Apply the normal distribution $N(\mu, \sigma^2)$ and the standardisation $Z = (X - \mu)/\sigma$ to compute normal probabilities and inverse probabilities, including the empirical 68-95-99.7 rule

What this dot point is asking

QCAA wants you to recognise the normal distribution, apply the standardisation transformation, use the empirical 68-95-99.7 rule for Paper 1 exact-value questions, and compute general normal probabilities and inverse probabilities using technology in Paper 2.

The normal distribution

A continuous random variable $X$ is normally distributed with mean $\mu$ and standard deviation $\sigma$:

The pdf is:

Properties:

• Symmetric about $\mu$.
• Bell-shaped, peak at $x = \mu$.
• Mean = median = mode = $\mu$.
• Standard deviation $\sigma$ controls the spread.

The standard normal IMATH_11

The standard normal is $Z \sim N(0, 1)$. Any normal can be converted to a standard normal by:

This is standardisation. The transformation preserves probabilities:

The empirical 68-95-99.7 rule

For any normal distribution:

• IMATH_13
• IMATH_14
• IMATH_15

Single-tail derivatives:

• IMATH_16
• IMATH_17
• IMATH_18

The rule is the workhorse for Paper 1 exact-value questions when the $x$-endpoints fall at $\mu \pm n \sigma$.

Computing normal probabilities

Paper 1 (exact-value). When the endpoints map cleanly to $\mu \pm n \sigma$ for $n = 1, 2, 3$, use the empirical rule.

Paper 2 (calculator-active).

1. State the distribution: $X \sim N(\mu, \sigma^2)$.
2. Standardise the endpoints: $z_1 = (a - \mu)/\sigma$, $z_2 = (b - \mu)/\sigma$.
3. Compute $P$ using calculator's normCdf: $P = \text{normCdf}(a, b, \mu, \sigma)$.

Inverse normal

Given a probability $p$, find $c$ such that $P(X \leq c) = p$:

1. Find $z = \text{invNorm}(p)$ such that $P(Z \leq z) = p$.
2. Convert back: $c = \mu + z \sigma$.

Common $z$ values:

For an "upper tail" question, $P(X > c) = 1 - p$, so use $z$ for the complementary $p$.

Applications

Quality control. Lengths or weights of manufactured items modelled as normal.

Test scores. Standardised test scores have a bell-curve distribution.

Biological measurements. Heights, blood pressure, gestation periods.

Modelling errors. Random measurement errors typically follow a normal distribution.

Worked example. Combined Paper 1 / Paper 2

A factory produces batteries with lifetime normally distributed, mean 200 hours, SD 20 hours.

(a) Paper 1 empirical-rule. Find $P(160 \leq X \leq 240)$.

$160 = 200 - 2 \cdot 20$ and $240 = 200 + 2 \cdot 20$. So this is $[\mu - 2\sigma, \mu + 2\sigma]$. Probability $\approx 0.95$.

(b) Paper 2 calculator-active. Find $P(X > 230)$.

Standardise: $z = (230 - 200)/20 = 1.5$.

$P(X > 230) = P(Z > 1.5) \approx 0.0668$.

(c) Paper 2 inverse. Find $c$ such that the longest 10 percent of batteries are warranted to last more than $c$ hours.

$P(X > c) = 0.10 \implies P(X \leq c) = 0.90$. $z = 1.2816$. $c = 200 + 1.2816 \times 20 \approx 225.6$ hours.

Common errors

Wrong sign on standardisation. $z = (x - \mu)/\sigma$, not $(\mu - x)/\sigma$.

Empirical rule misapplied. The rule is for $\mu \pm n \sigma$ specifically. For other endpoints, standardise and use a table or calculator.

Inverse for the wrong tail. "Top 10 percent" means $P(X > c) = 0.10$, so $P(X \leq c) = 0.90$. Read carefully.

Using $\sigma^2$ where $\sigma$ is asked. $z = (x - \mu)/\sigma$ uses the standard deviation, not the variance.

Calculator without set-up. Paper 2 expects the standardisation set-up shown explicitly with the calculator value at the end.

In one sentence

The normal distribution $N(\mu, \sigma^2)$ is the bell-shaped distribution with mean $\mu$ and standard deviation $\sigma$; standardisation via $Z = (X - \mu)/\sigma$ converts any normal probability question to a standard-normal question, the empirical 68-95-99.7 rule handles Paper 1 exact-value endpoints at $\mu \pm n \sigma$, and the inverse-normal function handles "find $c$ such that $P(X \leq c) = p$" questions on Paper 2.