Year 12 · Statistical analysis
Quick tips — memory joggers
The normal curve
z-scores
68–95–99.7 rule
Comparing & probability
Level 1 · Fluency
A normal distribution curve is symmetrical about which value?
The mean, \(\mu\) (the curve is symmetric about the mean).
A bell-shaped curve rises to a single peak and falls away evenly on both sides. What name is given to this distribution?
The normal distribution.
State one real-world variable that is usually approximately normally distributed.
e.g. adult heights (also birth weights or exam marks) — any one is acceptable.
Level 2 · Application
Give an example of a variable that is often approximately normally distributed, and describe the shape of its graph.
e.g. adult heights — the graph is a symmetric, bell-shaped curve centred on the mean, tapering off on both sides.
Sketch and describe the shape of a normal distribution, and state where its single peak occurs.
A symmetric, bell-shaped curve with a single peak at the mean \(\mu\), tapering evenly towards both tails.
Explain why a normal curve is described as “bell-shaped” and why the mean, median and mode all sit at the same point.
It rises to a single maximum in the middle and falls away symmetrically on both sides, like a bell. Because the curve is symmetric about the centre, the mean, median and mode all coincide at \(\mu\).
Level 3 · Further Application
Test scores are normally distributed with mean 60 and standard deviation 10.
70
(a) A bell curve centred at 60.
(b) \(60 + 10 =\) 70.
Reaction times are normally distributed with mean 40 ms and standard deviation 5 ms.
35 ms
(a) A bell curve centred at 40 ms.
(b) \(40 - 5 =\) 35 ms.
Package weights are normally distributed with mean 250 g and standard deviation 20 g.
290 g
(a) A bell curve centred at 250 g.
(b) \(250 + 2 \times 20 = 250 + 40 =\) 290 g.
Level 1 · Fluency
For a perfectly normal distribution, how do the mean and median compare?
They are equal (both at the centre of symmetry).
In a symmetric (normal) distribution the mean equals the median. What third measure of centre also equals them?
The mode (mean = median = mode at the centre).
If a dataset's mean and median are almost the same value, what does this suggest about its shape?
It is roughly symmetric, so it may be approximately normal.
Level 2 · Application
A dataset has mean 51 and median 50.5. State whether it is likely to be approximately normal and why.
Likely approximately normal — the mean and median are almost equal, indicating near symmetry.
A dataset has mean 42 and median 41.8. Is it likely to be approximately normal? Justify your answer.
Yes — the mean (42) and median (41.8) differ by only 0.2, so the data is nearly symmetric and consistent with a normal distribution.
A dataset has mean 30 and median 30.2. Comment on whether it could be approximately normal.
Yes — the mean and median differ by only 0.2, indicating near symmetry, so it could well be approximately normal.
Level 3 · Further Application
A dataset has mean 80 and median 65.
(a) Not normal.
(b) The mean (80) is well above the median (65), suggesting a positively (right) skewed distribution rather than a symmetric one.
A dataset has mean 45 and median 60.
(a) Not normal.
(b) The mean (45) is well below the median (60), suggesting a negatively (left) skewed distribution rather than a symmetric one.
A dataset has mean 120 and median 121.
(a) Likely approximately normal.
(b) The mean (120) and median (121) differ by only 1, so the data is nearly symmetric — consistent with a normal distribution.
Level 1 · Fluency
A value equals the mean. What is its z-score?
\(z = 0\).
A value lies exactly one standard deviation above the mean. What is its z-score?
\(z = 1\).
A value lies exactly two standard deviations below the mean. What is its z-score?
\(z = -2\).
Level 2 · Application
A dataset has mean 115 and standard deviation 12. Calculate the z-score of the value 139.
2
\(z = \dfrac{139 - 115}{12} =\) 2.
A dataset has mean 40 and standard deviation 5. Calculate the z-score of the value 52.
2.4
\(z = \dfrac{52 - 40}{5} = \dfrac{12}{5} =\) 2.4.
