The Normal distributionEdexcel International A Level Maths: Revision notes

A continuous random variable $X$ that is Normally distributed has the notation $X\sim\mathrm{N}(\mu,\sigma^2)$, where $\mu$ is the mean and $\sigma^2$ is the variance, so the standard deviation is $\sigma$. The graph of its probability density is a symmetrical bell shape, centred on $\mu$, with the mean, median and mode all equal to $\mu$. About 68% of values lie within one standard deviation of the mean and about 95% within two. The total area under the curve is 1, so $\mathrm{P}(X<\mu)=0.5$. Because $X$ is continuous, $\mathrm{P}(X=x)=0$. You do not need to know the density function or to derive the mean, variance or cumulative distribution function.

To use tables, standardise: $Z=\frac{X-\mu}{\sigma}\sim\mathrm{N}(0,1).$ $Z$ counts how many standard deviations $X$ is above ($z>0$) or below ($z<0$) the mean. The table of the cumulative distribution function gives $\Phi(z)=\mathrm{P}(Z<z)$ for positive $z$. Interpolation is not needed. Example: eggs with $X\sim\mathrm{N}(60,4^2)$ and $x=66$ give $z=\frac{66-60}{4}=1.5$.

Always sketch the curve and shade the required area, then use symmetry:

• $\mathrm{P}(Z<z)=\Phi(z)$.
• $\mathrm{P}(Z>z)=1-\Phi(z)$.
• $\mathrm{P}(Z<-z)=1-\Phi(z)$.
• $\mathrm{P}(a<Z<b)=\Phi(b)-\Phi(a)$.
• $\mathrm{P}(-z<Z<z)=2\Phi(z)-1$. Example: $T\sim\mathrm{N}(42,3^2)$. $\mathrm{P}(T>45)$ has $z=1$, so it is $1-0.8413=0.1587$. Example: heights $\mathrm{N}(164,6^2)$, $\mathrm{P}(158<X<173)$: $z=-1$ and $1.5$, so $0.9332-(1-0.8413)=0.7745$.

To find $x$ given a probability, work backwards. Use the table of percentage points (or the table of $\Phi$ in reverse) to find $z$ with $\Phi(z)=p$, then solve $x=\mu+z\sigma$. Useful values: $p=0.90$ gives $z=1.2816$; $p=0.95$ gives $z=1.6449$; $p=0.975$ gives $z=1.9600$; $p=0.99$ gives $z=2.3263$. Example: the tallest 5% of women, $\mathrm{N}(164,6^2)$: $z=1.6449$, so $x=164+6(1.6449)=173.9$ cm. For a lower tail with $p<0.5$, find $z$ for $1-p$ and make it negative.

When two probabilities are given and both $\mu$ and $\sigma$ are unknown, standardise each condition to form simultaneous equations. Example: 10% of bags are below 995 g and 5% are above 1010 g. $\frac{995-\mu}{\sigma}=-1.2816$ and $\frac{1010-\mu}{\sigma}=1.6449$. Rearrange: $995=\mu-1.2816\sigma$ and $1010=\mu+1.6449\sigma$. Subtract: $15=2.9265\sigma$, so $\sigma=5.13$, and then $\mu=1001.6$. If only one condition is given with $\sigma$ known (or $\mu$ known), a single equation is enough.

Write each step: the distribution with its parameters, the standardised $z$ value, the table reading and the final probability or value. Give probabilities to 4 decimal places from the tables, or 3 significant figures at the end. Remember the context: state whether a probability is for a mass, time or height, with units. A small probability does not mean impossible, because the Normal curve extends in both directions.