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 as areas under the bell curve, and find percentiles using the standard normal. You also need to interpret z-scores when comparing observations from different distributions. The single idea that ties it all together is that a probability for a normal variable is an area under its curve, and standardising lets you read every such area off one fixed curve, the standard normal.

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 probability density function is

You are never asked to integrate this by hand; what matters is what its graph looks like and how to read areas off it. Key features:

• The graph is a bell curve symmetric about $x = \mu$.
• $\mu$ is the mean, median and mode (all at the centre by symmetry).
• $\sigma$ controls the spread: a larger $\sigma$ gives a flatter, wider curve, a smaller $\sigma$ a taller, narrower one.
• The total area under the curve is $1$, so the area over any interval is the probability of landing in that interval.

The case $\mu = 0$, $\sigma = 1$ is the standard normal, denoted $Z \sim N(0, 1)$. Every normal calculation is turned into one about this single curve by standardising.

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 $Z = \frac{X - \mu}{\sigma} \sim N(0, 1)$. The two operations in the formula have a clear meaning: subtracting $\mu$ slides the value so the mean sits at $0$, and dividing by $\sigma$ rescales so one standard deviation becomes one unit. A value above the mean has a positive z-score, one below has a negative z-score, and the sign must be kept.

z-scores let you compare observations from different distributions on the same scale, because they strip out the original mean and spread. A higher z-score means "further above the mean in standard-deviation units", regardless of what the raw units were.

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 ($\mu \pm 3 \sigma$).

These are the central two-sided bands shown in the figure above. Because the curve is symmetric, halve them to get one-sided areas from the mean: $P(0 \le Z \le 1) \approx 0.34$, $P(0 \le Z \le 2) \approx 0.475$, $P(0 \le Z \le 3) \approx 0.4985$. Tail probabilities are the complement of half the band: $P(Z > 1) \approx 0.16$, $P(Z > 2) \approx 0.025$, $P(Z > 3) \approx 0.0015$.

Computing probabilities as areas

For $X \sim N(\mu, \sigma^2)$ and $a < b$, the probability of landing between $a$ and $b$ is the area under the curve over that interval, and standardising turns it into a standard-normal area:

In the exam, the empirical rule covers the common endpoints (whole numbers of standard deviations). For other endpoints, use the standard normal table or the calculator's normalcdf function. The picture is always the same: shade the region you want, then assemble it from the areas you know.

A single cut-off. The cumulative probability $P(Z \le z)$ is the whole area to the left of $z$. For example the area left of $z = 1$ is about $0.84$ (half the curve, $0.5$, plus the $0.34$ between $0$ and $1$).

Between two cut-offs. When the two endpoints straddle the mean, split the area at $0$ and add the two empirical-rule halves. For $z$ from $-1$ to $2$, the area is $P(-1 \le Z \le 0) + P(0 \le Z \le 2) \approx 0.34 + 0.475 = 0.815$.

Inverse problems (percentiles)

The reverse question gives you an area and asks for the cut-off value. To find $x$ such that $P(X \le x) = p$, find the corresponding $z$ from a table or invNorm, then untransform with $x = \mu + z \sigma$. The most-quoted z-scores are the $90$th percentile $z \approx 1.28$, the $95$th $z \approx 1.645$, and the $97.5$th $z \approx 1.96$. The diagram below shows the $95$th-percentile cut-off: the upper tail beyond $z = 1.645$ has area $\approx 0.05$, so $95\%$ of the area lies to its left.

How exam questions ask about the normal distribution

• "What percentage lie within / between ...?" Express the endpoints as whole numbers of standard deviations and read the empirical rule. If they straddle the mean, split at $0$.
• "Find the probability that $X$ is between $a$ and $b$." Standardise both endpoints, sketch and shade the region, then assemble it from empirical-rule halves (or normalcdf for awkward endpoints).
• "Find the probability that $X$ exceeds / is less than a value." A one-sided area: use a half plus or minus an empirical-rule piece, or the symmetry $P(Z \ge -z) = P(Z \le z)$.
• "Compare these two scores from different tests." Convert each to a z-score; the larger z-score is the better relative performance.
• "Find the value below which $p\%$ lie" or "the $p$th percentile." This is the inverse: get $z$ for that area, then $x = \mu + z\sigma$.

Edge cases worth knowing

• The variance versus the standard deviation. $N(\mu, \sigma^2)$ states the variance, but z-scores and the empirical rule use $\sigma$. If a question gives $\sigma^2 = 25$, then $\sigma = 5$ before you do anything else.
• A value that is not a whole number of standard deviations. The empirical rule only covers $1$, $2$, $3$ standard deviations. For $z = 1.5$ or $z = 0.7$ you need the standard normal table or normalcdf, not the rule.
• "At least" and "at most". For a continuous variable $P(X \ge a) = P(X > a)$ and $P(X \le a) = P(X < a)$, because a single point has zero probability, so do not fuss over strict versus non-strict inequalities.
• Symmetry shortcuts. $P(Z \ge -1) = P(Z \le 1) \approx 0.84$ and $P(Z < -1) = P(Z > 1) \approx 0.16$. Use the mirror image rather than computing a tail twice.