The normal distributionEdexcel A-Level Maths: Revision notes

Many continuous measurements, such as heights, masses and times, are symmetrical and bell-shaped, with most values near the mean. We model them with the Normal distribution, written $X\sim N(\mu,\sigma^2)$, where $\mu$ is the mean and $\sigma^2$ is the variance (so $\sigma$ is the standard deviation).

• The curve is symmetrical about $x=\mu$, so the mean, median and mode are equal and $P(X<\mu)=0.5$.
• The total area under the curve is 1, and probability is area. For a continuous variable $P(X=a)=0$, so $P(X<a)=P(X\le a)$.
• A histogram of continuous data that is roughly symmetrical and bell-shaped suggests that a Normal model is sensible; its mean and standard deviation estimate $\mu$ and $\sigma$. You do not need to know the formula for the probability density function.

The curve has its maximum at $x=\mu$ and the points of inflection (where it changes from curving downwards to curving upwards) are at $x=\mu\pm\sigma$. You do not have to derive this.

• A larger $\sigma$ gives a wider, flatter curve; a smaller $\sigma$ gives a narrower, taller one. The area stays 1.
• Changing $\mu$ slides the curve left or right without changing its shape. For $H\sim N(42,3^2)$ the maximum is at $h=42$ and the points of inflection are at $h=39$ and $h=45$.

Use the calculator's Normal cumulative distribution function with the correct $\mu$ and $\sigma$ (not $\sigma^2$).

• $P(a<X<b)$: enter both limits.
• $P(X>a)=1-P(X<a)$. For a lower or upper tail you can enter a very large or very small limit, or subtract.
• Symmetry: $P(X<\mu-k)=P(X>\mu+k)$, and $P(X<\mu)=0.5$. Example: $H\sim N(42,3^2)$. $P(38<H<47)=0.861$ and $P(H>45)=0.1587$. Always sketch the curve and shade the area first so you know whether the answer should be large or small.

To compare any Normal distribution with the standard Normal $Z\sim N(0,1^2)$, standardise: $Z=\frac{X-\mu}{\sigma}.$ So $P(X<x)=P\left(Z<\frac{x-\mu}{\sigma}\right)$. A $z$-value tells you how many standard deviations $x$ is from the mean. For $X\sim N(100,15^2)$, a score of 130 has $z=2$. For an inverse problem, where a probability is given and you must find a value, use the inverse Normal function (or the $z$-value from the table of percentage points): if $P(Z<z)=0.9$ then $z=1.2816$, so the score exceeded by 10% of people is $100+1.2816\times15=119.2$. Similarly $P(Z<z)=0.975$ gives $z=1.96$.

If one of $\mu,\sigma$ is unknown, standardise using a given probability and solve one equation. If both are unknown you are given two probabilities. Convert each to a $z$-value and form two equations. Example: $P(L<200)=0.1$ and $P(L>260)=0.2$.

• $P(L<200)=0.1$ gives $\frac{200-\mu}{\sigma}=-1.2816$, so $\mu-1.2816\sigma=200$.
• $P(L>260)=0.2$ means $P(L<260)=0.8$, so $\frac{260-\mu}{\sigma}=0.8416$ and $\mu+0.8416\sigma=260$. Subtract: $2.1232\sigma=60$, so $\sigma=28.3$; then $\mu=236$.