z-scores, standardisation and comparing normal distributions for HSC Maths Standard 2

Syllabus dot point

“Calculate z-scores and use them to compare values from different normal distributions and find probabilities”

HSCMaths Standard 2Year 12: Statistical Analysis12 min readUpdated 20 Jun 2026

Quick answer

A z-score $z = (x - \mu)/\sigma$ measures how many standard deviations a value is from the mean; compare observations from different normal distributions by their z-scores, with the higher z-score meaning further above the mean relative to the local spread, and reverse with $x = \mu + z\sigma$ to find the value at a given percentile.

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What this dot point is asking
The answer
Exam-style questions
Practice questions

What this dot point is asking

NESA wants you to do four things. Compute a z-score for any value from a normal distribution (a bell-shaped spread of data). Read that z-score as "how many standard deviations above or below the mean". Reverse the process to find a value from a given z-score or percentile. And use z-scores to compare observations that come from different normal distributions. The z-score is the bridge between a raw measurement and one common scale on which every normal distribution looks the same.

The answer

Note

Say two friends sit different tests and you want to know who really did better. You cannot just compare the marks, because one test might be much harder than the other. A z-score fixes this. It tells you how far above or below the average each person scored, measured in "steps" of the usual spread of marks. Once both scores are turned into z-scores, they sit on the same fair scale, so the bigger z-score is the better effort. The same trick lets you compare any results, even things measured in different units like centimetres and dollars.

The z-score formula

For a value $x$ from a normal distribution with mean $\mu$ and standard deviation $\sigma$ :

z = \frac{x - \mu}{\sigma}.

The z-score is the number of standard deviations $x$ lies from the mean. It is negative when $x$ is below the mean and positive when above. Subtracting $\mu$ shifts the data so the mean sits at $0$ . Dividing by $\sigma$ then rescales it so one standard deviation becomes one unit. The result is a pure number with no units. A height in cm and a salary in dollars both become plain z-scores, and that is exactly why they can then be compared.

Standardising a value, step by step

Standardising means carrying a raw value across to the common z-scale. The diagram below maps a test mark of $85$ (on a test with $\mu = 70$ , $\sigma = 12$ ) down to its z-score. The two parts of the formula are two moves on the picture. Subtracting the mean slides the marked value across relative to the centre. Dividing by $\sigma$ then measures that gap in standard-deviation steps.

Here $\dfrac{85 - 70}{12} = \dfrac{15}{12} = 1.25$ , so the mark sits $1.25$ standard deviations above the mean.

Key fact

The z-score $z = \dfrac{x - \mu}{\sigma}$ is the number of standard deviations between $x$ and the mean: subtract the mean, then divide by the standard deviation. It is signed (negative below the mean, positive above) and unitless, so values from completely different distributions become directly comparable. Reverse it with $x = \mu + z\sigma$ to recover the raw value at a given z-score.

Reverse: finding $x$ from a z-score

Rearranging the formula gives the raw value back:

x = \mu + z\sigma.

This is the move for percentile questions. Given a target z-score (often supplied, or read from the table), it returns the actual height, mark or salary at that position. The two forms are a matched pair, one going each way. The first formula, $z = (x-\mu)/\sigma$ , turns a value into a z-score. The second, $x = \mu + z\sigma$ , turns a z-score back into a value.

Interpretation

$z = 0$ : the value equals the mean.
$z = 1$ : one standard deviation above the mean (top $16\%$ of a normal distribution).
$z = -2$ : two standard deviations below the mean (bottom $2.5\%$ ).
$|z| > 3$ : a very extreme value; under $0.3\%$ of data sit this far from the mean.

A z-score immediately tells you how rare a value is. Anything past $\pm 2$ is uncommon, and past $\pm 3$ is rare. That reading is the same for every normal distribution. This works because the empirical rule percentages (the $68$ , $95$ and $99.7$ figures) attach to the z-scale, not to the raw values.

