Exam P · practice
Exam P — Continuous Distributions Practice Flashcards
Thirty-three exam-realistic multiple-choice problems spanning pdfs/cdfs, survival and hazard rates, percentiles and modes, and the uniform, exponential, gamma, normal, and beta distributions in the style of the SOA/CAS Exam P.
8 free sample33 total · in appFree · fact-checked · LaTeX math
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Common Traps Conditional Probability & Bayes Conditional Probability & Bayes Practice Continuous Distributions Covariance, Sums & CLT Covariance, Sums & CLT Practice
Browse all 33 problems as a list
- pdf/cdfA continuous random variable $X$ has probability density function $f(x)=k\,x(4-x)$ for $0\leq x\leq 4$, and $f(x)=0$ otherwise. Determine $k$. (A) $\dfrac{3}{128}$ (B) $\dfrac{1}{16}$ (C) $\dfrac{3}{32}$ (D) $\dfrac{1}{8}$ (E) $\dfrac{3}{16}$**Answer: (C).** A valid density integrates to $1$ over its support: $\int_{0}^{4} k\,x(4-x)\,dx = k\int_{0}^{4}(4x-x^{2})\,dx = k\left[2x^{2}-\frac{x^{3}}{3}\right]_{0}^{4}.$ Evaluate the bracket at $x=4$: $2(16)-\dfrac{64}{3}=32-\dfrac{64}{3}=\dfrac{96-64}{3}=\dfrac{32}{3}$. Setting $k\cdot\dfrac{32}{3}=1$ gives $k=\dfrac{3}{32}$.
- computing-probabilitiesLosses $X$ follow the density $f(x)=\dfrac{3}{32}\,x(4-x)$ for $0\leq x\leq 4$. Calculate $\Pr(X\leq 1)$. (A) $0.020$ (B) $0.156$ (C) $0.250$ (D) $0.313$ (E) $0.500$**Answer: (B).** $\Pr(X\leq 1)=\int_{0}^{1}\frac{3}{32}x(4-x)\,dx=\frac{3}{32}\int_{0}^{1}(4x-x^{2})\,dx.$ The antiderivative is $2x^{2}-\dfrac{x^{3}}{3}$, which at $x=1$ equals $2-\dfrac{1}{3}=\dfrac{5}{3}$. Thus $\Pr(X\leq 1)=\frac{3}{32}\cdot\frac{5}{3}=\frac{5}{32}=0.15625\approx 0.156.$
- pdf/cdfA random variable $X$ has cumulative distribution function $F(x)=0$ for $x<0$, $F(x)=\dfrac{x^{2}}{4}$ for $0\leq x<1$, $F(x)=\dfrac{x}{2}-\dfrac14$ for $1\leq x<2.5$, and $F(x)=1$ for $x\geq 2.5$. Calculate $\Pr(0.5\leq X\leq 2)$. (A) $0.563$ (B) $0.625$ (C) $0.688$ (D) $0.750$ (E) $0.813$**Answer: (C).** For a continuous cdf, $\Pr(0.5\leq X\leq 2)=F(2)-F(0.5)$. Since $2$ lies in $[1,2.5)$: $F(2)=\dfrac{2}{2}-\dfrac14=1-\dfrac14=\dfrac34$. Since $0.5$ lies in $[0,1)$: $F(0.5)=\dfrac{(0.5)^{2}}{4}=\dfrac{0.25}{4}=0.0625$. Therefore $\Pr(0.5\leq X\leq 2)=0.75-0.0625=0.6875\approx 0.688.$
- pdf/cdfA nonnegative loss random variable $X$ has survival function $S(x)=\left(\dfrac{10}{10+x}\right)^{3}$ for $x\geq 0$. Calculate $E[X]$. (A) $3.33$ (B) $5.00$ (C) $6.67$ (D) $10.00$ (E) $15.00$**Answer: (B).** For a nonnegative variable, $E[X]=\displaystyle\int_{0}^{\infty} S(x)\,dx$. $E[X]=\int_{0}^{\infty}\left(\frac{10}{10+x}\right)^{3}dx = 10^{3}\int_{0}^{\infty}(10+x)^{-3}\,dx.$ With $u=10+x$, $=1000\left[\frac{(10+x)^{-2}}{-2}\right]_{0}^{\infty}=1000\cdot\frac{1}{2}\cdot 10^{-2}=\frac{1000}{200}=5.00.$
- hazardThe lifetime $T$ (in years) of a component has hazard rate $h(t)=\dfrac{2t}{100}$ for $t\geq 0$. Calculate the probability that the component survives at least $5$ years. (A) $0.607$ (B) $0.779$ (C) $0.882$ (D) $0.905$ (E) $0.951$**Answer: (B).** The survival function is recovered from the hazard rate by $S(t)=\exp\!\left(-\int_{0}^{t} h(u)\,du\right).$ Here $\int_{0}^{5}\frac{2u}{100}\,du=\frac{u^{2}}{100}\Big|_{0}^{5}=\frac{25}{100}=0.25.$ Thus $S(5)=e^{-0.25}\approx 0.779.$
- hazardA nonnegative random variable $X$ has survival function $S(x)=e^{-(x/\theta)^{2}}$ for $x\geq 0$ (a Weibull). Its hazard rate $h(x)=\dfrac{f(x)}{S(x)}$ equals which of the following? (A) $\dfrac{x}{\theta^{2}}$ (B) $\dfrac{2x}{\theta^{2}}$ (C) $\dfrac{2x}{\theta}$ (D) $\dfrac{x^{2}}{\theta^{2}}$ (E) $\dfrac{2}{\theta^{2}}$**Answer: (B).** The hazard rate can be obtained directly as $h(x)=-\dfrac{d}{dx}\ln S(x)$. $\ln S(x)=-\left(\frac{x}{\theta}\right)^{2}=-\frac{x^{2}}{\theta^{2}},\qquad \frac{d}{dx}\ln S(x)=-\frac{2x}{\theta^{2}}.$ Therefore $h(x)=-\left(-\frac{2x}{\theta^{2}}\right)=\frac{2x}{\theta^{2}}.$
- computing-probabilitiesAnnual maintenance cost $X$ (in thousands) has density $f(x)=\dfrac{1}{8}x$ for $0\leq x\leq 4$, and $f(x)=0$ otherwise. Calculate $E[X]$. (A) $2.00$ (B) $2.40$ (C) $2.67$ (D) $3.00$ (E) $3.20$**Answer: (C).** $E[X]=\int_{0}^{4} x\cdot\frac{1}{8}x\,dx=\frac{1}{8}\int_{0}^{4}x^{2}\,dx=\frac{1}{8}\cdot\frac{x^{3}}{3}\Big|_{0}^{4}=\frac{1}{8}\cdot\frac{64}{3}=\frac{64}{24}=\frac{8}{3}\approx 2.67.$
- percentilesA loss random variable has density $f(x)=3x^{2}$ on $0\leq x\leq 1$. Determine the $80$th percentile of $X$. (A) $0.512$ (B) $0.737$ (C) $0.800$ (D) $0.928$ (E) $0.964$**Answer: (D).** The cdf is $F(x)=\displaystyle\int_{0}^{x}3t^{2}\,dt=x^{3}$ on $[0,1]$. Set $F(x_{0.8})=0.80$: $x^{3}=0.80\Rightarrow x=0.80^{1/3}=0.9283\approx 0.928.$ The distractor $0.512=0.8^{3}$ comes from cubing instead of taking the cube root.