A dataset has mean 200 and standard deviation 25. Calculate the z-score of the value 150.
−2
\(z = \dfrac{150 - 200}{25} = \dfrac{-50}{25} =\) −2.
Level 3 · Further Application
IQ scores are normal with mean 100 and standard deviation 15.
\(-0.6\)
(a) \(z = \dfrac{130 - 100}{15} = 2\).
(b) \(z = \dfrac{91 - 100}{15} =\) −0.6.
Annual rainfall is normal with mean 800 mm and standard deviation 50 mm.
\(-0.4\)
(a) \(z = \dfrac{900 - 800}{50} = 2\).
(b) \(z = \dfrac{780 - 800}{50} =\) −0.4.
Cholesterol readings are normal with mean 5.0 and standard deviation 0.8.
\(-1\)
(a) \(z = \dfrac{6.6 - 5.0}{0.8} = \dfrac{1.6}{0.8} = 2\).
(b) \(z = \dfrac{4.2 - 5.0}{0.8} = \dfrac{-0.8}{0.8} =\) −1.
Level 1 · Fluency
In \(z = \dfrac{x - \mu}{\sigma}\), what does \(\sigma\) represent?
The standard deviation.
In \(z = \dfrac{x - \mu}{\sigma}\), what does \(\mu\) represent?
The mean.
Rearranged, the formula becomes \(x = \mu + z\sigma\). What does it let you find?
The actual value (raw score) \(x\) from a known z-score.
Level 2 · Application
A test has mean 58 and standard deviation 6. Ravi’s z-score is 2. Find his actual mark.
70
\(x = \mu + z\sigma = 58 + 2 \times 6 =\) 70.
A test has mean 64 and standard deviation 7. Mia’s z-score is \(-1\). Find her actual mark.
57
\(x = \mu + z\sigma = 64 + (-1) \times 7 =\) 57.
A dataset has mean 150 and standard deviation 20. A value has a z-score of 1.5. Find the value.
180
\(x = \mu + z\sigma = 150 + 1.5 \times 20 =\) 180.
Level 3 · Further Application
Heights are normal with mean 170 cm and standard deviation 8 cm.
158 cm
(a) \(x = 170 + (-1.5)(8) = 170 - 12 =\) 158 cm.
(b) \(z = \dfrac{182 - 170}{8} = 1.5\).
A machine fills bottles with a mean of 500 mL and standard deviation 40 mL.
480 mL
(a) \(x = 500 + (-0.5)(40) = 500 - 20 =\) 480 mL.
(b) \(z = \dfrac{620 - 500}{40} = 3\).
Daily temperatures are normal with mean 25°C and standard deviation 4°C.
34°C
(a) \(x = 25 + 2.25 \times 4 = 25 + 9 =\) 34°C.
(b) \(z = \dfrac{19 - 25}{4} = -1.5\).
Level 1 · Fluency
A value has \(z = 3\). How many standard deviations above the mean is it?
3 standard deviations above the mean.
A value has \(z = -1\). How many standard deviations from the mean is it, and on which side?
1 standard deviation below the mean.
A value has \(z = 0\). Where does it lie relative to the mean?
Exactly at the mean (0 standard deviations away).
Level 2 · Application
A result has a z-score of \(-2\). Describe its position relative to the mean.
It is 2 standard deviations below the mean.
A result has a z-score of 2.5. Describe its position relative to the mean.
It is 2.5 standard deviations above the mean.
Two values have z-scores \(-0.5\) and 1.5. Describe each relative to the mean.
The first is 0.5 standard deviations below the mean; the second is 1.5 standard deviations above the mean.
Level 3 · Further Application
In a normal distribution, a value has \(z = 1.2\).
56
(a) 1.2 standard deviations above the mean.
(b) \(x = 50 + 1.2 \times 5 =\) 56.
In a normal distribution, a value has \(z = -1.8\).
182
(a) 1.8 standard deviations below the mean.
