What is the normal distribution?

The normal distribution is the bell-shaped model written X N(μ, σ^2). It is symmetric about the mean μ, and σ controls its spread.

TASMath MethodsSyllabus dot point

How do we describe continuous random variables and work with the normal distribution?

Use probability density functions and the normal distribution, including standardisation to z-scores, to find probabilities.

A continuous random variable is described by a probability density function whose area gives probability; the normal distribution is the key model, and standardising to z-scores lets you compute its probabilities.

Generated by Claude Opus 4.79 min answerUpdated 2026-05-29

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

A continuous random variable can take any value in an interval, so individual values have probability zero and we work with intervals instead. Its distribution is given by a probability density function $f(x)$ .

Because area is what matters, $P(X = c) = 0$ for any single value, and therefore $P(a \le X \le b) = P(a < X < b)$ ; the endpoints do not change a continuous probability.

The mean and variance are the continuous analogues of the discrete formulas:

\mu = \int_{-\infty}^{\infty} x\,f(x)\,dx, \qquad \sigma^2 = \int_{-\infty}^{\infty} (x-\mu)^2 f(x)\,dx.

The normal distribution

The normal distribution is the bell-shaped model written $X \sim N(\mu, \sigma^2)$ . It is symmetric about the mean $\mu$ , and $\sigma$ controls its spread.

Standardising to z-scores

Every normal distribution can be converted to the standard normal $Z \sim N(0,1)$ by standardising.

Summary

Treat continuous probability as area under the pdf, computed by integration, with total area $1$ . For normal problems, standardise with $z = \dfrac{x-\mu}{\sigma}$ , look up the lower-tail probability, then adjust for the region you need: subtract from $1$ for an upper tail, or subtract two lower-tail values for a between-values range. Use the 68-95-99.7 rule as a quick sanity check on your answer.

Exam-style practice questions

Practice questions written in the style of TASC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2024 TASC5 marksA random variable X is normally distributed with a mean of 10 and a standard deviation of 2. Using the approximate values for areas of the normal distribution, find: a) P(6 <= X <= 14); b) P(6 <= X <= 8); c) the value of x so that P(X >= x) = 0.0015.

Show worked answer →

Use the empirical (68-95-99.7) rule. With mu = 10 and sigma = 2: one SD spans 8 to 12, two SDs span 6 to 14, three SDs span 4 to 16.

a) 6 and 14 are mu - 2 sigma and mu + 2 sigma, so P(6 <= X <= 14) is about 95% = 0.95.

b) 8 = mu - sigma and 6 = mu - 2 sigma. The area between one and two SDs below the mean is (95% - 68%)/2 = 27%/2 = 13.5% = 0.135.

c) P(X >= x) = 0.0015 corresponds to the upper tail beyond three standard deviations (since about 99.7% lies within 3 SDs, leaving 0.15% in each tail). So x = mu + 3 sigma = 10 + 6 = 16.

2023 TASC6 marksForearm length is normally distributed with a mean of 40 cm and a standard deviation of 10 cm. a) What is the probability a randomly selected person has a forearm length greater than 45 cm? b) What forearm length would put a person in the longest 20% of the population? c) If 10 people are selected at random, what is the probability that at least 2 will have a forearm length greater than 45 cm?

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a) Standardise: z = (45 - 40)/10 = 0.5. P(X > 45) = P(Z > 0.5) = 0.3085, about 0.309.

b) The longest 20% means P(X > x) = 0.20, so x is the 80th percentile. The z-value with 20% above is z = 0.8416. Then x = 40 + 0.8416 times 10 = 48.42 cm.

c) Let Y be the number out of 10 with forearm > 45 cm. Y is binomial with n = 10 and p = 0.3085.
P(Y >= 2) = 1 - P(Y = 0) - P(Y = 1).
P(Y = 0) = (0.6915)^10 = 0.0247.
P(Y = 1) = 10 times 0.3085 times (0.6915)^9 = 0.1102.
P(Y >= 2) = 1 - 0.0247 - 0.1102 = 0.865. This part links the normal probability from a) into a binomial calculation.

2024 TASC4 marksA normal distribution has 10% above 30 and 25% below 10. Find mu and sigma to two decimal places.

Show worked answer →

Convert each statement into a standardised equation. From standard normal tables: P(Z > z) = 0.10 gives z = 1.2816, and P(Z < z) = 0.25 gives z = -0.6745.

So 30 = mu + 1.2816 sigma and 10 = mu - 0.6745 sigma.

Subtract the second from the first: 20 = (1.2816 + 0.6745) sigma = 1.9561 sigma, so sigma = 20/1.9561 = 10.22.

Substitute back: mu = 10 + 0.6745 times 10.22 = 10 + 6.89 = 16.89.
So mu = 16.89 and sigma = 10.22 (to 2 d.p.). The skill being tested is setting up two simultaneous equations from two given tail areas and solving them.