Comparing values from different distributions

This is the headline use of z-scores and a favourite exam question. Two raw scores from different tests cannot be compared head to head, because the tests have different means and spreads. Convert each to a z-score and the comparison becomes easy. The higher z-score is the better result relative to its own group. Watch the comparison happen in three stages below.

Stage 1, the raw marks are not comparable. Anika scores $76$ on a test with $\mu = 68$ , $\sigma = 6$ ; Ben scores $82$ on a test with $\mu = 76$ , $\sigma = 8$ . Ben's raw mark is higher, but the two tests sit on different scales, so the raw numbers cannot settle who did better.

Stage 2, standardise each within its own distribution. Convert each mark using its own mean and standard deviation. Anika: $\dfrac{76 - 68}{6} = \dfrac{8}{6} \approx 1.33$ . Ben: $\dfrac{82 - 76}{8} = \dfrac{6}{8} = 0.75$ . Each z-score says how many standard deviations the mark is above its own mean.

Stage 3, compare on one common z-scale. Place both z-scores on a single axis. Anika at $1.33$ sits to the right of Ben at $0.75$ , so despite the lower raw mark, Anika performed better relative to her cohort. On the z-scale, further right always means better relative standing.

The same logic compares salaries across industries, sports results against different reference groups, or a child's height against age-specific charts. In each case, standardise within each distribution first, then compare the z-scores.

Z-scores and percentiles

Common percentile-to-z-score values from the standard normal table:

Percentile	z-score
50th	$0$
75th	$0.67$
90th	$1.28$
95th	$1.65$
97.5th	$1.96$
99th	$2.33$

Because the bell curve is symmetric, a lower percentile uses the negative of the matching upper z-score. The 10th percentile is at $z = -1.28$ , the 5th at $z = -1.65$ , and the 2.5th at $z = -1.96$ . So once you can do the top tail, the bottom tail is just a sign change.

Z-scores and probabilities

To find the probability that an observation is less than some value $x$ :

Compute $z = (x - \mu)/\sigma$ .
Look up $\Phi(z)$ , the area to the left of $z$ , in the standard normal table provided in the HSC paper.
$P(X \le x) = \Phi(z)$ .

For the other directions:

$P(X > x) = 1 - \Phi(z)$ (the right tail).
$P(a \le X \le b) = \Phi(z_b) - \Phi(z_a)$ (subtract the two left-areas).

When the endpoints are whole standard deviations from the mean ( $z = \pm 1, 2, 3$ ), you can skip the table and read the percentage straight off the empirical rule.

How exam questions ask about z-scores

The wording varies; the method does not. Translate the question:

"Calculate the z-score / standardise the value." Apply $z = (x - \mu)/\sigma$ directly and state the answer with its sign.
"Who performed better relative to their group?" A comparison: compute a z-score for each, then say the larger z-score is better and why ("further above its own mean").
"Find the value at the $p$ th percentile" or "... the top $5\%$ cutoff". Reverse the formula: get the z-score for that percentile (from the table or given), then $x = \mu + z\sigma$ .
"What percentage / probability lie above (or below) $x$ ?" Standardise, then use $\Phi(z)$ from the table, or the empirical rule if $z$ is a whole number.
"Between what two values do the middle $90\%$ lie?" Symmetric percentiles: the middle $90\%$ runs from $z = -1.65$ to $z = +1.65$ , so compute $\mu \pm 1.65\sigma$ .
"Is this value unusual?" Read the size of $|z|$ : beyond $2$ is uncommon, beyond $3$ is rare.

Exam technique

Quote the formula $z = \dfrac{x - \mu}{\sigma}$ before substituting; markers award a method mark for the correct formula even if the arithmetic slips. Keep the sign: a value below the mean has a negative z-score, and dropping the minus is the most common error. For a comparison question, compute both z-scores, then write an explicit sentence naming who did better and why ("higher z-score, so further above the mean"); the conclusion carries its own mark. For a percentile or cutoff, use $x = \mu + z\sigma$ and give the answer to the accuracy the data warrants (usually one decimal place), with units.

