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SAMath MethodsSyllabus dot point

What are the properties of the normal distribution and why is it so widely used?

The normal distribution is a symmetric bell-shaped density defined by its mean and standard deviation, with predictable probabilities given by the empirical rule.

The shape and properties of the normal distribution, the role of the mean and standard deviation, and the 68-95-99.7 empirical rule for estimating probabilities.

Generated by Claude Opus 4.77 min answer

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  1. What this dot point is asking
  2. Properties of the bell curve
  3. The empirical rule (68-95-99.7)
  4. Common errors
  5. Why it matters

What this dot point is asking

The normal distribution is the most important continuous distribution in statistics. Many natural and measured quantities - heights, exam scores, measurement errors - follow it closely, and it is the limiting shape of the binomial distribution for large nn.

Properties of the bell curve

The notation XN(μ,σ2)X\sim N(\mu,\sigma^2) means XX is normally distributed with mean μ\mu and variance σ2\sigma^2 (so standard deviation σ\sigma).

The empirical rule (68-95-99.7)

For any normal distribution, fixed proportions of the data fall within whole numbers of standard deviations of the mean:

Because the curve is symmetric, each tail beyond ±1σ\pm 1\sigma holds about 16%16\%, beyond ±2σ\pm 2\sigma about 2.5%2.5\%, and beyond ±3σ\pm 3\sigma about 0.15%0.15\%.

Common errors

Why it matters

The normal distribution underpins the rest of Topic 5 (z-scores and exact probabilities) and all of Topic 6 (sampling distributions and confidence intervals), where sample means are approximately normal. Recognising when data is normal and using the empirical rule for quick estimates is a core Stage 2 skill.

Exam-style practice questions

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

2024 SACE Stage 22 marksLet X be a normally distributed random variable with a mean of 54 and a standard deviation of 6. Determine (i) Pr(50 < X < 52) and (ii) Pr(X > 58).
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Use the normal distribution with mean 54 and standard deviation 6 (calculator's normal CDF, or standardise to z = (x - 54) / 6).

(i) Pr(50 < X < 52): standardising, z goes from (50 - 54)/6 = -0.667 to (52 - 54)/6 = -0.333. The probability is the area between these, Pr(-0.667 < Z < -0.333) = 0.117 (to three significant figures). (1 mark)

(ii) Pr(X > 58): z = (58 - 54)/6 = 0.667, so Pr(Z > 0.667) = 1 - 0.7475 = 0.252 (to three significant figures). (1 mark)

Markers accept direct calculator values to three significant figures. A useful check: both 50-52 and 58 lie within one standard deviation of the mean, so neither probability is extreme.

2023 SACE Stage 21 marksThe daily milk production of a randomly chosen cow is normally distributed with a mean of 21.9 litres and a standard deviation of 3.26 litres. Determine the probability that the daily milk production of a randomly selected cow is less than 23 litres.
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Standardise the value 23 using z = (x - mean) / sd:

z = (23 - 21.9) / 3.26 = 1.1 / 3.26 = 0.337.

Then Pr(X < 23) = Pr(Z < 0.337) = 0.632 (to three significant figures), read from the standard normal distribution or computed directly with the calculator's normal CDF.

So the probability is 0.632. The single mark is for the correct probability. Because 23 is just above the mean of 21.9, the answer should be a little more than 0.5, which it is.