(b) \(x = 200 + (-1.8)(10) = 200 - 18 =\) 182.
In a normal distribution, a value has \(z = 2.4\).
66
(a) 2.4 standard deviations above the mean.
(b) \(x = 60 + 2.4 \times 2.5 = 60 + 6 =\) 66.
Level 1 · Fluency
After converting a dataset to z-scores, what is the mean of the z-scores?
0.
After standardising a dataset, what is the standard deviation of the z-scores?
1.
State the mean and standard deviation of a standardised (z-score) distribution.
Mean 0 and standard deviation 1.
Level 2 · Application
State the mean and standard deviation of a set of z-scores, and explain why standardising is useful.
Mean 0 and standard deviation 1. It puts different datasets on the same scale, so values from different distributions can be compared fairly.
Explain what it means that standardised scores always have mean 0 and standard deviation 1.
The original mean maps to \(z = 0\), and every value is measured in standard-deviation units, so one step of \(z\) equals one standard deviation. This turns any dataset into the same standard scale.
Why does standardising to z-scores let you compare marks from two different tests?
Both tests are rescaled to mean 0 and standard deviation 1, so a z-score measures how far a mark is above or below its own test average in a common unit — allowing a fair comparison.
Level 3 · Further Application
Two subjects have different means and standard deviations.
(a) The subjects have different scales (means and spreads), so a raw mark of, say, 80 means different things in each.
(b) Converting to z-scores (mean 0, sd 1) lets you compare relative performance across the two subjects.
A Maths test has mean 60 and standard deviation 10; a Science test has mean 70 and standard deviation 5.
(a) In Maths, 75 is \(z = \dfrac{75-60}{10} = 1.5\) standard deviations above the mean, but in Science it is only \(z = \dfrac{75-70}{5} = 1\) — so 75 is relatively better in Maths.
(b) Converting to z-scores (mean 0, sd 1) puts both subjects on the same scale, allowing a fair comparison.
A class's marks are converted to z-scores.
(a) Mean 0 and standard deviation 1.
(b) A z-score of 0 is a mark exactly equal to the class mean — neither above nor below average.
Level 1 · Fluency
Two z-scores are 1.5 and 0.8. Which result is relatively better?
The one with \(z = 1.5\) (further above its mean).
Two z-scores are \(-0.5\) and 0.9. Which result is relatively better?
The one with \(z = 0.9\) (above its mean, whereas \(-0.5\) is below).
Two z-scores are \(-1.2\) and \(-0.3\). Which result is relatively better?
The one with \(z = -0.3\) (closer to the mean, so less far below it).
Level 2 · Application
In Maths (mean 70, sd 8) a student scores 82; in English (mean 65, sd 5) they score 74. In which subject did they perform relatively better?
English
Maths: \(z = \dfrac{82 - 70}{8} = 1.5\). English: \(z = \dfrac{74 - 65}{5} = 1.8\). Relatively better in English (higher \(z\)).
In Physics (mean 55, sd 10) a student scores 70; in Chemistry (mean 60, sd 8) they score 76. In which subject did they perform relatively better?
Chemistry
Physics: \(z = \dfrac{70 - 55}{10} = 1.5\). Chemistry: \(z = \dfrac{76 - 60}{8} = 2\). Relatively better in Chemistry (higher \(z\)).
In test A (mean 50, sd 12) Sam scores 68; in test B (mean 72, sd 6) he scores 81. In which test did he perform relatively better?
Equally well (both \(z = 1.5\))
Test A: \(z = \dfrac{68 - 50}{12} = 1.5\). Test B: \(z = \dfrac{81 - 72}{6} = 1.5\). The z-scores are equal, so he performed equally well in both.
Level 3 · Further Application
A student scores in three subjects (all normal): History 80 (mean 65, sd 6), Biology 78 (mean 66, sd 8), Geography 72 (mean 60, sd 12).
(a) History: \(\dfrac{80 - 65}{6} = 2.5\); Biology: \(\dfrac{78 - 66}{8} = 1.5\); Geography: \(\dfrac{72 - 60}{12} = 1.0\).
