4.12 Standardising normal variables

Every normal distribution has the same shape. Two normal curves differ only in where they are centred and how widely they are spread, so measuring every value by its distance from the mean, in units of standard deviations, puts all of them on one common scale. This process is called standardising, and the result is a $z$-value.

Standardising has two main uses. It makes values from different distributions comparable, such as a score in one test and a score in another test marked in a different way. It also turns a question about an unknown mean or standard deviation into a linear equation, which is how a manufacturer sets a machine so that only a small fraction of its products are underweight.

The z-value

Measuring distance from the mean in standard deviations

A score of 72 means little without its context. If the scores of a large group are modelled by $X \sim N(60, 8^2)$, then 72 is 12 above the mean, and since each standard deviation is 8, it is $\frac{12}{8} = 1.5$ standard deviations above the mean. That number, $1.5$, is the z-value of 72.

The formula is in the formula booklet. Its numerator is the distance from the mean, with a sign, and dividing by $\sigma$ converts that distance into standard deviations. A $z$-value has no units: the units of $x - \mu$ and $\sigma$ cancel.

The last two rows come from the 68–95–99.7 rule (4.9). A $z$-value therefore gives a quick sense of how ordinary or extreme a value is, whatever the original units.

Going back from z to x

Rearranging the definition gives the value that has a particular $z$-value:

In words, start at the mean and move $z$ standard deviations: to the right if $z$ is positive, to the left if it is negative. This form is the one used whenever $z$ is known and $x$ is wanted, including every inverse normal problem later in the subtopic.

Comparing values from different distributions

Two values measured on different scales cannot be compared directly, but their $z$-values can. The value with the larger $z$-value is further above its own mean, measured in its own standard deviations, so it is the better performance relative to its group.

The standard normal distribution

Standardising a normal variable

Standardising can be applied to the whole random variable as well as to single values. If $X \sim N(\mu, \sigma^2)$, define

Subtracting $\mu$ moves every value left by $\mu$, a translation, so the new mean is $0$ and the shape and spread are unchanged. Dividing by $\sigma$ then shrinks every distance from the mean by the factor $\sigma$, so the new standard deviation is $1$. A translation and a scaling keep the shape of the bell, so $Z$ is still normally distributed.

The graph of the standard normal distribution is centred on $0$, has points of inflexion at $z = \pm 1$, and in practice lies between $z = -3$ and $z = 3$. Its $z$-values are measured in standard deviations, so the 68–95–99.7 rule reads directly off its axis.

Why standardising keeps probabilities

The event $X < x$ happens exactly when $\frac{X - \mu}{\sigma} < \frac{x - \mu}{\sigma}$, because subtracting $\mu$ and dividing by the positive number $\sigma$ do not change the direction of an inequality. The two events are the same event, so they have the same probability:

On a diagram, the shaded area under the curve of $X$ to the left of $x$ is equal to the shaded area under the standard normal curve to the left of $z$.

With a GDC the two methods give identical answers, so standardising is not needed for probabilities with a known mean and standard deviation. It becomes necessary in two situations: when a probability for $Z$ is given in the question instead of a GDC being available, and when $\mu$ or $\sigma$ is unknown.

Inverse normal calculations with z

From a probability to z, then to x

The inverse normal function of a GDC works for any normal distribution, so a boundary can be found directly. It can also be found in two steps through $Z$: first find the $z$-value with the given area, using $\mu = 0$ and $\sigma = 1$, then convert it with $x = \mu + z\sigma$. The two-step method is the one that extends to an unknown $\mu$ or $\sigma$, so it is worth practising first where the answer can be checked.

Useful z-values

A few $z$-values occur so often that it helps to recognise them. Each is found from the inverse normal function of $Z \sim N(0, 1)$, and the symmetry of the curve gives the negative values from the positive ones.

The middle 95% of any normal distribution lies within $1.96$ standard deviations of the mean. The value $2$ in the 68–95–99.7 rule is a rounded form of this.

Finding an unknown mean or standard deviation

The method

When $\mu$ or $\sigma$ is unknown, the GDC's normal functions cannot be used directly, because they need both values. Standardising gets round this, because the inverse normal function of $Z \sim N(0, 1)$ needs neither. The method has four steps.

1. Sketch the curve, mark the given boundary and shade the given probability.

2. Write the probability as an area to the left, and find the matching $z$-value from $Z \sim N(0, 1)$.

3. Write the equation $\frac{x - \mu}{\sigma} = z$, with the known values substituted.

4. Solve for the unknown, and check the answer with the normal cdf.

An unknown mean

If $\sigma$ is known and $\mu$ is not, the equation $\frac{x - \mu}{\sigma} = z$ is linear in $\mu$. A common application is setting a filling machine. The standard deviation depends on the machine's precision and is fixed, while the mean can be adjusted, and the question is how high to set it so that few packets fall below the stated mass.

In this setting, the extra 9.3 g in each packet is the cost of the machine's imprecision. A machine with a smaller standard deviation could be set to a lower mean with the same 1% guarantee, which is one reason manufacturers invest in more precise equipment.

