Skip to main content
ExamExplained

The normal distribution and z-scores (SACE Stage 2 General Mathematics)

Syllabus dot point

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

8 min read

Quick answer

The normal distribution is the symmetric bell curve. The 68-95-99.7 rule gives the proportion of data within 1, 2 and 3 standard deviations of the mean, and a z-score z=x−μσz = \dfrac{x - \mu}{\sigma} measures how many standard deviations a value sits from the mean.

Jump to a section
  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. The inverse problem: from proportion to value
  6. Comparing groups with different spreads
  7. Finding proportions and percentiles
  8. Exam-style questions

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.

The 68-95-99.7 rule

For a normal distribution, about 68% of values lie within 1 standard deviation of the mean, about 95% within 2 standard deviations, and about 99.7% within 3 standard deviations.

Worked example

Applying the empirical rule

Adult resting heart rates are normally distributed with mean μ=70\mu = 70 and standard deviation σ=8\sigma = 8 beats per minute. What proportion lie between 54 and 86?

The interval 5454 to 8686 is 70±16=70±2(8)70 \pm 16 = 70 \pm 2(8), exactly 2 standard deviations either side of the mean. By the rule, about 95% of adults have a resting heart rate between 54 and 86 beats per minute.

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}

Z-score

z=x−μσz = \frac{x - \mu}{\sigma}

where xx is the value, μ\mu is the mean and σ\sigma is the standard deviation. A positive zz is above the mean, a negative zz is below.

Worked example

Calculating a z-score

A student scores 82 on a test with mean 70 and standard deviation 6. Find the z-score.

z=82−706=126=2z = \frac{82 - 70}{6} = \frac{12}{6} = 2

The score is exactly 2 standard deviations above the mean, a strong result that beats about 97.5% of students (since 95% lie within 2 SDs, leaving 2.5% above z=2z = 2).

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."

Worked example

Which result is better?

Mia scores 78 in English (mean 70, SD 8) and 65 in Maths (mean 55, SD 5). Relative to her class, which is the stronger result?

zEng=78−708=1.0,zMaths=65−555=2.0z_{\text{Eng}} = \frac{78 - 70}{8} = 1.0, \qquad z_{\text{Maths}} = \frac{65 - 55}{5} = 2.0

Her Maths z-score (2.0) is higher than her English z-score (1.0), so relative to each class her Maths result is the stronger performance, even though the raw English mark is higher.

The inverse problem: from proportion to value

Many SACE questions run the standardisation backwards: you are told a proportion or percentile and must find the corresponding value. Use the inverse normal to get the zz-score for that area, then unstandardise with x=μ+zσx = \mu + z\sigma. For example, the cut-off for the top 10% of a distribution has z≈1.2816z \approx 1.2816, and substituting the mean and standard deviation gives the actual boundary value. Watch the direction: "the slowest 2%" or "the fastest 2%" both refer to a lower-tail area, while "the top 5%" is an upper-tail area, so the sign of zz must match the side of the distribution the question describes.

Comparing groups with different spreads

A subtle but examinable point is that a larger standard deviation puts more of a distribution into the tails. Two groups with similar means can have very different proportions beyond a high threshold if their spreads differ, with the more variable group having more extreme values at both ends. This is why a candle type with a lower mean burn time can still have a greater proportion burning beyond a long target, as in the candle question above. Comparing zz-scores at the threshold, rather than comparing means, is the reliable way to settle such claims.

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.

Sketch the curve first

Draw the bell, mark the mean and shade the region you want before calculating. A sketch stops you from finding the area on the wrong side of the value, which is the most common slip in these questions.

Common mistake

Do not forget the order in the z-score formula: it is x−μx - \mu, not μ−x\mu - x. Getting this backwards flips the sign and turns an above-average score into a below-average one. Also remember the 68-95-99.7 percentages apply to the region within those standard deviations; to find a one-sided tail you must halve the leftover.

Exam-style questions

Questions in the style of SACE Board exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

SACE 2022-style2 marks
Calculator-assumed. Tennis service times are normally distributed with mean 20.320.3 seconds and standard deviation 2.42.4 seconds. Calculate the longest service time within the fastest 2% of service times.
Show worked answer →

The "fastest 2%" is the shortest 2%, so find the value with 0.020.02 of the area to its left (the 2nd percentile).

The zz-score with 0.020.02 below it is z=−2.054z = -2.054. Unstandardise with x=μ+zσx = \mu + z\sigma:

x=20.3+(−2.054)(2.4)=20.3−4.93≈15.4 seconds.x = 20.3 + (-2.054)(2.4) = 20.3 - 4.93 \approx 15.4 \text{ seconds}.

Marks: one for the inverse-normal boundary, one for the value ≈15.4\approx 15.4 seconds.

SACE 2021-style2 marks
Calculator-assumed. Brand X battery life is normally distributed with mean 16.316.3 hours and standard deviation 2.42.4 hours. If 834 batteries were tested, how many would be expected to last less than 15 hours?
Show worked answer →

Standardise: z=15−16.32.4=−1.32.4≈−0.542z = \dfrac{15 - 16.3}{2.4} = \dfrac{-1.3}{2.4} \approx -0.542, so P(X<15)≈0.2939P(X < 15) \approx 0.2939. [1 mark]

Multiply by the number tested: 0.2939×834≈2450.2939 \times 834 \approx 245 batteries. [1 mark]

So about 245 batteries would be expected to last less than 15 hours.

SACE 2023-style3 marks
Calculator-assumed. Vanilla candles burn for a mean of 252 minutes (SD 20). Rose candles burn for a mean of 245 minutes (SD 25). A company claims a greater proportion of rose than vanilla candles burn longer than 300 minutes. Test the claim with calculations.
Show worked answer →

Vanilla: z=300−25220=2.4z = \dfrac{300 - 252}{20} = 2.4, so P(X>300)≈0.0082P(X > 300) \approx 0.0082 (about 0.82%).

Rose: z=300−24525=2.2z = \dfrac{300 - 245}{25} = 2.2, so P(X>300)≈0.0139P(X > 300) \approx 0.0139 (about 1.39%). [2 marks]

Since 1.39%>0.82%1.39\% > 0.82\%, a greater proportion of rose candles burn beyond 300 minutes, so the claim is accurate. Despite the lower mean, rose's larger standard deviation puts more of its distribution beyond 300. [1 mark]

Practise this

Sources & how we know this

Based on the official SACE Board subject outline. Not yet individually reviewed by a person; always check SACE Board and your school's materials for current rules.

ExamExplained