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Normalise a density and compute probabilities and moments by integration. Integrate on the correct branch and keep a cumulative distribution continuous.
A density f is nonnegative and integrates to one over its support. Probabilities are areas: \(P(a<X<b)=\int_a^b f(x)\,dx\). A density may be larger than 1; a probability cannot.
Use \(E(X)=\int xf(x)\,dx\), \(E(X^2)=\int x^2f(x)\,dx\) and \(\operatorname{Var}(X)=E(X^2)-E(X)^2\). A median m solves \(\int_{-\infty}^m f(x)\,dx=1/2\).
Normalise a piecewise density by adding its separate integrals. The cumulative distribution \(F(x)=P(X\le x)\) must include all area to the left, including completed earlier pieces.
For an absolutely continuous variable, F is continuous, nondecreasing, zero below the support and one above it. At a point where f changes formula, match the two CDF expressions rather than resetting the accumulated probability.
X has density \(f(x)=kx^{2}\) for \(0\le x\le 3\) and zero elsewhere.
\(k\frac{3^{3}}{3}=1\implies k=\frac{1}{9}\)
Answer: \(\frac{1}{9}\).
\(\int_0^L xf(x)\,dx=\frac{9}{4}\)
Answer: \(\frac{9}{4}\).
\(E(X^2)-E(X)^2=\frac{27}{80}\)
Answer: \(\frac{27}{80}\).
\((m/3)^{3}=1/2\implies m=32^{-1/3}\)
Answer: \(2.3811015779523\).
The density is \(f(x)=kx\ (0\le x\le 4),\quad f(x)=k(8-x)\ (4<x\le 8)\), and zero elsewhere.
\(k(4^2/2+4^2/2)=1\implies k=\frac{1}{16}\)
Answer: \(\frac{1}{16}\).
\(E(X)=4\) by symmetry about the centre of the support.
Answer: \(4\).
\(\int(x-E(X))^2f(x)\,dx=\frac{8}{3}\)
Answer: \(\frac{8}{3}\).
\(\int_0^1x/16\,dx=\frac{1}{32}\)
Answer: \(\frac{1}{32}\).
X has density \(f(x)=kx^{1}\) for \(0\le x\le 5\) and zero elsewhere.
\(k\frac{5^{2}}{2}=1\implies k=\frac{2}{25}\)
Answer: \(\frac{2}{25}\).
\(\int_0^L xf(x)\,dx=\frac{10}{3}\)
Answer: \(\frac{10}{3}\).
\(E(X^2)-E(X)^2=\frac{25}{18}\)
Answer: \(\frac{25}{18}\).
\((m/5)^{2}=1/2\implies m=52^{-1/2}\)
Answer: \(3.53553390593274\).
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Normalise a density and compute probabilities and moments by integration.
Try the problem & review your workingIntegrate on the correct branch and keep a cumulative distribution continuous.
Try the problem & review your workingApply the method to recent source-labelled questions from this chapter.
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