(b) Strongest: History (\(z = 2.5\)); weakest: Geography (\(z = 1.0\)).
A student scores in three subjects (all normal): Maths 85 (mean 70, sd 10), Music 78 (mean 62, sd 8), PE 63 (mean 60, sd 3).
(a) Maths: \(\dfrac{85 - 70}{10} = 1.5\); Music: \(\dfrac{78 - 62}{8} = 2.0\); PE: \(\dfrac{63 - 60}{3} = 1.0\).
(b) Strongest: Music (\(z = 2.0\)); weakest: PE (\(z = 1.0\)).
A student scores in three subjects (all normal): Economics 72 (mean 60, sd 6), Legal 68 (mean 50, sd 12), Business 76 (mean 68, sd 8).
(a) Economics: \(\dfrac{72 - 60}{6} = 2.0\); Legal: \(\dfrac{68 - 50}{12} = 1.5\); Business: \(\dfrac{76 - 68}{8} = 1.0\).
(b) Strongest: Economics (\(z = 2.0\)); weakest: Business (\(z = 1.0\)).
Level 1 · Fluency
What percentage of normally distributed data lies within one standard deviation of the mean?
About 68%.
What percentage of normally distributed data lies within two standard deviations of the mean?
About 95%.
What percentage of normally distributed data lies within three standard deviations of the mean?
About 99.7%.
Level 2 · Application
For normal data, what percentage lies within two standard deviations of the mean, and hence what percentage lies beyond?
About 95% lies within 2 sd, so about 5% lies beyond (2.5% in each tail).
For normal data, what percentage lies within three standard deviations of the mean, and hence what percentage lies beyond?
About 99.7% lies within 3 sd, so about 0.3% lies beyond (0.15% in each tail).
For normal data, what percentage lies between one and two standard deviations above the mean?
13.5%
Within 2 sd is 95% and within 1 sd is 68%, so 27% lies between 1 sd and 2 sd across both sides. By symmetry the upper side is half of this: \(\dfrac{27}{2} =\) 13.5%.
Level 3 · Further Application
Heights are normal with mean 170 cm and standard deviation 6 cm.
2.5%
(a) \(\mu \pm 2\sigma = 170 \pm 12 =\) 158 cm to 182 cm.
(b) 182 cm is \(\mu + 2\sigma\), so about \(\dfrac{100 - 95}{2} =\) 2.5% are taller.
Exam marks are normal with mean 60 and standard deviation 12.
2.5%
(a) \(\mu \pm 1\sigma = 60 \pm 12 =\) 48 to 72.
(b) 84 is \(\mu + 2\sigma\), so about \(\dfrac{100 - 95}{2} =\) 2.5% score above it.
Adult weights are normal with mean 70 kg and standard deviation 5 kg.
2.5%
(a) \(\mu \pm 3\sigma = 70 \pm 15 =\) 55 kg to 85 kg.
(b) 60 kg is \(\mu - 2\sigma\), so about \(\dfrac{100 - 95}{2} =\) 2.5% weigh below it.
Level 1 · Fluency
About what percentage of normal data lies below the mean?
50% (the distribution is symmetric).
About what percentage of normal data lies above the mean?
50% (the distribution is symmetric).
About what percentage of normal data lies between the mean and one standard deviation above it?
34%
Within 1 sd is 68%, split equally either side of the mean: \(\dfrac{68}{2} =\) 34%.
Level 2 · Application
Weights are normal with mean 500 g and standard deviation 20 g. What percentage of items weigh between 480 g and 520 g?
68%
480–520 g is \(\mu \pm 1\sigma\), so about 68%.
Heights are normal with mean 160 cm and standard deviation 6 cm. What percentage of people are between 148 cm and 172 cm tall?
95%
148 cm \(= \mu - 2\sigma\) and 172 cm \(= \mu + 2\sigma\), so about 95%.