Worked example

Compute and interpret a z-score

A student scores $85$ on a test with $\mu = 70$ , $\sigma = 12$ .

Apply the z-score formula:

z = \frac{x - \mu}{\sigma} = \frac{85 - 70}{12} = \frac{15}{12} = 1.25.

Interpret the z-score: The student is $1.25$ standard deviations above the mean. From the standard normal table, $\Phi(1.25) \approx 0.8944$ , so about $89.4\%$ of students scored lower: the student is at roughly the $89$ th percentile.

Answer: $z = 1.25$ , at about the $89$ th percentile.

Compare salaries across two industries

A Sydney accountant earns $ $135\,000$ in an industry with mean $ $110\,000$ and SD $ $20\,000$ . A Melbourne consultant earns $ $120\,000$ in an industry with mean $ $95\,000$ and SD $ $15\,000$ . Who earns more relative to their own industry?

Compute a z-score for each, using its own distribution:

z_{\text{accountant}} = \frac{135\,000 - 110\,000}{20\,000} = \frac{25\,000}{20\,000} = 1.25.

z_{\text{consultant}} = \frac{120\,000 - 95\,000}{15\,000} = \frac{25\,000}{15\,000} \approx 1.67.

Compare on the common z-scale: The consultant's z-score is higher, near the $95$ th percentile of their industry, while the accountant is near the $89$ th. Despite the smaller dollar salary, the consultant ranks higher relative to their own cohort.

Answer: the consultant ( $z \approx 1.67$ ) ranks higher than the accountant ( $z = 1.25$ ).

Find the cutoff for the top 5 percent

For exam marks that are normally distributed with $\mu = 65$ , $\sigma = 12$ , find the cutoff score for the top $5\%$ .

Identify the z-score for the percentile: The top $5\%$ is the $95$ th percentile. From the table, $z = 1.65$ .

De-standardise with $x = \mu + z\sigma$ :

x = 65 + 1.65 \times 12 = 65 + 19.8 = 84.8.

Answer: the cutoff for the top $5\%$ is a mark of about $84.8$ .

Find a symmetric pair of percentiles

Heights of Year 12 girls are normally distributed with $\mu = 164$ cm, $\sigma = 6$ cm. Find the heights at the 90th and 10th percentiles (use $z = 1.28$ for the 90th).

The 90th percentile:

x = \mu + z\sigma = 164 + 1.28 \times 6 = 164 + 7.68 = 171.68 \approx 171.7 \text{ cm}.

The 10th percentile by symmetry: The 10th percentile is the mirror image, so $z = -1.28$ :

x = 164 + (-1.28)(6) = 164 - 7.68 = 156.32 \approx 156.3 \text{ cm}.

Answer: the 90th percentile is about $171.7$ cm and the 10th about $156.3$ cm.

Common traps

Sign error: $z = \dfrac{x - \mu}{\sigma}$ , not the reverse, and a value below the mean gives a negative z-score. Keep the minus.
Confusing $\sigma$ and $\sigma^2$: Use the standard deviation, not the variance. Square-root the variance first if that is what you are given.
Forgetting both tails: $P(|Z| > 1)$ is both tails: $2 \times 16\% = 32\%$ , not $16\%$ . Read whether one tail or two is wanted.
Misreading the table: $\Phi(z)$ is the area to the left, $P(Z \le z)$ . For "greater than", use $1 - \Phi(z)$ .
Comparing raw scores instead of z-scores: Across different distributions, the bigger raw number is not necessarily the better result; standardise first.
Using z-scores on non-normal data: Z-scores and the table percentages assume an approximately normal distribution; they do not apply to clearly skewed data.

Exam-style questions

Questions in the style of NESA 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.

