The normal distribution

40 min

The bell curve

Measure the heights of many pupils of the same age, the masses of many bags filled by one machine, or the errors of many careful measurements of the same length, and draw a histogram. Again and again the same shape appears: high in the middle, symmetric, falling away smoothly on both sides. The curve that fits it is the normal (or Gaussian) distribution, studied by Abraham de Moivre in the 1730s and by Carl Friedrich Gauss in his work on errors of measurement. It also appears whenever many small independent effects add up, which is the content of the central limit theorem, and it approximates the binomial distribution when $n$ is large.

Definition

A continuous random variable $X$ has the normal distribution with mean $μ \mu$ and variance $\sigma^2$ ( $\sigma \gt 0$ ), written $X \sim N(\mu, \sigma^2)$ , if its density is

f(x) = \frac{1}{\sigma\sqrt{2\pi}} \, e^{-\frac{(x - \mu)^2}{2\sigma^2}} , \qquad x \in \mathbb{R} .

Then $E(X) = \mu$ and $\operatorname{Var}(X) = \sigma^2$ . The case $\mu = 0$ , $\sigma = 1$ is the standard normal distribution; its variable is called $Z$ and its cdf $\Phi(z) = P(Z \le z)$ . Example: if pupils' heights follow $N (165, 8^{2})$ in cm, the mean is 165 cm and the standard deviation 8 cm.

Important

Properties of the normal curve. It is bell-shaped and symmetric about $x = \mu$ , where it is highest; it has points of inflection at $\mu \pm \sigma$ ; it approaches the $x$ -axis on both sides without touching it; the total area under it is 1.

Standardising. If $X \sim N(\mu, \sigma^2)$ , then

Z = \frac{X - \mu}{\sigma} \sim N(0, 1) , \qquad P(X \le x) = \Phi\left(\frac{x - \mu}{\sigma}\right) .

By symmetry, $\Phi(-z) = 1 - \Phi(z)$ . The integral of $f$ has no formula in elementary functions, so $\Phi$ is read from tables or computed by a calculator.

The normal curve and the 68–95–99.7 rule. Areas computed from the standard normal density; the bands are drawn from it exactly.

Some values of $\Phi(z)$ , to 4 decimal places:

$z$	0	0.5	1	1.25	1.5	1.645	1.9	1.96	2	2.5	3
$\Phi(z)$	0.5000	0.6915	0.8413	0.8944	0.9332	0.9500	0.9713	0.9750	0.9772	0.9938	0.9987

The 68–95–99.7 rule follows: $P(|X - \mu| \lt \sigma) = 2\Phi(1) - 1 \approx 0.683$ , and similarly 0.954 for $2\sigma$ and 0.997 for $3\sigma$ .

Worked example 1: probabilities by standardising

In an illustrative model, the heights of S6 pupils follow $N (165, 8^{2})$ in cm. Find the proportion of pupils (a) shorter than 175 cm; (b) between 155 cm and 175 cm.

Solution. (a) $z = \frac{175 - 165}{8} = 1.25$ , so $P(X \lt 175) = \Phi(1.25) \approx 0.8944$ : about 89%.

(b) For 155 cm, $z = \frac{155 - 165}{8} = -1.25$ . $P(155 \lt X \lt 175) = \Phi(1.25) - \Phi(-1.25) = 2\Phi(1.25) - 1 \approx 0.7887$ : about 79%.

Worked example 2: working backwards

(a) With the heights of worked example 1, find the height below which 90% of pupils lie. (b) In an illustrative exam, 10% of candidates score below 40 and 5% above 80. Assuming the marks are normal, find $μ \mu$ and $\sigma$ .

Solution. (a) $\Phi(z) = 0.9$ gives $z \approx 1.2816$ , so $h = 165 + 1.2816 \times 8 \approx 175.3$ cm.

(b) $\Phi(z_1) = 0.10$ gives $z_1 \approx -1.2816$ , and $\Phi(z_2) = 0.95$ gives $z_2 \approx 1.6449$ . So

\frac{40 - \mu}{\sigma} = -1.2816 , \qquad \frac{80 - \mu}{\sigma} = 1.6449 .

Subtracting: $\frac{40}{\sigma} = 2.9265$ , so $\sigma \approx 13.67$ and $\mu = 40 + 1.2816 \times 13.67 \approx 57.5$ .