An unknown standard deviation

If $\mu$ is known and $\sigma$ is not, the equation is $\frac{x - \mu}{\sigma} = z$ with $\sigma$ in the denominator. Multiply by $\sigma$ and divide by $z$:

Both $x - \mu$ and $z$ have the same sign (both positive when $x$ is above the mean, both negative when it is below), so $\sigma$ comes out positive, as it must.

Using symmetry to find the mean

When two tails of equal probability are given, one on each side, symmetry gives the mean without any calculation: the mean is the midpoint of the two boundaries. Only the standard deviation then needs a $z$-value.

The same idea links a normal distribution to its quartiles and percentiles (4.2). The lower quartile has area $0.25$ to its left and the upper quartile has area $0.75$, so the quartiles are symmetric about the mean and lie $0.6745$ standard deviations either side of it. The interquartile range of any normal distribution is therefore $2 \times 0.6745\sigma \approx 1.35\sigma$.

Finding both the mean and the standard deviation

Two conditions, two equations

When both $\mu$ and $\sigma$ are unknown, one probability is not enough: it gives one equation in two unknowns. Two probabilities at two different boundaries give two equations, and because each equation $\frac{x - \mu}{\sigma} = z$ can be rewritten as

they form a pair of linear simultaneous equations in $\mu$ and $\sigma$. These can be solved by elimination, by substitution or with the simultaneous equation solver of a GDC.

A quick check on the signs: in the figure, 30 is below the mean and 50 is above it, so the $z$-values must be negative and positive respectively, and the answer for $\mu$ must lie between 30 and 50. The value $37.9$ does, and it is closer to 30 because the left tail (0.2) is larger than the right tail (0.1).

Two unknowns in context

In context, the two conditions are usually given as percentages of a population, often as classes such as "small" and "large". Each percentage becomes an area to the left, then a $z$-value, then an equation, exactly as in the abstract case.

Exam-style questions on standardising

Questions on standardising usually combine a context, a pair of given percentages and a follow-on part that uses the new parameters. The equations must be written out, because a GDC cannot find $\mu$ and $\sigma$ in one step, and "show that" parts often ask for exactly those equations.

The second question gives the $z$-value in the question, as happens on papers where a GDC is not allowed. The method is the same; only the inverse normal step is replaced by the given value.

Self-check

• To do:

  I can calculate a $z$-value with $z = \frac{x - \mu}{\sigma}$ and interpret it as a number of standard deviations from the mean.

• To do:

  I can convert a $z$-value back to a value with $x = \mu + z\sigma$.

• To do:

  I can use $z$-values to compare values from different normal distributions.

• To do:

  I can state that $Z = \frac{X - \mu}{\sigma} \sim N(0, 1)$ and explain why $P(X < x) = P(Z < z)$

• To do:

  I can find a $z$-value from a probability, choosing the correct sign.

• To do:

  I can find an unknown mean or an unknown standard deviation from one probability.

• To do:

  I can use symmetry to find the mean when two equal tails are given.

• To do:

  I can set up and solve simultaneous equations to find both $\mu$ and $\sigma$.

(b) $z = \frac{48 - 60}{8} = \frac{-12}{8} = -1.5$: one and a half standard deviations below the mean, the mirror image of 72.

(c) $z = \frac{50 - 60}{8} = \frac{-10}{8} = -1.25$: one and a quarter standard deviations below the mean.

(d) $z = \frac{60 - 60}{8} = 0$: exactly at the mean.

For the second probability, $\frac{50 - 60}{8} = -1.25$ and $\frac{70 - 60}{8} = 1.25$:

(b) The normal cdf with lower boundary $-10^{99}$, upper boundary $72$, $\mu = 60$ and $\sigma = 8$ gives $0.93319\ldots$, the same value.

(b) $\frac{48 - 60}{8} = -1.5$, so you need $P(-1.5 < Z < 1.5)$. By symmetry the left tail $P(Z < -1.5)$ equals the right tail $P(Z > 1.5) = 0.0668$:

Convert back to $x$:

The direct inverse normal calculation with $\mu = 60$, $\sigma = 8$ and area $0.9$ gives the same value.

Using the lower boundary instead, $\frac{20 - 28}{\sigma} = -1.03643\ldots$, gives the same value.

Write each condition as $\mu + z\sigma = x$:

Subtract (1) from (2) to eliminate $\mu$:

Substitute into (1):

Check with the normal cdf: with $\mu = 37.928$ and $\sigma = 9.4199$, $P(X < 30) = 0.200$ and $P(X > 50) = 0.100$.

(b) $P(X > 510) = 0.1$, so the area to the left of 510 is $0.9$, and $z = 1.28155\ldots$:

(c) Subtract the equation in (a) from the equation in (b), using the unrounded $z$-values:

(d) With $\mu = 503.43\ldots$ and $\sigma = 5.1257\ldots$,

(e) Now $P(X < 495) = 0.01$, so $z = -2.32634\ldots$, with $\mu = 503.43\ldots$:

(f) Each bottle contains less than 495 ml with probability $0.01$, independently. Let $Y \sim B(12, 0.01)$ be the number of such bottles in a pack:

(c) The region from 14 to 26 has area $1 - 0.2 - 0.2 = 0.6$, and the mean 20 splits it in half: