Use the normal distribution, z-scores, the empirical rule and the standard normal table to find probabilities and percentiles

What this dot point is asking

NESA wants you to standardise normally distributed data using z-scores, apply the $68$-$95$-$99.7$ empirical rule, find probabilities and percentiles using the standard normal, and interpret z-scores when comparing observations from different distributions.

The answer

The normal distribution

A continuous random variable $X$ is normally distributed with mean $\mu$ and standard deviation $\sigma$, written $X \sim N(\mu, \sigma^2)$, if its pdf is

Key features:

• The graph is a bell curve symmetric about $x = \mu$.
• IMATH_11 is the mean, median and mode.
• IMATH_12 controls the spread: larger $\sigma$ gives a flatter, wider curve.
• The total area under the curve is $1$.

The case $\mu = 0$, $\sigma = 1$ is the standard normal, denoted $Z \sim N(0, 1)$.

z-scores

The z-score of a value $x$ measures how many standard deviations it is from the mean:

If $X \sim N(\mu, \sigma^2)$, then $Z = \frac{X - \mu}{\sigma} \sim N(0, 1)$. Standardising turns any normal calculation into one about the standard normal.

z-scores let you compare observations from different distributions on the same scale. A higher z-score is "further above the mean in standard deviation units".

The empirical rule (68-95-99.7)

For any normal distribution,

• about $68\%$ of values lie within $1$ standard deviation of the mean ($\mu \pm \sigma$),
• about $95\%$ within $2$ standard deviations ($\mu \pm 2 \sigma$),
• about $99.7\%$ within $3$ standard deviations ($\mu \pm 3 \sigma$).

By symmetry, $P(0 \le Z \le 1) \approx 0.34$, $P(0 \le Z \le 2) \approx 0.475$, $P(0 \le Z \le 3) \approx 0.4985$.

Tail probabilities are the complement: $P(Z > 1) \approx 0.16$, $P(Z > 2) \approx 0.025$, $P(Z > 3) \approx 0.0015$.

Computing probabilities

For $X \sim N(\mu, \sigma^2)$ and $a < b$:

In the exam, the empirical rule covers the common endpoints. For other endpoints, use the standard normal table or the calculator's normalcdf function.

Inverse problems (percentiles)

To find the value $x$ such that $P(X \le x) = p$, find the corresponding $z$ from a table or invNorm, then transform: $x = \mu + z \sigma$. The 90th percentile of $Z$ is $z \approx 1.28$, the 95th is $z \approx 1.645$, the 97.5th is $z \approx 1.96$.

Worked examples

Direct use of the empirical rule

Heights of adult males in a city are normally distributed with $\mu = 175$ cm and $\sigma = 7$ cm. About what percentage of men are taller than $189$ cm?

$z = \frac{189 - 175}{7} = 2$. So we need $P(Z > 2) \approx 0.025$, or $2.5\%$.

Two-sided interval

For the same distribution, what percentage are between $168$ and $182$ cm?

These are $\mu \pm \sigma$, so by the empirical rule about $68\%$.

Mixed empirical-rule interval

For $X \sim N(50, 100)$ (so $\sigma = 10$), find $P(30 < X < 60)$.

$z_1 = \frac{30 - 50}{10} = -2$, $z_2 = \frac{60 - 50}{10} = 1$.

$P(-2 \le Z \le 1) = P(-2 \le Z \le 0) + P(0 \le Z \le 1) \approx 0.475 + 0.34 = 0.815$.

Comparing two distributions with z-scores

Two students sit different tests. Alex scores $82$ on a test with $\mu = 70$, $\sigma = 8$. Sam scores $75$ on a test with $\mu = 60$, $\sigma = 10$. Who performed better relative to their cohort?

Alex: $z = \frac{82 - 70}{8} = 1.5$. Sam: $z = \frac{75 - 60}{10} = 1.5$.

Same z-score, so they performed equally well relative to their cohorts.

Inverse normal

For $X \sim N(100, 225)$ (so $\sigma = 15$), find the value below which $95\%$ of the data lies.

The 95th percentile of $Z$ is $z \approx 1.645$, so $x = 100 + 1.645 \cdot 15 \approx 124.7$.

Common traps

Standardising with the wrong sign. $z = \frac{x - \mu}{\sigma}$. A value below the mean has a negative z-score. Do not drop the sign.

Confusing $\sigma$ and $\sigma^2$. $N(\mu, \sigma^2)$ uses the variance, but the empirical rule and z-score use $\sigma$. If a question gives $\sigma^2 = 25$, then $\sigma = 5$.

Forgetting symmetry. $P(Z \ge -1) = P(Z \le 1) \approx 0.84$. Use the symmetry of the bell curve rather than computing tails twice.

Adding empirical rule pieces incorrectly. $P(-1 \le Z \le 1) \approx 0.68$ is for the full two-sided interval. The one-sided half is $\frac{0.68}{2} = 0.34$. Do not double-count the central area.

Applying the empirical rule to non-normal data. The $68$-$95$-$99.7$ rule is specific to the normal distribution. For other shapes you must use other methods.

In one sentence

For $X \sim N(\mu, \sigma^2)$, standardise with $z = (x - \mu) / \sigma$ to convert to the standard normal, then use the empirical rule, a table, or normalcdf or invNorm on a calculator to find probabilities and percentiles.