2022 HSC-style4 marks

Two students sit different tests. Anika scores

76

on a test with

\mu = 68

and

\sigma = 6

. Ben scores

82

on a test with

\mu = 76

and

\sigma = 8

. Which student performed better relative to their cohort?

Show worked answer →

Compute each z-score: $z = \frac{x - \mu}{\sigma}$ .

Anika: $z = \frac{76 - 68}{6} = \frac{8}{6} \approx 1.33$ .

Ben: $z = \frac{82 - 76}{8} = \frac{6}{8} = 0.75$ .

Anika's z-score is higher, so she performed better relative to her cohort (further above the mean in standard-deviation units), even though Ben's raw score is higher.

Markers reward both z-scores, comparison, and the final conclusion with the reason ("higher z-score means further above the mean").

2023 HSC-style4 marks

Heights of Year 12 girls at a school are normally distributed with mean

164

cm and standard deviation

6

cm. Find the height that corresponds to (a) the 90th percentile, and (b) the 10th percentile. (Use

z = 1.28

for the 90th percentile.)

Show worked answer →

(a) For the 90th percentile, $z = 1.28$ . Use $x = \mu + z\sigma$ .

$x = 164 + 1.28 \times 6 = 164 + 7.68 = 171.68$ cm. Round to $171.7$ cm.

(b) By symmetry, the 10th percentile has $z = -1.28$ .

$x = 164 + (-1.28) \times 6 = 164 - 7.68 = 156.32$ cm. Round to $156.3$ cm.

Markers reward the formula rearrangement, the symmetric z-score for the lower tail, and answers to one decimal place.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation1 mark

A runner's

100

m time is

11.2

seconds. The times for her squad are normally distributed with mean

\mu = 12.0

seconds and standard deviation

\sigma = 0.4

seconds. Calculate her z-score.

Show worked solution →

Apply the z-score formula. Substitute $x = 11.2$ , $\mu = 12.0$ and $\sigma = 0.4$ into $z = \frac{x - \mu}{\sigma}$ :

z = \frac{11.2 - 12.0}{0.4} = \frac{-0.8}{0.4} = -2.00.

Keep the sign. Her time is below the mean, so the z-score is negative. (For a race, a faster time below the mean is a good result.)

Answer: $z = -2.00$ .

foundation2 marks

A student's NAPLAN-style numeracy score is

620

. The scores are normally distributed with mean

\mu = 550

and standard deviation

\sigma = 50

. (a) Calculate the z-score. (b) State how many standard deviations the score is above the mean.

Show worked solution →

Apply the z-score formula (a). Substitute $x = 620$ , $\mu = 550$ and $\sigma = 50$ into $z = \frac{x - \mu}{\sigma}$ :

z = \frac{620 - 550}{50} = \frac{70}{50} = 1.40.

Interpret the size (b). A z-score of $1.40$ means the score sits $1.40$ standard deviations above the mean.

Answer: (a) $z = 1.40$ ; (b) $1.40$ standard deviations above the mean.

foundation2 marks

A class test is normally distributed with mean

\mu = 72

and standard deviation

\sigma = 8

. A mark has a z-score of

z = 1.5

. Find the mark.

Show worked solution →

Choose the reverse formula. A z-score is given and the raw mark is wanted, so de-standardise with $x = \mu + z\sigma$ .

Substitute the values. Use $\mu = 72$ , $z = 1.5$ and $\sigma = 8$ :

x = 72 + 1.5 \times 8 = 72 + 12 = 84.

Answer: the mark is $84$ .

foundation2 marks

IQ scores are normally distributed with mean

\mu = 100

and standard deviation

\sigma = 15

. (a) What z-score corresponds to the 50th percentile? (b) Use

x = \mu + z\sigma

to find the IQ score at the 50th percentile.

Show worked solution →

Recall the 50th percentile z-score (a). The 50th percentile is the middle of a symmetric normal distribution, which sits exactly at the mean, so $z = 0$ .

