Normal Distribution

AS · S1 · 3 min

The normal distribution models continuous quantities that cluster symmetrically around a mean: heights, masses, measurement errors, exam marks. It is the single most examined topic in S1.

y = (1/sqrt(2 pi)) exp(-x^2 / 2)

Definition

If $X \sim N(\mu, \sigma^2)$ then $X$ is normally distributed with mean $\mu$ and variance $\sigma^2$ (so standard deviation $\sigma$ ). The curve is bell-shaped and symmetric about $\mu$ , and the total area under it is $1$ .

Watch out

The second parameter is the variance, not the standard deviation. $N(50, 16)$ has $\sigma = 4$ . Read the question carefully: "standard deviation 4" and "variance 16" describe the same distribution.

Standardising

Every normal distribution is converted to the standard normal $Z \sim N(0, 1)$ so that one set of tables covers all cases.

Key result

Z = \frac{X - \mu}{\sigma}

$\Phi(z) = P(Z < z)$ is read from the tables (for $z \ge 0$ ). Everything else comes from symmetry:

$P(Z > z) = 1 - \Phi(z)$
$P(Z < -z) = 1 - \Phi(z)$
$P(a < Z < b) = \Phi(b) - \Phi(a)$

Method for finding a probability

Write the probability in terms of $X$ .
Standardise each boundary: $z = \dfrac{x - \mu}{\sigma}$ , to 3 decimal places.
Draw a quick sketch and shade the region you want.
Use the tables and symmetry to get the area.

A single boundary

The masses of eggs are $N(58, 4^2)$ grams. Find the probability that an egg has mass greater than $63$ g.

Solution

P(X > 63) = P\left(Z > \frac{63 - 58}{4}\right) = P(Z > 1.25) = 1 - \Phi(1.25) = 1 - 0.8944 = 0.1056

Between two values

With the same distribution, find $P(52 < X < 60)$ .

Solution

P(52 < X < 60) = P\left(\frac{52 - 58}{4} < Z < \frac{60 - 58}{4}\right) = P(-1.5 < Z < 0.5)

= \Phi(0.5) - \big(1 - \Phi(1.5)\big) = 0.6915 - (1 - 0.9332) = 0.6915 - 0.0668 = 0.6247

Working backwards

If a probability is given and you need $x$ , $\mu$ or $\sigma$ , find the $z$ -value first from the inverse table (the critical values), then un-standardise.

Finding a value from a probability

Lengths are $N(20, 2.5^2)$ cm. Find the length exceeded by the longest $10\%$ .

Solution

$P(X > x) = 0.1$ , so $P(Z > z) = 0.1$ and $\Phi(z) = 0.9$ , giving $z = 1.282$ .

\frac{x - 20}{2.5} = 1.282 \quad\Rightarrow\quad x = 20 + 2.5 \times 1.282 = 23.2 \text{ cm (3 s.f.)}

Finding the mean and standard deviation

$X \sim N(\mu, \sigma^2)$ . Given $P(X < 30) = 0.2$ and $P(X > 50) = 0.1$ , find $\mu$ and $\sigma$ .

Solution

$P(X < 30) = 0.2 \Rightarrow \dfrac{30 - \mu}{\sigma} = -0.842$ (since $\Phi(0.842) = 0.8$ and the $z$ is on the left).

$P(X > 50) = 0.1 \Rightarrow \dfrac{50 - \mu}{\sigma} = 1.282$ .

So $30 - \mu = -0.842\sigma$ and $50 - \mu = 1.282\sigma$ . Subtracting: $20 = 2.124\sigma$ , giving $\sigma = 9.42$ and then $\mu = 30 + 0.842 \times 9.42 = 37.9$ (3 s.f.).

Tip

Two unknowns always means two simultaneous equations from two $z$ -values. Watch the sign of each $z$ : a value below the mean gives a negative $z$ .

Continuity correction

When the normal approximates a discrete distribution (the binomial, see Binomial to Normal Approximation), widen each integer to its half-unit interval:

$P(X \le 7) \to P(X < 7.5)$
$P(X < 7) \to P(X < 6.5)$
$P(X \ge 7) \to P(X > 6.5)$
$P(X = 7) \to P(6.5 < X < 7.5)$

No correction is needed when the original variable is already continuous.

Exam tip

Give $z$ to 3 d.p. and final probabilities to 4 d.p. or 3 s.f. as the question asks. A quick sketch with the shaded area is worth drawing every time: it earns method marks even when the arithmetic slips, and it stops the classic $\Phi(z)$ versus $1 - \Phi(z)$ mistake.

Practice

Question

$X \sim N(100, 15^2)$ . Find $P(X > 120)$ and $P(85 < X < 110)$ .
$X \sim N(40, 9)$ . Find $x$ such that $P(X < x) = 0.95$ .
Times are normally distributed with mean $12$ minutes; $5\%$ of times exceed $15$ minutes. Find the standard deviation.
$P(X < 12) = 0.3$ and $P(X < 20) = 0.8$ for $X \sim N(\mu, \sigma^2)$ . Find $\mu$ and $\sigma$ .

Answers

$1 - \Phi(1.333) = 0.0912$ ; $\Phi(0.667) - (1 - \Phi(1)) = 0.7475 - 0.1587 = 0.5888$
$z = 1.645$ , $x = 40 + 3(1.645) = 44.9$
$\dfrac{3}{\sigma} = 1.645 \Rightarrow \sigma = 1.82$
$z$ -values $-0.524$ and $0.842$ ; $\sigma = 5.86$ , $\mu = 15.1$

Standardising

Method for finding a probability

Working backwards

Continuity correction

Practice

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