Mean, variance, mode, median and quartilesEdexcel International A Level Further Maths: Revision notes

For a continuous random variable with pdf $f(x)$, the mean is an integral over the whole range: $E(X)=\int_{-\infty}^{\infty}xf(x)\,dx,\qquad E(X^2)=\int_{-\infty}^{\infty}x^2f(x)\,dx.$ In practice, integrate only over the interval where $f$ is not 0. The variance is $\text{Var}(X)=E(X^2)-[E(X)]^2.$ Example: $f(x)=\frac{3x^2}{8}$ on $[0,2]$. $E(X)=\frac{3}{8}\int_0^2x^3\,dx=\frac32$, $E(X^2)=\frac38\int_0^2x^4\,dx=\frac{12}{5}$, so $\text{Var}(X)=\frac{12}{5}-\frac94=\frac3{20}$.

The mode is the value of $x$ at which $f(x)$ is greatest. Look for a maximum of $f$:

• If $f$ has a turning point inside the interval, solve $f'(x)=0$ and check it is a maximum.
• If $f$ is increasing (or decreasing) throughout, the mode is at an end of the interval. Example: $f(x)=12x^2(1-x)$ on $[0,1]$ has $f'(x)=12x(2-3x)=0$ at $x=0$ and $x=\frac23$. At $x=0$, $f$ is 0, so the mode is $\frac23$.

The median $m$ splits the probability in half: $F(m)=\frac12,\qquad\text{equivalently}\qquad\int_{-\infty}^mf(x)\,dx=\frac12.$ Find $F(x)$ first, then solve $F(m)=\frac12$ for $m$. Example: $f(y)=\frac y2$ on $[0,2]$ gives $F(y)=\frac{y^2}{4}$, so $\frac{m^2}{4}=\frac12$ and $m=\sqrt2$.

The lower quartile $Q_1$ and upper quartile $Q_3$ satisfy $F(Q_1)=\frac14,\qquad F(Q_3)=\frac34.$ The interquartile range is $Q_3-Q_1$. For $F(y)=\frac{y^2}{4}$: $Q_1=1$, $Q_3=\sqrt3$, and the interquartile range is $\sqrt3-1$. Compare the mean and the median to describe skew. For $f(t)=\frac{3t^2}{64}$ on $[0,4]$ the mean is $3$ and the median is $3.17$. The median is above the mean, which shows negative skew (a longer tail on the left).