This note covers standardisation of normal variables using \(z\)-values, and how to find an unknown mean or standard deviation from given probabilities. It lets you compare results from different scales, such as two tests with different averages, and work backwards from tolerance or percentile information.
Key ideas
Every normal distribution has the same bell shape. Only the centre and the spread change. Standardising converts any value \(x\) from \(X \sim N(\mu, \sigma^2)\) into a \(z\)-value that says how many standard deviations \(x\) is from the mean:
\[ z = \frac{x - \mu}{\sigma}. \]- A positive \(z\) means \(x\) is above the mean. A negative \(z\) means it is below.
- \(z = 0\) means \(x\) equals the mean.
- \(z = 2\) means \(x\) is two standard deviations above the mean.
The standardised variable \(Z = \dfrac{X - \mu}{\sigma}\) follows the standard normal distribution \(Z \sim N(0, 1)\), with mean 0 and standard deviation 1. For any \(x\), \(P(X \lt x) = P\!\left(Z \lt \dfrac{x - \mu}{\sigma}\right)\).
Comparing different distributions. A raw score on its own does not say how good a result is. Standardise each score using its own mean and standard deviation, then compare the \(z\)-values. The larger \(z\) is the better result relative to its group.
Unknown \(\mu\) or \(\sigma\). Sometimes a probability is given but the mean or standard deviation is not. The method is:
- Sketch the curve and turn the given probability into an area to the left.
- Use the inverse normal function on your GDC with mean 0 and standard deviation 1 to find the \(z\)-value for that area.
- Write the standardising equation \(z = \dfrac{x - \mu}{\sigma}\) with the known numbers filled in.
- Solve for the unknown.
If both \(\mu\) and \(\sigma\) are unknown, two probabilities are given. You get two \(z\)-values and two equations, and you solve them simultaneously. A neat way is to write each equation as \(x = \mu + z\sigma\) and subtract one from the other to eliminate \(\mu\).
Two special cases are quick. If \(P(X \lt a) = 0.5\), then \(z = 0\) and \(\mu = a\). If a probability is below 0.5 for "less than", then \(z\) is negative.
Reading a \(z\)-value. Because \(z\) is measured in standard deviations, it is a fair way to compare results on different scales. A \(z\)-value near 0 is typical, and a \(z\)-value beyond 2 or below \(-2\) is unusual. Standardising also explains why a single table or function for \(N(0,1)\) is enough for every normal distribution: you convert first, then look up the area.
The sketch is especially helpful in unknown-parameter questions. Mark the known value on the axis, shade the area given, and decide whether the value lies above or below the mean. That tells you the sign of \(z\) before you calculate anything.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(z = \dfrac{x - \mu}{\sigma}\) | Standardising: the number of standard deviations \(x\) lies from the mean. |
| \(x = \mu + z\sigma\) | Rearranged form, useful for setting up equations for unknown \(\mu\) or \(\sigma\). |
| \(Z \sim N(0, 1)\) | Standard normal distribution. Every normal variable can be converted to this one. |
| \(P(X \lt x) = P\!\left(Z \lt z\right)\) | The area to the left is unchanged by standardising. |
| Inverse normal with \(N(0,1)\) | Given an area to the left, the GDC returns the \(z\)-value. |
Worked examples
Example 1 In a maths test the scores are normally distributed with mean 62 and standard deviation 8. In a physics test the mean is 70 and the standard deviation is 5. Aisha scores 74 in maths and 78 in physics. In which subject did she do relatively better?
Maths: \(z = \dfrac{74 - 62}{8} = \dfrac{12}{8} = 1.5\).
Physics: \(z = \dfrac{78 - 70}{5} = \dfrac{8}{5} = 1.6\).
The physics \(z\)-value is larger, so her physics score is further above the mean in standard-deviation units.
Answer: she did relatively better in physics (\(z = 1.6\) compared with \(z = 1.5\)).
Example 2 The time \(X\) minutes that a delivery takes is normally distributed with mean 70 and unknown standard deviation \(\sigma\). It is known that \(P(X \gt 82) = 0.1\). Find \(\sigma\).
The area to the right of 82 is 0.1, so the area to the left is \(P(X \lt 82) = 0.9\).
Use the inverse normal function with area 0.9, mean 0 and standard deviation 1: \(z = 1.2815\dots\).
Standardise: \(\dfrac{82 - 70}{\sigma} = 1.2815\dots\).
So \(\sigma = \dfrac{12}{1.2815\dots} = 9.363\dots\).
Answer: \(\sigma \approx 9.36\) minutes.
Example 3 The lengths of a type of steel rod, \(X\) cm, are normally distributed with unknown mean \(\mu\) and standard deviation \(\sigma\). It is known that 20% of rods are shorter than 40 cm and 10% of rods are longer than 60 cm. Find \(\mu\) and \(\sigma\).
\(P(X \lt 40) = 0.2\), so the \(z\)-value is the inverse normal of 0.2: \(z_1 = -0.8416\dots\).
\(P(X \gt 60) = 0.1\), so \(P(X \lt 60) = 0.9\), and \(z_2 = 1.2815\dots\).
Write each as \(x = \mu + z\sigma\):
\(40 = \mu - 0.8416\sigma\) and \(60 = \mu + 1.2815\sigma\).
Subtract the first from the second: \(20 = (1.2815 + 0.8416)\sigma = 2.1232\sigma\), so \(\sigma = 9.4198\dots\).
Substitute into the second equation: \(\mu = 60 - 1.2815\times 9.4198 = 47.93\dots\).
Check: \(47.93 - 0.8416\times 9.42 = 40.0\). It matches.
Answer: \(\mu \approx 47.9\) cm and \(\sigma \approx 9.42\) cm.
Common errors
- Common error: using the area to the right in the inverse normal function. Always convert to an area to the left first.
- Common error: forgetting that a probability below 0.5 for "less than" gives a negative \(z\)-value. A wrong sign gives a negative standard deviation or a mean in the wrong place.
- Common error: putting the variance into \(\dfrac{x - \mu}{\sigma}\). The denominator is the standard deviation.
- Common error: rounding the \(z\)-values early in a simultaneous-equation problem. Keep full values until the end.
- Common error: comparing raw scores from different distributions instead of comparing \(z\)-values.
Exam tip: For unknown-parameter questions, write the \(z\)-value on its own line, then the equation \(\dfrac{x - \mu}{\sigma} = z\). Making that step visible keeps the algebra tidy and the signs correct.
Check yourself
- \(X \sim N(60, 5^2)\). Find the \(z\)-value for \(x = 45\).
Answer
\(z = \dfrac{45 - 60}{5} = -3\). - \(X \sim N(\mu, 4^2)\) and \(P(X \lt 30) = 0.1\). Find \(\mu\).
Answer
Inverse normal of 0.1 gives \(z = -1.2816\). So \(\dfrac{30 - \mu}{4} = -1.2816\), giving \(\mu = 30 + 5.1262 = 35.1\). - \(X \sim N(120, 10^2)\). Find the value of \(x\) with \(z = 1.8\).
Answer
\(x = \mu + z\sigma = 120 + 1.8\times 10 = 138\). - \(X \sim N(\mu, \sigma^2)\) with \(P(X \lt 10) = 0.5\) and \(P(X \gt 16) = 0.1\). Find \(\mu\) and \(\sigma\).
Answer
An area of 0.5 means \(z = 0\), so \(\mu = 10\). Then \(P(X \lt 16) = 0.9\), \(z = 1.2816\), so \(\sigma = \dfrac{16 - 10}{1.2816} = 4.68\).