Important

Normal approximation to the binomial. If $X \sim B(n, p)$ with $np \gt 5$ and $nq \gt 5$ , then $X$ is approximately $N (n p, n p q)$ . Because $X$ takes whole-number values and the normal variable is continuous, use a continuity correction: the value $k$ is treated as the interval from $k - 0.5$ to $k + 0.5$ . For example, $P(X \ge 60) \approx P(Y \gt 59.5)$ and $P(X \le 50) \approx P(Y \lt 50.5)$ .

Worked example 3: 100 tosses of a coin

A fair coin is tossed 100 times. Estimate the probability of at least 60 heads.

Solution. $X \sim B(100, 0.5)$ : $np = nq = 50 \gt 5$ , $\mu = 50$ , $\sigma = \sqrt{100 \times 0.5 \times 0.5} = 5$ .

P(X \ge 60) \approx P(Y \gt 59.5) = 1 - \Phi\left(\frac{59.5 - 50}{5}\right) = 1 - \Phi(1.9) \approx 0.0287 .

The exact binomial value is 0.0284. Without the correction, $1 - \Phi(2) \approx 0.0228$ , noticeably worse.

Warning

Divide by $\sigma$ , not by $\sigma^2$ . In $N (165, 64)$ the second number is the variance; the standard deviation is 8.

Draw a sketch before using the table. Tables give the area to the LEFT of $z$ . For "more than", subtract from 1; for a negative $z$ , use $\Phi(-z) = 1 - \Phi(z)$ . A sketch shows at once whether the answer should be more or less than 0.5.

Tip

Remember three numbers: about 68% of values lie within $1\sigma$ of the mean, 95% within $2\sigma$ (more exactly $1.96\sigma$ ) and 99.7% within $3\sigma$ . They give a quick check of any normal calculation, and a value more than $3\sigma$ from the mean is rare enough to be worth checking.

StatQuest with Josh Starmer: The Normal Distribution, Clearly Explained!!! (about 5 min). Watch how the mean sets the centre of the curve and the standard deviation sets its width, and where the 95% of values lie.

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Practice

Questions 1 to 3 are routine; 4 to 6 need more care; 7 asks for a proof.

$Z \sim N(0, 1)$ . Find $P(Z \lt 1.5)$ , $P(Z \gt -0.5)$ and $P(-1 \lt Z \lt 2)$ .
$X \sim N(50, 4^2)$ . Find $P(X \gt 58)$ and $P(46 \lt X \lt 54)$ .
Find $z$ such that $P(Z \lt z) = 0.975$ .
Bags of flour are filled with masses that follow $N (25, {0.2}^{2})$ in kg (illustrative). What proportion of bags weigh less than 24.7 kg?
$X \sim N(100, \sigma^2)$ and $P(X \gt 120) = 0.05$ . Find $\sigma$ .
$X \sim B(200, 0.3)$ . Use a normal approximation with continuity correction to estimate $P(X \le 50)$ .
Prove that, for every $X \sim N(\mu, \sigma^2)$ , $P(|X - \mu| \lt 2\sigma)$ has the same value, and give it.

Show answersHide answers

0.9332; $\Phi(0.5) = 0.6915$ ; $0.9772 - (1 - 0.8413) = 0.8185$ (more exactly 0.8186).
$1 - \Phi(2) \approx 0.0228$ ; $2\Phi(1) - 1 \approx 0.6827$ .
$z \approx 1.96$
$z = \frac{24.7 - 25}{0.2} = -1.5$ : $1 - 0.9332 = 0.0668$ , about 6.7%.
$\frac{20}{\sigma} = 1.6449$ , $\sigma \approx 12.16$ .
$\mu = 60$ , $\sigma = \sqrt{42} \approx 6.481$ ; $P(Y \lt 50.5) = \Phi(-1.466) \approx 0.071$ (exact binomial 0.070).
$P(|X - \mu| \lt 2\sigma) = P(-2 \lt Z \lt 2)$ with $Z = \frac{X - \mu}{\sigma} \sim N(0, 1)$ , whatever $μ \mu$ and $\sigma$ : $2\Phi(2) - 1 \approx 0.9545$ .

1The bell curve

The bell curve