The normal distribution: VCE Math Methods Unit 4

What this dot point is asking

VCAA wants you to recognise the normal distribution, apply the empirical 68/95/99.7 rule, standardise to the standard normal $Z$, and compute normal probabilities and inverse probabilities by hand (for Paper 1 empirical-rule questions) and by calculator (for Paper 2). The dot point is the bridge between the abstract pdf concept and applied probability questions.

The normal distribution

A continuous random variable $X$ is normally distributed if its pdf is:

Notation: $X \sim N(\mu, \sigma^2)$.

Properties:

• Symmetric about the mean. The pdf is bell-shaped, with peak at $x = \mu$.
• Mean = median = mode = $\mu$.
• Standard deviation = $\sigma$. Two-thirds of probability mass within $\pm \sigma$ of $\mu$.
• Support is $(-\infty, \infty)$. $f$ is positive but extremely small far from $\mu$.

The pdf itself is rarely computed by hand in VCE Methods; it is provided in the formula sheet and integrated by technology.

The empirical 68/95/99.7 rule

For any normal distribution:

• $P(\mu - \sigma \leq X \leq \mu + \sigma) \approx 0.68$
• $P(\mu - 2 \sigma \leq X \leq \mu + 2 \sigma) \approx 0.95$
• $P(\mu - 3 \sigma \leq X \leq \mu + 3 \sigma) \approx 0.997$

Equivalently, in $Z$:

• $P(-1 \leq Z \leq 1) \approx 0.68$
• $P(-2 \leq Z \leq 2) \approx 0.95$
• $P(-3 \leq Z \leq 3) \approx 0.997$

These are the values expected by hand on Paper 1. Use them whenever the question's $x$-values land neatly at $\mu \pm n \sigma$ for $n = 1, 2, 3$.

By symmetry, the probabilities in each tail beyond $\mu + n \sigma$ are:

• Beyond $\mu + \sigma$: $\frac{1 - 0.68}{2} = 0.16$
• Beyond $\mu + 2\sigma$: $\frac{1 - 0.95}{2} = 0.025$
• Beyond $\mu + 3\sigma$: $\frac{1 - 0.997}{2} = 0.0015$

Standardisation: $Z = (X - \mu) / \sigma$

The standard normal $Z \sim N(0, 1)$ has mean 0 and standard deviation 1. Any normal random variable $X$ can be converted to a standard normal by the transformation:

This is called standardisation. The transformation preserves probabilities:

The standardisation is the key skill: it converts any normal probability question to a question about $Z$, which has tabulated values and is supported on every CAS.

Worked standardisation

$X \sim N(50, 4^2)$. Find $P(X \leq 56)$.

$z = \frac{56 - 50}{4} = 1.5$.

$P(X \leq 56) = P(Z \leq 1.5) \approx 0.9332$ (from table or calculator).

Inverse normal: from probability to $x$-value

Given a probability $p$, find the $x$-value $c$ such that $P(X \leq c) = p$ (or $P(X > c) = 1 - p$).

Procedure:

1. Find $z$ such that $P(Z \leq z) = p$. Use the inverse-normal function on calculator: $z = \text{invNorm}(p)$, or use a table.
2. Convert back to $x$: $c = \mu + z \sigma$.

Common values:

These appear in confidence-interval questions (covered in the confidence-intervals dot point).

Worked inverse normal

$X \sim N(50, 0.4^2)$. Find $c$ such that $P(X > c) = 0.10$.

$P(X \leq c) = 0.90$. So $z = 1.2816$.

$c = 50 + 1.2816 \times 0.4 = 50.513$.

Calculator workflow (Paper 2)

For probability questions (find $P(a \leq X \leq b)$): use normCdf(a, b, mu, sigma) on TI-Nspire or equivalent on ClassPad.

For inverse questions (find $c$ given $P(X \leq c) = p$): use invNorm(p, mu, sigma).

State the set-up explicitly in the working: "$X \sim N(50, 0.4^2)$; $P(X \leq c) = 0.90$; from inverse-normal, $c \approx 50.513$."

Standard contexts

Quality control

The lengths or weights of manufactured items are often modelled as normal with mean = target and standard deviation = tolerance. Questions ask for the proportion outside specification.

Examinations and IQ

Scores on standardised tests are typically modelled as normal.

Biology and medicine

Heights, blood pressure, gestation periods, biomarker levels are often approximately normal.

Finance

Returns on a portfolio over a fixed period are sometimes modelled as normal (in basic Methods contexts; real finance uses heavier tails).

Examples in context

Example 1. Quality control using the empirical rule. A bottling line fills to a mean of $500$ mL with standard deviation $4$ mL, normally distributed. The interval $[492, 508]$ is $\mu \pm 2\sigma$, so by the empirical rule about $95\%$ of bottles fall in this range. The proportion above $508$ mL (one tail beyond $\mu + 2\sigma$) is $\frac{1 - 0.95}{2} = 0.025$, or $2.5\%$.

Example 2. Cut-off score for a scholarship. Test scores are $N(60, 12^2)$. To award scholarships to the top $5\%$, find the cut-off $c$ with $P(X > c) = 0.05$, i.e. $P(X \le c) = 0.95$. The inverse-normal gives $z = 1.645$, so $c = 60 + 1.645 \times 12 = 60 + 19.7 = 79.7$. Students scoring above about $80$ qualify.

Try this

Q1. For $X \sim N(100, 15^2)$, use the empirical rule to estimate $P(85 \le X \le 115)$. [2 marks]

• Cue. $[85, 115] = \mu \pm \sigma$, so $\approx 0.68$.

Q2. For $X \sim N(20, 2^2)$, find $P(X \le 23)$ by standardising. [3 marks]

• Cue. $z = \frac{23 - 20}{2} = 1.5$; $P(Z \le 1.5) \approx 0.9332$.

Q3. Scores are $N(50, 8^2)$. Find the score exceeded by only the top $10\%$. [3 marks]

• Cue. $z = 1.2816$; $c = 50 + 1.2816 \times 8 = 60.3$.