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Normalise a density and compute probabilities and moments by integration.
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 \(\int_{-\infty}^m f(x)\,dx=1/2\).
X has density \(f(x)=kx^{3}\) for \(0\le x\le 4\) and zero elsewhere.
\(k\frac{4^{4}}{4}=1\implies k=\frac{1}{64}\)
Answer: \(\frac{1}{64}\).
\(\int_0^L xf(x)\,dx=\frac{16}{5}\)
Answer: \(\frac{16}{5}\).
\(E(X^2)-E(X)^2=\frac{32}{75}\)
Answer: \(\frac{32}{75}\).
\((m/4)^{4}=1/2\implies m=42^{-1/4}\)
Answer: \(3.36358566101486\).
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: \(3.53553390593274\).
X has density \(f(x)=kx^{2}\) for \(0\le x\le 6\) and zero elsewhere.
\(k\frac{6^{3}}{3}=1\implies k=\frac{1}{72}\)
Answer: \(\frac{1}{72}\).
\(\int_0^L xf(x)\,dx=\frac{9}{2}\)
Answer: \(\frac{9}{2}\).
\(E(X^2)-E(X)^2=\frac{27}{20}\)
Answer: \(\frac{27}{20}\).
\((m/6)^{3}=1/2\implies m=62^{-1/3}\)
Answer: \(4.7622031559046\).
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Normalise a density and compute probabilities and moments by integration.
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