Probability Using z-scores & Applications
Master probabilities using z-scores for NSW Year 12 Mathematics Standard 2. Standardise any value with \(z=\dfrac{x-\mu}{\sigma}\), then read the probability of a region — the shaded area under the bell curve — straight from the empirical rule (68-95-99.7) whenever the boundary sits at a whole standard deviation.
You will learn to find more-than, less-than and between probabilities, symmetric-tail probabilities, the expected number of items in a group \((\text{probability}\times n)\), and how to judge how unusual a value is — core Standard 2 skills for test scores, heights, masses, lifetimes and quality control.
Theory
A z-score rewrites a value as the number of standard deviations it is from the mean, \(z=\dfrac{x-\mu}{\sigma}\). This Year 12 Standard 2 (NSW) guide shows how to use it to find probabilities from the empirical rule — more-than, less-than and between regions, symmetric tails, the expected number in a batch, and a judgement about how unusual an outcome is.
A z-score measures how many standard deviations a value sits above or below the mean: \(z = \dfrac{x-\mu}{\sigma}\). A positive \(z\) is above the mean, a negative \(z\) is below it, and \(z=0\) is the mean itself.
For normally distributed data the probability of a region is the area under the bell curve over that region. When a boundary lands on a whole z-score (\(z=\pm1,\pm2,\pm3\)) you can read that area straight from the empirical rule (68-95-99.7) and the curve’s symmetry — no z-table needed.
This Year 12 Standard 2 (NSW) guide shows how to find more-than, less-than and between probabilities, symmetric-tail probabilities, the expected number in a group of \(n\) items \((\text{probability}\times n)\), and how to judge how unusual a value is.
Standardise a value \(x\) to a z-score using the mean \(\mu\) and standard deviation \(\sigma\):
At whole z-boundaries the empirical rule gives the area (probability) directly:
The percentage in each half-segment (one side of the mean) is fixed:
| Region | \(0\) to \(1\) SD | \(1\) to \(2\) SD | \(2\) to \(3\) SD |
|---|---|---|---|
| Percentage | 34% | 13.5% | 2.35% |
Each half of the curve holds \(50\%\). A one-sided tail beyond a whole z-boundary is:
Finding a probability with z-scores
- Standardise each boundary value: \(z = \dfrac{x-\mu}{\sigma}\). Check it is a whole number \(\pm1\), \(\pm2\) or \(\pm3\).
- Sketch a bell curve and shade the region the question asks for (less-than, more-than or between).
- Read the area. Add \(34\%\), \(13.5\%\), \(2.35\%\) segments and \(50\%\) halves; for a one-sided tail use \(\dfrac{100\%-\text{within}}{2}\).
- Apply it. For an expected count multiply the probability by \(n\); for a judgement, a small tail (a few percent or less) means the outcome is unusual.
Standardise \(700\), then read the lower tail.
| \(z\) | \(=\) | \(\dfrac{700-800}{50} = -2\) |
| \(\text{within }\pm2\,\text{SD}\) | \(=\) | \(95\%\) |
| \(\text{below }z=-2\) | \(=\) | \(\dfrac{100\%-95\%}{2} = 2.5\%\) |
About 2.5% of globes last less than \(700\) hours.
Convert each boundary to a z-score, then add the segments the region covers.
| \(z_1\) | \(=\) | \(\dfrac{162-170}{8} = -1\) |
| \(z_2\) | \(=\) | \(\dfrac{186-170}{8} = 2\) |
| \(P(-1\le z\le 0)\) | \(=\) | \(34\%\) |
| \(P(0\le z\le 2)\) | \(=\) | \(47.5\%\) |
| \(\text{total}\) | \(=\) | \(34\%+47.5\% = 81.5\%\) |
About 81.5% of adults are between \(162\) and \(186\) cm tall.
Standardise, read the upper tail, then scale to the group.
| \(z\) | \(=\) | \(\dfrac{190-150}{20} = 2\) |
| \(\text{above }z=2\) | \(=\) | \(\dfrac{100\%-95\%}{2} = 2.5\%\) |
| \(\text{expected}\) | \(=\) | \(2.5\% \times 2000 = 50\) |
About 50 of the \(2000\) apples are heavier than \(190\) g.
Find the z-score, then the tail above it.
| \(z\) | \(=\) | \(\dfrac{96-60}{12} = 3\) |
| \(\text{within }\pm3\,\text{SD}\) | \(=\) | \(99.7\%\) |
| \(\text{above }z=3\) | \(=\) | \(\dfrac{100\%-99.7\%}{2} = 0.15\%\) |
Yes — \(96\) is \(3\) SD above the mean and only about 0.15% of students score higher, so it is very unusual.
Common pitfalls
Frequently asked questions
How do I turn a value into a z-score?
Subtract the mean and divide by the standard deviation: z equals x minus the mean, all over the standard deviation. A z-score tells you how many standard deviations the value is above or below the mean, so z equals 2 means the value is two standard deviations above the mean.
How does a z-score give a probability?
For normally distributed data the probability of a region is the area under the bell curve over that region. When the boundary is a whole z-score (plus or minus 1, 2 or 3), you read that area straight from the empirical rule, 68% within one standard deviation, 95% within two and 99.7% within three, together with the curve's symmetry.
How do I find a 'more than' or 'less than' probability?
A one-sided region splits at the mean, where 50% lies on each side. For a symmetric tail beyond a whole z-boundary, take 100% minus the 'within' percentage and halve it. For example, more than z equals 2 is (100% minus 95%) divided by 2, which is 2.5%; less than z equals minus 2 is the same 2.5% by symmetry.
How do I find a 'between' probability?
Standardise both boundaries to whole z-scores, then add up the segments the region covers: 34% from the mean to one standard deviation, 13.5% from one to two, and 2.35% from two to three, using 50% for a whole half. For example, between z equals minus 1 and z equals 2 is 34% plus 47.5%, which is 81.5%.
How do I work out how many items fall in a range?
Find the probability of the range with the empirical rule, then multiply by the total number of items n. For example, if 2.5% of apples weigh more than 190 g, then out of 2000 apples about 0.025 times 2000, which is 50 apples, are expected to be heavier than 190 g.
How do I decide whether a value is unusual?
Find its z-score and the tail probability beyond it. A value more than two standard deviations from the mean lies in a tail of about 2.5% or less, and beyond three standard deviations only about 0.15% of data is further out, so such values are rare and count as unusual.