean, leaving $5\%$ in the two tails combined. Because the normal distribution is symmetric, each tail holds the same proportion.

By symmetry each tail holds $2.5\%$ , so $P(X>26) \approx 0.025$ .

Final answer: about $2.5\%$ of seedlings exceed $26$ cm.

Building probabilities from the rule

The empirical rule gives more than the three headline figures. Because the curve is symmetric, you can combine bands to answer many calculator-free questions. The region from the mean to $1\sigma$ holds half of $68\%$ , that is $34\%$ . The region between $1\sigma$ and $2\sigma$ holds half of $(95-68)\%$ , that is $13.5\%$ . Beyond $2\sigma$ each tail holds $2.5\%$ , and beyond $3\sigma$ each holds about $0.15\%$ . Sketching the bell and labelling these slices lets you assemble probabilities such as $P(\mu+\sigma\le X\le\mu+2\sigma)\approx 0.135$ without a calculator.

Combining empirical-rule bands

IQ scores are normal with $\mu=100$ and $\sigma=15$ . Estimate the proportion scoring between $115$ and $130$ .

Step 1: Express both boundaries in standard deviations

To use the empirical rule, convert both limits to multiples of $\sigma$ from the mean.

$115 = 100 + 1\times 15 = \mu + \sigma$ and $130 = 100 + 2\times 15 = \mu + 2\sigma$ .

Step 2: Find the area between the two bands by subtraction

The empirical rule gives total areas within $1\sigma$ and $2\sigma$ . The band between $1\sigma$ and $2\sigma$ on one side is found by subtracting the central proportions and halving for symmetry.

The region within $2\sigma$ holds $95\%$ and within $1\sigma$ holds $68\%$ , so the two-sided gap is $(95-68)\% = 27\%$ . Half of this lies on the positive side (between $\mu+\sigma$ and $\mu+2\sigma$ ).

P(115\le X\le 130)\approx \frac{27\%}{2} = 0.135

Final answer: about $13.5\%$ of scores fall in this band.

Comparing distributions

Because $\sigma$ sets the width, two normal distributions with the same mean but different standard deviations look very different: the smaller $\sigma$ concentrates values near the mean. The empirical rule helps compare them: a value $2\sigma$ above the mean is in the top $2.5\%$ of either distribution, regardless of the actual numbers.

Exam-style practice questions

Practice questions written in the style of SCSA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

WACE 20215 marksCalculator-free. The masses of apples are normally distributed with mean

\mu=150

g and standard deviation

\sigma=10

g. Using the empirical rule, find (a)

P(140\le X\le 160)

, (b)

P(X>170)

and (c) the proportion between

130

g and

150

Show worked answer →

A calculator-free empirical-rule question.

(a) $140=\mu-\sigma$ and $160=\mu+\sigma$ , so this is the central $1\sigma$ band: $P\approx 0.68$ .

(b) $170=\mu+2\sigma$ . Beyond $2\sigma$ is $5\%$ split into two tails, so the upper tail is $P(X>170)\approx 0.025$ .

(c) $130=\mu-2\sigma$ and $150=\mu$ . From $-2\sigma$ to the mean is half of the central $95\%$ band, so $P\approx 0.475$ .

Markers reward expressing each boundary as a multiple of $\sigma$ and using symmetry.

WACE 20234 marksCalculator-free. A normal distribution has

P(X<\mu)=0.5

and

\sigma=4

. Given

\mu=30

, find (a)

P(X>34)

using the empirical rule and (b) the value

3

standard deviations above the mean.

Show worked answer →

A short symmetry-and-parameters question.

(a) $34=\mu+\sigma=30+4$ . Within $1\sigma$ is $68\%$ , leaving $32\%$ in the two tails, so the upper tail is $P(X>34)\approx 0.16$ .

(b) $\mu+3\sigma=30+3(4)=42$ .

Markers reward identifying $34$ as $1\sigma$ above the mean, halving the remaining $32\%$ , and the $3\sigma$ value.