HL only. A density is non-negative and its total area is one. Integrate each piece over its own interval; density height is not the probability of a single value. Example: f(x)= kx0 <= x 1, k(2-x)1<x 2, 0otherwise. 1= _01 kx,dx+ _12 k(2-x),dx = k2+ k2=k With k=1, P(X<0.5)= _00.5x,dx= 18. Symmetry gives mean and median 1; the density is greatest at 1, so this is also the mode. E(X2)= _01x3,dx+ _12x2(2-x),dx= 76. Var(X)= 76-12= 16, and SD(X)=sqrt(16). For Y=3X-2, E(Y)=1 and Var(Y)= 32.

HL only. A density is non-negative and its total area is one. Integrate each piece over its own interval; density height is not the probability of a single value.

Example: \(\mathrm{f}(x)=\begin{cases}kx&0\le x\le1,\\k(2-x)&1<x\le2,\\0&\text{otherwise}.\end{cases}\)

\(\begin{aligned}1&=\int_0^1 kx\,\mathrm{d}x+\int_1^2 k(2-x)\,\mathrm{d}x\\&=\dfrac{k}{2}+\dfrac{k}{2}=k\end{aligned}\)

With \(k=1\), \(P(X<0.5)=\int_0^{0.5}x\,\mathrm{d}x=\dfrac18\). Symmetry gives mean and median \(1\); the density is greatest at \(1\), so this is also the mode.

\(E(X^2)=\int_0^1x^3\,\mathrm{d}x+\int_1^2x^2(2-x)\,\mathrm{d}x=\dfrac76\).

\(\operatorname{Var}(X)=\dfrac76-1^2=\dfrac16\), and \(\operatorname{SD}(X)=\sqrt{\dfrac16}\). For \(Y=3X-2\), \(E(Y)=1\) and \(\operatorname{Var}(Y)=\dfrac32\).

Continuous Random Variables

International Baccalaureate Mathematics: Analysis and Approaches Higher Level

Piecewise Densities and Moments

Piecewise Densities and Moments