Marks are normal with mean 55 and standard deviation 10. What percentage of marks lie between 25 and 85?
99.7%
25 \(= \mu - 3\sigma\) and 85 \(= \mu + 3\sigma\), so about 99.7%.
Level 3 · Further Application
Battery life is normal with mean 40 hours and standard deviation 4 hours.
50 batteries
(a) 32–48 h is \(\mu \pm 2\sigma\), so about 95%.
(b) Above 48 h (\(\mu + 2\sigma\)) is about 2.5%, so \(0.025 \times 2000 =\) 50 batteries.
Bags of rice are normal with mean 1000 g and standard deviation 15 g.
800 bags
(a) 985–1015 g is \(\mu \pm 1\sigma\), so about 68%.
(b) Below 985 g (\(\mu - 1\sigma\)) is \(\dfrac{100 - 68}{2} = 16\%\), so \(0.16 \times 5000 =\) 800 bags.
Light-bulb life is normal with mean 1200 hours and standard deviation 100 hours.
20 bulbs
(a) 1000–1400 h is \(\mu \pm 2\sigma\), so about 95%.
(b) Above 1400 h (\(\mu + 2\sigma\)) is about 2.5%, so \(0.025 \times 800 =\) 20 bulbs.
Level 1 · Fluency
On a normal curve, which region represents the top 2.5% of values?
The shaded tail beyond \(\mu + 2\sigma\) (to the right of two standard deviations above the mean).
On a normal curve, which region represents the bottom 16% of values?
The tail below \(\mu - \sigma\) (to the left of one standard deviation below the mean).
On a normal curve, which region represents the middle 95% of values?
The central region between \(\mu - 2\sigma\) and \(\mu + 2\sigma\).
Level 2 · Application
Describe the region you would shade to show the middle 68% of a normal distribution.
Shade the area between \(\mu - \sigma\) and \(\mu + \sigma\) (one standard deviation either side of the mean).
Describe the region you would shade to show the top 16% of a normal distribution.
Shade the right-hand tail beyond \(\mu + \sigma\) (everything more than one standard deviation above the mean).
Describe the region you would shade to show the middle 99.7% of a normal distribution.
Shade the central area between \(\mu - 3\sigma\) and \(\mu + 3\sigma\) (three standard deviations either side of the mean).
Level 3 · Further Application
A normal distribution has mean 100 and standard deviation 15.
(a) \(70 = \mu - 2\sigma\), so shade the left tail below \(\mu - 2\sigma\).
(b) About 2.5%.
A normal distribution has mean 50 and standard deviation 8.
(a) \(66 = \mu + 2\sigma\), so shade the right tail beyond \(\mu + 2\sigma\).
(b) About 2.5%.
A normal distribution has mean 200 and standard deviation 20.
(a) \(180 = \mu - \sigma\) and \(220 = \mu + \sigma\), so shade the central region between \(\mu - \sigma\) and \(\mu + \sigma\).
(b) About 68%.
Level 1 · Fluency
A value has \(z = 2\). Using the empirical rule, what percentage of values are more extreme (further from the mean on that side)?
About 2.5% (the tail beyond \(z = 2\)).
A value has \(z = 3\). Using the empirical rule, what percentage of values are more extreme on that side?
About 0.15%
Within 3 sd is 99.7%, so 0.3% lies beyond \(\pm 3\sigma\); half of that is in one tail: \(\dfrac{0.3}{2} =\) 0.15%.
A value has \(z = -2\). Using the empirical rule, what percentage of values lie below it?
About 2.5% (the tail below \(z = -2\)).
Level 2 · Application
A result has \(z = 1\). Using the empirical rule, estimate the percentage of results greater than it.
16%
Within 1 sd is 68%, so 32% lies beyond \(\pm 1\sigma\); half of that is above, giving about 16%.
A result has \(z = -1\). Using the empirical rule, estimate the percentage of results less than it.
16%
Within 1 sd is 68%, so 32% lies beyond \(\pm 1\sigma\); half of that is below, giving about 16%.
