The normal distribution

5.5 The normal distribution

The normal curve

Many continuous quantities, such as heights and masses, follow a symmetric bell-shaped curve, written $X \sim \mathrm{N}(\mu, \sigma^2)$. The mean $\mu$ sets its centre, and the standard deviation $\sigma$ sets how spread out it is.

• The total area under the curve is 1, and an area gives a probability.
• For a continuous variable, $\mathrm{P}(X = a) = 0$, so $\leq$ and $<$ give the same result.
• About 68% of values lie within $\sigma$ of the mean, and about 95% within $2\sigma$.

Standardising

Any normal variable can be converted to the standard normal $Z \sim \mathrm{N}(0, 1)$, which is what the tables give.

Finding probabilities

Sketch the curve, shade the area you want, and use the symmetry:

$\sigma = 4$, so $z = \dfrac{55 - 50}{4} = 1.25$ and $\Phi(1.25) = 0.8944$.

$\mathrm{P}(X > 55) = 1 - 0.8944 = 0.1056$

$\mathrm{P}(45 < X < 55) = 2(0.8944) - 1 = 0.7888$

Going backwards from a probability

Given a probability, use the tables in reverse to find $z$, and then $x$.

$\Phi(z) = 0.9$ gives $z = 1.282$. So $a = 50 + 1.282 \times 4 = 55.1$.

Finding $\mu$ or $\sigma$

$\mathrm{P}(X < 80) = 0.9$, so $\dfrac{80 - 70}{\sigma} = 1.282$ and $\sigma = 7.80$.

Two unknowns need two equations.

$\frac{30 - \mu}{\sigma} = -1.282 \qquad \frac{60 - \mu}{\sigma} = 0.842$

Subtracting gives $30 = 2.124\sigma$, so $\sigma = 14.1$, and then $\mu = 30 + 1.282 \times 14.1 = 48.1$.

Normal approximation to the binomial

A binomial with a large $n$ has a bell-shaped distribution, so it can be approximated by a normal curve.

The continuity correction

The binomial is discrete and the normal is continuous, so each whole number is treated as the interval reaching 0.5 either side of it.

$np = 40$ and $np(1 - p) = 24$, both of which allow this approximation, so use $Y \sim \mathrm{N}(40, 24)$.

$\mathrm{P}(Y > 44.5): \quad z = \frac{44.5 - 40}{\sqrt{24}} = 0.919$

$1 - \Phi(0.919) = 1 - 0.8209 = 0.179$