The Normal Distribution

A random variable $X$ follows a normal distribution with mean $\mu$ and variance $\sigma^2$ if its probability density function has the characteristic bell shape. We write $X \sim N(\mu, \sigma^2)$.

Normal distribution bell curve with empirical rule: 68% within $\mu \pm \sigma$ (green), 95% within $\mu \pm 2\sigma$ (orange). Inflection points at $\mu \pm \sigma$.

Key properties:

• Symmetric about the mean $\mu$, left half mirrors the right
• Mean = Median = Mode all three coincide at the peak
• Asymptotic the curve approaches but never touches the horizontal axis
• Total area = 1 the entire area represents 100% of probability
• Inflection points at $\mu - \sigma$ and $\mu + \sigma$, where the curve changes from concave down to concave up

The parameter $\mu$ centres the curve. The parameter $\sigma$ controls spread: larger $\sigma$ → wider, flatter bell; smaller $\sigma$ → narrower, taller bell.

Real-world examples modelled by the normal distribution:

• Heights of adult males in a homogeneous population
• Measurement errors in laboratory experiments
• IQ scores (standardised to $\mu = 100$, $\sigma = 15$)
• Exam scores in large, well-designed assessments
• Manufacturing tolerances when a process is in control

$X \sim N(\mu, \sigma^2)$: mean $\mu$, variance $\sigma^2$, standard deviation $\sigma$; Symmetric bell curve: Mean = Median = Mode; asymptotic to the $x$-axis

Pause, copy the notation $X \sim N(\mu, \sigma^2)$, the three-in-one property (Mean = Median = Mode $= \mu$), and the fact that inflection points occur at $\mu \pm \sigma$ into your book.

We just saw that a normal distribution is fully characterised by $\mu$ and $\sigma$, and that inflection points occur at $\mu \pm \sigma$. That raises a question: can we calculate specific probabilities without calculus, just by knowing where a value sits relative to $\mu$? This card answers it → the 68–95–99.7 rule, which gives exact percentages for intervals within 1, 2, and 3 standard deviations of the mean.

For any normal distribution, the empirical rule gives the approximate percentage of data within 1, 2, and 3 standard deviations of the mean:

HSC precision note: The exact percentages are 68.27%, 95.45%, and 99.73%. In the HSC, "approximately 68%", "approximately 95%", and "approximately 99.7%" are all acceptable.

One-sided probabilities from symmetry:

• $P(X < \mu) = 50\%$, half the distribution is below the mean
• $P(X < \mu + \sigma) \approx 84\%$ (50% + 34%)
• $P(X < \mu + 2\sigma) \approx 97.5\%$ (50% + 47.5%)
• $P(X > \mu + 2\sigma) \approx 2.5\%$, upper tail only

Empirical rule: 68% in $\mu \pm \sigma$; 95% in $\mu \pm 2\sigma$; 99.7% in $\mu \pm 3\sigma$; Always check: is the given value at $\mu \pm k\sigma$ for $k = 1, 2,$ or $3$?

Pause, copy the three intervals (68%: $\mu \pm \sigma$; 95%: $\mu \pm 2\sigma$; 99.7%: $\mu \pm 3\sigma$) and the one-sided derivations (e.g., $P(X < \mu + 2\sigma) \approx 97.5\%$) into your book.

We just saw the empirical rule works only when values happen to land exactly at $\mu \pm k\sigma$. That raises a question: what if the value doesn't land on a whole-SD boundary, or we need to compare scores from two different distributions? This card answers it → the z-score $z = (x-\mu)/\sigma$, which standardises any normal variable to $Z \sim N(0,1)$ so any probability can be looked up from tables.

Every normal distribution can be transformed into the standard normal distribution $Z \sim N(0, 1)$ using the z-score formula:

The z-score tells you how many standard deviations $x$ is above or below the mean:

• $z = 0$ means $x = \mu$ (exactly average)
• $z = 1$ means $x$ is one SD above the mean
• $z = -2$ means $x$ is two SDs below the mean
• $|z| > 3$ is typically considered unusual or an outlier

Why standardise? Converting to z-scores lets us compare values from completely different normal distributions on the same scale. A z-score of 1.5 means the same thing whether you are talking about IQ scores, exam results, or manufacturing dimensions.

Example, are these performances equally good?

In a Maths exam: $\mu = 72$, $\sigma = 12$; student scores 84. In Physics: $\mu = 65$, $\sigma = 10$; same student scores 75.

Maths: $z = \dfrac{84 - 72}{12} = \dfrac{12}{12} = 1.0$; Physics: $z = \dfrac{75 - 65}{10} = \dfrac{10}{10} = 1.0$

Both scores are exactly 1 SD above their respective means. The performances are equally good relative to each cohort.

Comparing two different students:

Student A scores 78 in Chemistry ($\mu = 70$, $\sigma = 8$). Student B scores 82 in Biology ($\mu = 75$, $\sigma = 6$). Who performed better relative to their class?

Student A: $z = \dfrac{78-70}{8} = 1.0$; Student B: $z = \dfrac{82-75}{6} = \dfrac{7}{6} \approx 1.17$

Student B's z-score is higher, Student B performed better relative to their class.

z-score formula: $z = \dfrac{x - \mu}{\sigma}$; reverse: $x = \mu + z\sigma$; z-score = number of standard deviations from the mean (+ above, − below)

Pause, copy the z-score formula $z = \dfrac{x-\mu}{\sigma}$ and its reverse $x = \mu + z\sigma$, plus the comparison rule (higher z-score = better relative performance regardless of distribution) into your book.