De-standardise (b). Substitute $\mu = 100$ , $z = 0$ and $\sigma = 15$ into $x = \mu + z\sigma$ :

x = 100 + 0 \times 15 = 100.

Answer: (a) $z = 0$ ; (b) the 50th percentile IQ score is $100$ (the mean).

core4 marks

Two swimmers each win their own event at a carnival, where a lower time is better. Mia swims the

50

m event in

26.8

seconds; that event has mean

\mu = 28.0

s and standard deviation

\sigma = 0.8

s. Zoe swims the

100

m event in

55.2

seconds; that event has mean

\mu = 57.0

s and standard deviation

\sigma = 1.0

s. Whose swim was better relative to her own event?

Show worked solution →

Standardise Mia's time. Use $z = \frac{x - \mu}{\sigma}$ with her own event's mean and standard deviation:

z_{\text{Mia}} = \frac{26.8 - 28.0}{0.8} = \frac{-1.2}{0.8} = -1.50.

Standardise Zoe's time. Use Zoe's event:

z_{\text{Zoe}} = \frac{55.2 - 57.0}{1.0} = \frac{-1.8}{1.0} = -1.80.

Compare, remembering lower is better. Both times are below their means (good for a race). Zoe's z-score of $-1.80$ is further below the mean than Mia's $-1.50$ , so Zoe is more standard deviations faster than her field.

Answer: Zoe's swim ( $z = -1.80$ ) was better relative to her event than Mia's ( $z = -1.50$ ).

core3 marks

Marks on a chemistry exam are normally distributed with mean

\mu = 68

and standard deviation

\sigma = 10

. A scholarship is offered to students in the top

25\%

. Using

z = 0.67

for the 75th percentile, find the lowest mark that earns the scholarship.

Show worked solution →

Identify the percentile (a). The top $25\%$ begins at the 75th percentile, which has $z = 0.67$ .

De-standardise with $x = \mu + z\sigma$ . Substitute $\mu = 68$ , $z = 0.67$ and $\sigma = 10$ :

x = 68 + 0.67 \times 10 = 68 + 6.7 = 74.7.

Interpret. A mark of $74.7$ marks the boundary, so a student needs about $75$ to be safely in the top $25\%$ .

Answer: the cutoff for the top $25\%$ is a mark of about $74.7$ .

core3 marks

Daniel scores

46

on a Paper 1 that is normally distributed with mean

\mu = 40

and standard deviation

\sigma = 4

. His teacher wants the equivalent mark on Paper 2, which is normally distributed with mean

\mu = 60

and standard deviation

\sigma = 10

. (a) Find Daniel's z-score on Paper 1. (b) Find the Paper 2 mark with the same z-score.

Show worked solution →

Standardise the Paper 1 mark (a). Use $z = \frac{x - \mu}{\sigma}$ with $x = 46$ , $\mu = 40$ , $\sigma = 4$ :

z = \frac{46 - 40}{4} = \frac{6}{4} = 1.50.

De-standardise onto Paper 2 (b). The equivalent mark keeps the same z-score, so use $x = \mu + z\sigma$ with Paper 2's $\mu = 60$ , $\sigma = 10$ and $z = 1.50$ :

x = 60 + 1.50 \times 10 = 60 + 15 = 75.

Answer: (a) $z = 1.50$ ; (b) the equivalent Paper 2 mark is $75$ .

core3 marks

The resting heart rates of a group of athletes are normally distributed with mean

\mu = 72

beats per minute and standard deviation

\sigma = 4

beats per minute. Using

z = \pm 1.65

for the middle

90\%

, find the two heart rates between which the middle

90\%

of athletes lie.

Show worked solution →

Set up the symmetric pair. The middle $90\%$ runs from $z = -1.65$ to $z = +1.65$ , so apply $x = \mu + z\sigma$ at each end with $\mu = 72$ and $\sigma = 4$ .