A result has \(z = 2\). Using the empirical rule, estimate the percentage of results lying between the mean and it.
47.5%
Within 2 sd is 95%, split equally either side of the mean: \(\dfrac{95}{2} =\) 47.5%.
Level 3 · Further Application
Resting heart rates are normal with mean 72 bpm, sd 8 bpm.
2.5%
(a) \(z = \dfrac{88 - 72}{8} = 2\).
(b) Beyond \(z = 2\) is about \(\dfrac{100 - 95}{2} =\) 2.5%.
Aptitude scores are normal with mean 500, sd 100.
2.5%
(a) \(z = \dfrac{300 - 500}{100} = -2\).
(b) Below \(z = -2\) is about \(\dfrac{100 - 95}{2} =\) 2.5%.
Commute times are normal with mean 30 min, sd 5 min.
0.15%
(a) \(z = \dfrac{45 - 30}{5} = 3\).
(b) Beyond \(z = 3\) is about \(\dfrac{100 - 99.7}{2} =\) 0.15%.
Level 1 · Fluency
A table gives \(P(0 < Z < 0.5) = 0.1915\). What is \(P(Z > 0)\)?
0.5 (half the distribution lies above the mean).
A table gives \(P(0 < Z < 1) = 0.3413\). What is \(P(Z < 0)\)?
0.5 (half the distribution lies below the mean).
A table gives \(P(0 < Z < 0.5) = 0.1915\). What is \(P(-0.5 < Z < 0.5)\)?
0.383
By symmetry \(P(-0.5 < Z < 0.5) = 2 \times P(0 < Z < 0.5) = 2 \times 0.1915 =\) 0.383.
Level 2 · Application
A table gives \(P(0 < Z < 0.4) = 0.1554\). Show that \(P(Z > 0.4) = 0.3446\).
0.3446
\(P(Z > 0.4) = P(Z > 0) - P(0 < Z < 0.4) = 0.5 - 0.1554 =\) 0.3446.
A table gives \(P(0 < Z < 1.2) = 0.3849\). Find \(P(Z > 1.2)\).
0.1151
\(P(Z > 1.2) = P(Z > 0) - P(0 < Z < 1.2) = 0.5 - 0.3849 =\) 0.1151.
A table gives \(P(0 < Z < 0.7) = 0.2580\). Find \(P(Z < 0.7)\).
0.7580
\(P(Z < 0.7) = P(Z < 0) + P(0 < Z < 0.7) = 0.5 + 0.2580 =\) 0.7580.
Level 3 · Further Application
Newborn lengths are normal with mean 50 cm and standard deviation 2.5 cm. A table gives \(P(0 < Z < 0.6) = 0.2257\).
219 newborns
(a) \(z = \dfrac{51.5 - 50}{2.5} = 0.6\).
(b) \(P(Z > 0.6) = 0.5 - 0.2257 = 0.2743\); \(0.2743 \times 800 \approx\) 219 newborns.
Marks are normal with mean 60 and standard deviation 10. A table gives \(P(0 < Z < 0.8) = 0.2881\).
106 students
(a) \(z = \dfrac{68 - 60}{10} = 0.8\).
(b) \(P(Z > 0.8) = 0.5 - 0.2881 = 0.2119\); \(0.2119 \times 500 \approx\) 106 students.
Adult weights are normal with mean 80 kg and standard deviation 5 kg. A table gives \(P(0 < Z < 1.4) = 0.4192\).
552 people
(a) \(z = \dfrac{87 - 80}{5} = 1.4\).
(b) \(P(Z < 1.4) = 0.5 + 0.4192 = 0.9192\); \(0.9192 \times 600 \approx\) 552 people.
Level 1 · Fluency
A calculator gives \(P(Z < 1) \approx 0.8413\). What is \(P(Z > 1)\)?
0.1587
\(1 - 0.8413 =\) 0.1587.
