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SAGeneral MathematicsSyllabus dot point

How do we describe and compare values within a bell-shaped distribution?

Use the normal distribution, the 68-95-99.7 rule and z-scores to describe data and compare values from different distributions.

How to apply the 68-95-99.7 rule, calculate z-scores to standardise values, find proportions and compare results from different normal distributions.

Generated by Claude Opus 4.77 min answer

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  1. What this dot point is asking
  2. The shape of the normal distribution
  3. Z-scores: standardising a value
  4. Comparing values from different distributions
  5. Finding proportions and percentiles

What this dot point is asking

You must apply the 68-95-99.7 rule, compute and interpret z-scores, find proportions, and use z-scores to compare values from different distributions.

The shape of the normal distribution

A normal distribution is symmetric about its mean μ\mu, with most data clustered near the centre and the curve tailing off either side. Its spread is set by the standard deviation σ\sigma. The mean, median and mode all sit at the centre.

Z-scores: standardising a value

A z-score rescales any value so it can be compared on a common scale. It tells you how many standard deviations the value lies above (positive) or below (negative) the mean.

z=x−μσz = \frac{x - \mu}{\sigma}

Comparing values from different distributions

Z-scores let you compare results measured on different scales, because they convert everything to "standard deviations from the mean."

Finding proportions and percentiles

Beyond the round numbers of the empirical rule, a calculator gives the proportion below any z-score, which is the percentile. For example, z=1.5z = 1.5 corresponds to about the 93rd percentile, meaning roughly 93% of values fall below it.

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.

2019 SACE Stage 22 marksTennis service times are normally distributed with a mean of 20.3 seconds and a standard deviation of 2.4 seconds. Calculate the longest service time that would be within the fastest 2% of service times.
Show worked answer →

The "fastest 2%" means the shortest 2% of service times, so we need the value with 2% of the area to its left (the 2nd percentile).

The z-score with 0.02 of the area below it is z = -2.054 (from the inverse normal).

Convert back using x = mean + z times standard deviation:
x = 20.3 + (-2.054)(2.4) = 20.3 - 4.93 = 15.37 seconds.

So the longest service time within the fastest 2% is about 15.4 seconds. Award 1 mark for finding the correct boundary using the inverse normal (mean 20.3, sd 2.4, lower 2%), and 1 mark for the answer of approximately 15.4 seconds.

2021 SACE Stage 22 marksBrand X smartphone battery life is normally distributed with a mean of 16.3 hours and a standard deviation of 2.4 hours. If 834 batteries were tested, how many could be expected to last less than 15 hours?
Show worked answer →

First find the proportion lasting less than 15 hours.

z = (15 - 16.3) / 2.4 = -1.3 / 2.4 = -0.542.

The area to the left of z = -0.542 is about 0.2939 (using the normal distribution, P(t < 15)).

Then multiply by the number tested:
0.2939 times 834 = 245.1.

So about 245 batteries would be expected to last less than 15 hours. Award 1 mark for the correct proportion (about 0.294) and 1 mark for multiplying by 834 to get approximately 245 batteries (rounded to a whole number).

2022 SACE Stage 23 marksVanilla candles burn for a mean of 252 minutes (sd 20). Rose candles burn for a mean of 245 minutes (sd 25). The company claims a greater proportion of rose candles than vanilla candles will burn longer than 300 minutes. Is the claim accurate? Support your answer with appropriate calculations and reasoning.
Show worked answer →

Find P(burn time > 300) for each candle using z-scores.

Vanilla: z = (300 - 252) / 20 = 48 / 20 = 2.4, so P(X > 300) = 0.0082 (about 0.82%).

Rose: z = (300 - 245) / 25 = 55 / 25 = 2.2, so P(X > 300) = 0.0139 (about 1.39%).

Since 1.39% (rose) is greater than 0.82% (vanilla), a greater proportion of rose candles burn longer than 300 minutes. The claim is accurate.

Award marks for both correct z-scores and probabilities (2 marks) and the comparison and conclusion that the claim is supported (1 mark). Even though rose has a lower mean, its larger standard deviation puts more of its distribution beyond 300 minutes.