Lower bound ( $z = -1.65$ ).

x = 72 + (-1.65)(4) = 72 - 6.6 = 65.4.

Upper bound ( $z = +1.65$ ).

x = 72 + 1.65 \times 4 = 72 + 6.6 = 78.6.

Answer: the middle $90\%$ of athletes have resting heart rates between $65.4$ and $78.6$ beats per minute.

exam4 marks

In her HSC trials, Priya scores

82

in Mathematics, which is normally distributed with mean

\mu = 70

and standard deviation

\sigma = 9

. She scores

75

in English, which is normally distributed with mean

\mu = 66

and standard deviation

\sigma = 6

. (a) Calculate her z-score in each subject, correct to two decimal places. (b) State the subject in which she performed better relative to her cohort, and justify your answer.

Show worked solution →

Standardise the Mathematics mark (a). Use $z = \frac{x - \mu}{\sigma}$ with $x = 82$ , $\mu = 70$ , $\sigma = 9$ :

z_{\text{Maths}} = \frac{82 - 70}{9} = \frac{12}{9} \approx 1.33.

Standardise the English mark (a). Use $x = 75$ , $\mu = 66$ , $\sigma = 6$ :

z_{\text{English}} = \frac{75 - 66}{6} = \frac{9}{6} = 1.50.

Compare and justify (b). English has the higher z-score ( $1.50$ against $1.33$ ), so Priya sits more standard deviations above the mean in English. She performed better in English relative to her cohort, even though her raw Mathematics mark is higher.

Answer: (a) Maths $z \approx 1.33$ , English $z = 1.50$ ; (b) English, because its higher z-score puts her further above her cohort's mean.

exam5 marks

Annual salaries in finance are normally distributed with mean $

104\,000

and standard deviation $

16\,000

. Salaries in technology are normally distributed with mean $

98\,000

and standard deviation $

14\,000

. Sam earns $

128\,000

in finance; Lee earns $

121\,000

in technology. (a) Find each z-score, to two decimal places. (b) State who earns more relative to their own industry. (c) Given

\Phi(1.64) = 0.9495

, find the percentage of technology workers who earn more than Lee.

Show worked solution →

Standardise Sam's salary (a). Use $z = \frac{x - \mu}{\sigma}$ for finance:

z_{\text{Sam}} = \frac{128\,000 - 104\,000}{16\,000} = \frac{24\,000}{16\,000} = 1.50.

Standardise Lee's salary (a). Use technology:

z_{\text{Lee}} = \frac{121\,000 - 98\,000}{14\,000} = \frac{23\,000}{14\,000} \approx 1.64.

Compare (b): Lee's z-score ( $1.64$ ) is higher than Sam's ( $1.50$ ), so Lee earns more relative to his own industry despite the smaller dollar salary.
Use the table for the tail (c): The area below Lee is $\Phi(1.64) = 0.9495$ , so the proportion above is $1 - 0.9495 = 0.0505$ .
Answer: (a) Sam $z = 1.50$ , Lee $z \approx 1.64$ ; (b) Lee; (c) about $5.05\%$ of technology workers earn more than Lee.

exam6 marks

The heights of adult men in a town are normally distributed with mean

\mu = 178

cm and standard deviation

\sigma = 7

cm. (a) Calculate the z-score of a man who is

192

cm tall. (b) Using

z = 1.65

for the 95th percentile, find the height that a man must exceed to be in the tallest

5\%

. (c) Given

\Phi(1.00) = 0.8413

, find the probability that a randomly chosen man is taller than

185

cm.

Show worked solution →

Standardise the height (a). Use $z = \frac{x - \mu}{\sigma}$ with $x = 192$ , $\mu = 178$ , $\sigma = 7$ :

z = \frac{192 - 178}{7} = \frac{14}{7} = 2.00.