A calculator gives \(P(Z < 0.5) \approx 0.6915\). What is \(P(Z > 0.5)\)?
0.3085
\(1 - 0.6915 =\) 0.3085.
A calculator gives \(P(Z < 2) \approx 0.9772\). What is \(P(Z > 2)\)?
0.0228
\(1 - 0.9772 =\) 0.0228.
Level 2 · Application
Using technology, \(P(Z < 1.2) \approx 0.8849\). Find the probability a value lies more than 1.2 standard deviations above the mean.
0.1151
\(P(Z > 1.2) = 1 - 0.8849 =\) 0.1151.
Using technology, \(P(Z < 1.5) \approx 0.9332\). Find the probability a value lies more than 1.5 standard deviations above the mean.
0.0668
\(P(Z > 1.5) = 1 - 0.9332 =\) 0.0668.
Using technology, \(P(Z < 0.75) \approx 0.7734\). Find \(P(Z > 0.75)\).
0.2266
\(P(Z > 0.75) = 1 - 0.7734 =\) 0.2266.
Level 3 · Further Application
Times are normal with mean 50 min and sd 6 min. Using technology, \(P(Z < 0.83) \approx 0.7967\).
0.2033
(a) \(z = \dfrac{55 - 50}{6} \approx 0.83\).
(b) \(P(Z > 0.83) = 1 - 0.7967 =\) 0.2033.
Heights are normal with mean 165 cm and sd 8 cm. Using technology, \(P(Z < 1.25) \approx 0.8944\).
0.1056
(a) \(z = \dfrac{175 - 165}{8} = 1.25\).
(b) \(P(Z > 1.25) = 1 - 0.8944 =\) 0.1056.
Masses are normal with mean 20 g and sd 4 g. Using technology, \(P(Z < 1.5) \approx 0.9332\).
0.0668
(a) \(z = \dfrac{26 - 20}{4} = 1.5\).
(b) \(P(Z > 1.5) = 1 - 0.9332 =\) 0.0668.
Level 1 · Fluency
A value has a z-score of 3.5. Is this an ordinary or an unusual result?
Unusual — it is 3.5 standard deviations from the mean, far into the tail.
A value has a z-score of 0.2. Is this an ordinary or an unusual result?
Ordinary — it is only 0.2 standard deviations from the mean, well within the typical range.
A value has a z-score of \(-3\). Is this an ordinary or an unusual result?
Unusual — it is 3 standard deviations below the mean, far into the lower tail.
Level 2 · Application
A manufacturer expects lifetimes to be normal. A batch has a mean z-score of \(-2.4\) against the specification. Comment on the batch.
A z-score of \(-2.4\) is well below the expected mean, so the batch is performing unusually poorly — likely a fault worth investigating.
A quality inspector finds a sample has a z-score of \(+0.3\) against the target. Comment on the sample.
A z-score of \(+0.3\) is very close to the target (well within one standard deviation), so the sample is within normal variation and acceptable.
A student’s coursework mark has a z-score of \(-2.6\) relative to the cohort. Comment on the result.
A z-score of \(-2.6\) is well below the cohort mean — an unusually low result that likely warrants support or further investigation.
Level 3 · Further Application
A student’s exam result has a z-score of 2.8 relative to the state.
(a) Very strongly — 2.8 standard deviations above the state mean.
(b) Beyond \(z = 2.8\) is well under the 2.5% that lies beyond \(z = 2\), so fewer than about 2.5% did better — an outstanding result.
An athlete’s race time has a z-score of \(-2.2\) relative to the field (a lower time is faster).
(a) 2.2 standard deviations below the mean time — a very fast result.
(b) Beyond \(z = 2\) lies only about 2.5%, so fewer than about 2.5% of the field were this fast — an exceptional performance.
A product’s measured dimension has a z-score of 0.5 against the specification.
(a) Half a standard deviation above the target — slightly above average.
(b) Within 1 sd holds about 68% of values, so a z-score of 0.5 is well inside the ordinary range — not unusual.