Find the top $5\%$ cutoff (b). The tallest $5\%$ start at the 95th percentile, $z = 1.65$ . De-standardise with $x = \mu + z\sigma$ :

x = 178 + 1.65 \times 7 = 178 + 11.55 = 189.55 \text{ cm}.

Find the right-tail probability (c). First standardise $185$ cm: $z = \frac{185 - 178}{7} = \frac{7}{7} = 1.00$ . The area below is $\Phi(1.00) = 0.8413$ , so the area above is $1 - 0.8413 = 0.1587$ .

Answer: (a) $z = 2.00$ ; (b) a man must exceed about $189.55$ cm; (c) $P(X > 185) = 0.1587$ , about $15.87\%$ .

exam5 marks

The life of a brand of phone battery is normally distributed with mean

\mu = 18

hours and standard deviation

\sigma = 2.5

hours. Use the values

\Phi(0.80) = 0.7881

and

\Phi(2.00) = 0.9772

, and recall that

\Phi(-z) = 1 - \Phi(z)

. (a) Find the z-score of a battery lasting

13

hours and of one lasting

20

hours, each to two decimal places. (b) Find the probability that a battery lasts between

13

and

20

hours. (c) Hence find the probability that a battery lasts either under

13

hours or over

20

hours. (d) In a delivery of

500

batteries, about how many would you expect to last between

13

and

20

hours?

Show worked solution →

Standardise both endpoints (a). Apply $z = \frac{x - \mu}{\sigma}$ with $\mu = 18$ and $\sigma = 2.5$ at each value:

z_{13} = \frac{13 - 18}{2.5} = \frac{-5}{2.5} = -2.00, \qquad z_{20} = \frac{20 - 18}{2.5} = \frac{2}{2.5} = 0.80.

Convert the lower tail with the symmetry rule (b). The interval $13 \le X \le 20$ becomes $-2.00 \le z \le 0.80$ . The area below the upper z is $\Phi(0.80) = 0.7881$ . For the lower z, use $\Phi(-2.00) = 1 - \Phi(2.00) = 1 - 0.9772 = 0.0228$ .

Subtract to get the between-area (b). The probability between the two values is the difference of the two areas:

P(13 \le X \le 20) = 0.7881 - 0.0228 = 0.7653.

Take the complement for the two outer tails (c). Lasting under $13$ or over $20$ hours is everything outside the interval, so subtract from $1$ :

P(\text{outside}) = 1 - 0.7653 = 0.2347.

Scale to the delivery (d). Multiply the between-probability by the number of batteries: $500 \times 0.7653 = 382.65$ , so about $383$ batteries.

Answer: (a) $z_{13} = -2.00$ , $z_{20} = 0.80$ ; (b) $P(13 \le X \le 20) = 0.7653$ (about $76.53\%$ ); (c) $0.2347$ (about $23.47\%$ ); (d) about $383$ batteries.

exam6 marks

A machine stamps out metal washers whose thickness (in microns) is normally distributed with unknown mean

\mu

and unknown standard deviation

\sigma

. Quality records show that the 92nd percentile thickness is

505

microns (for which

z = 1.40

) and the 31st percentile thickness is

486

microns (for which

z = -0.50

). (a) By writing

x = \mu + z\sigma

at each percentile, set up two equations and find

\sigma

. (b) Hence find

\mu

. (c) Using

z = \pm 1.96

for the middle

95\%

, find the two thicknesses between which the middle

95\%

of washers lie.

Show worked solution →

Write an equation at each percentile (a). Substitute each known thickness and z-score into $x = \mu + z\sigma$ :

505 = \mu + 1.40\,\sigma, \qquad 486 = \mu - 0.50\,\sigma.

Eliminate $\mu$ by subtracting (a). Subtracting the second equation from the first cancels $\mu$ :

505 - 486 = (1.40 - (-0.50))\,\sigma \implies 19 = 1.90\,\sigma \implies \sigma = \frac{19}{1.90} = 10.

Back-substitute for the mean (b). Put $\sigma = 10$ into the first equation:

\mu = 505 - 1.40 \times 10 = 505 - 14 = 491.

A quick check with the second equation agrees: $486 + 0.50 \times 10 = 491$ .

Build the symmetric middle- $95\%$ interval (c). The central $95\%$ runs from $z = -1.96$ to $z = +1.96$ , so apply $x = \mu + z\sigma$ at each end with $\mu = 491$ and $\sigma = 10$ :

x_{\text{low}} = 491 + (-1.96)(10) = 491 - 19.6 = 471.4,

x_{\text{high}} = 491 + 1.96 \times 10 = 491 + 19.6 = 510.6.

Answer: (a) $\sigma = 10$ microns; (b) $\mu = 491$ microns; (c) the middle $95\%$ of washers have thicknesses between $471.4$ and $510.6$ microns.

exam6 marks

In a course, a student scores

89

on the Major Project, which is normally distributed with mean

\mu = 72

and standard deviation

\sigma = 8

. The same student scores

86

on the Written Exam, which is normally distributed with mean

\mu = 65

and standard deviation

\sigma = 12

. (a) Calculate the student's z-score on each task, to two decimal places. (b) State which task was the stronger result relative to the cohort, and justify your answer. (c) Given

\Phi(1.75) = 0.9599

, find the probability that a randomly chosen student scores higher than

86

on the Written Exam, and hence, in a cohort of

240

students, about how many score above this student on that task. (d) The student claims the equal-looking marks mean the two performances were equally good. Explain why this reasoning is flawed.

Show worked solution →

Standardise the Major Project mark (a). Apply $z = \frac{x - \mu}{\sigma}$ with $x = 89$ , $\mu = 72$ , $\sigma = 8$ :

z_{\text{Project}} = \frac{89 - 72}{8} = \frac{17}{8} \approx 2.13.

Standardise the Written Exam mark (a). Apply the formula with $x = 86$ , $\mu = 65$ , $\sigma = 12$ :

z_{\text{Exam}} = \frac{86 - 65}{12} = \frac{21}{12} = 1.75.

Compare the relative standings (b): The Major Project has the higher z-score ( $2.13$ against $1.75$ ), so the student sits further above the mean, in standard-deviation units, on the Major Project. That was the stronger result relative to the cohort.
Find the right-tail probability (c): Scoring above $86$ on the Written Exam means $z > 1.75$ . The area below is $\Phi(1.75) = 0.9599$ , so the area above is $1 - 0.9599 = 0.0401$ . In a cohort of $240$ , the expected number above is $240 \times 0.0401 = 9.624$ , so about $10$ students.
Explain the flaw (d): Raw marks from two different tasks are not comparable, because the tasks have different means and standard deviations. Equal-looking raw marks can sit at different distances above their own means once standardised, so the z-scores ( $2.13$ and $1.75$ ) show the performances were not equally good.
Answer: (a) Project $z \approx 2.13$ , Exam $z = 1.75$ ; (b) the Major Project, because its higher z-score places the student further above the cohort mean; (c) $P(X > 86) = 0.0401$ (about $4.01\%$ ), so roughly $10$ of the $240$ students; (d) raw marks from different distributions are not directly comparable, and the z-scores show the Major Project was the better result.

Practise this

Sources & how we know this

Mathematics Standard Stage 6 Syllabus (2017) — NESA
Mathematics Standard HSC exam papers — NESA

What this dot point is asking

The answer

The z-score formula

Standardising a value, step by step

Reverse: finding xxx from a z-score

Interpretation

Comparing values from different distributions

Z-scores and percentiles

Z-scores and probabilities

How exam questions ask about z-scores

Compute and interpret a z-score

Compare salaries across two industries

Find the cutoff for the top 5 percent

Find a symmetric pair of percentiles

Exam-style questions

Practice questions

Practise this

Sources & how we know this

Reverse: finding $x$ from a z-score