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S2 2020 Jan Q6

A Level / Edexcel / S2

IAL 2020 Jan Paper · Question 6

题目

Problem

The random variable XX has probability density function f(x)f(x) where

 f(x)= { 18(x2+2x+1) −1⩽x<1 14 1⩽x⩽113 0 otherwise\begin{align*} &\,f(x)=\\ &\,\begin{cases} \begin{aligned} &\,\dfrac{1}{8}(x^2+2x+1)\\[-1mm] &\,\quad -1 \leqslant x < 1 \end{aligned}\\[3mm] \begin{aligned} &\,\dfrac{1}{4}\\[-1mm] &\,\quad 1 \leqslant x \leqslant \dfrac{11}{3} \end{aligned}\\[3mm] \begin{aligned} &\,0\\[-1mm] &\,\quad \text{otherwise} \end{aligned} \end{cases} \end{align*}

Given that E(X)=3118E(X)=\dfrac{31}{18}

(a) use algebraic integration to find Var⁡(X)\operatorname{Var}(X)

(4)

(b) Calculate

 P(X<−12) +P(X>12)\begin{align*} &\,P\big(X<-\tfrac12\big)\\ &\,\quad +P\big(X>\tfrac12\big) \end{align*}
(4)

(Total for Question 6 is 8 marks)

题目中文翻译

随机变量 XX 的概率密度函数 f(x)f(x) 为

 f(x)= { 18(x2+2x+1) −1⩽x<1 14 1⩽x⩽113 0 otherwise\begin{align*} &\,f(x)=\\ &\,\begin{cases} \begin{aligned} &\,\dfrac{1}{8}(x^2+2x+1)\\[-1mm] &\,\quad -1 \leqslant x < 1 \end{aligned}\\[3mm] \begin{aligned} &\,\dfrac{1}{4}\\[-1mm] &\,\quad 1 \leqslant x \leqslant \dfrac{11}{3} \end{aligned}\\[3mm] \begin{aligned} &\,0\\[-1mm] &\,\quad \text{otherwise} \end{aligned} \end{cases} \end{align*}

已知 E(X)=3118E(X)=\dfrac{31}{18}。

(a) 使用代数积分求 Var(X)\mathrm{Var}(X)。

(4)

(b) 计算

 P(X<−12) +P(X>12)\begin{align*} &\,P\big(X<-\tfrac12\big)\\ &\,\quad +P\big(X>\tfrac12\big) \end{align*}
(4)

（第 6 题共 8 分）

解答

(a)

解法一

思路

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先按概率密度函数的两个区间分别积分，求出 E(X2)E(X^2)。再用二阶矩减去均值的平方，得到方差。

答题过程

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Let the two contributions to E(X2)E(X^2) be AA and BB.

An antiderivative of x4+2x3+x2x^4+2x^3+x^2 is

G(x)=x55+x42+x33.G(x)=\frac{x^5}{5}+\frac{x^4}{2}+\frac{x^3}{3}. A= 18∫−11x2(x+1)2 dx= 18[G(1)−G(−1)]= 18(15+12+13) −18(−15+12−13)= 215.\begin{align*} A=&\,\frac18\int_{-1}^{1}x^2(x+1)^2\,\mathrm{d}x\\[4mm] =&\,\frac18\big[G(1)-G(-1)\big]\\[4mm] =&\,\frac18\Big(\frac15+\frac12+\frac13\Big)\\[4mm] &\,\hspace{2pt}-\frac18\Big(-\frac15+\frac12-\frac13\Big)\\[4mm] =&\,\frac{2}{15}. \end{align*} B= 14∫111/3x2 dx= [x312]111/3= 112(133127−1)= 32681.\begin{align*} B=&\,\frac14\int_{1}^{11/3}x^2\,\mathrm{d}x\\[4mm] =&\,\Big[\frac{x^3}{12}\Big]_{1}^{11/3}\\[4mm] =&\,\frac1{12}\Big(\frac{1331}{27}-1\Big)\\[4mm] =&\,\frac{326}{81}. \end{align*}

Therefore,

E(X2)= A+B= 215+32681= 1684405.\begin{align*} E(X^2)=&\,A+B\\[4mm] =&\,\frac{2}{15}+\frac{326}{81}\\[4mm] =&\,\frac{1684}{405}. \end{align*}  Var⁡(X)= E(X2)−[E(X)]2= 1684405−(3118)2= 19311620= 1.19197…\begin{align*} &\,\operatorname{Var}(X)\\[4mm] =&\,E(X^2)-[E(X)]^2\\[4mm] =&\,\frac{1684}{405}-\Big(\frac{31}{18}\Big)^2\\[4mm] =&\,\frac{1931}{1620}\\[4mm] =&\,1.19197\ldots \end{align*}

Hence

Var⁡(X)=1.19(3 s.f.).\boxed{\operatorname{Var}(X)=1.19}\quad\text{(3 s.f.)}.

(b)

解法一

思路

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把所求概率拆成左右两尾。左尾只涉及第一段密度；右尾跨过 x=1x=1，所以需要按密度函数的分段点拆成两个积分，再把三部分相加。

答题过程

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For the lower tail,

 P(X<−12)= ∫−1−1/218(x+1)2 dx= [(x+1)324]−1−1/2= 1192.\begin{align*} &\,P\big(X<-\tfrac12\big)\\[4mm] =&\,\int_{-1}^{-1/2}\frac18(x+1)^2\,\mathrm{d}x\\[4mm] =&\,\Big[\frac{(x+1)^3}{24}\Big]_{-1}^{-1/2}\\[4mm] =&\,\frac{1}{192}. \end{align*}

For the upper tail,

 P(X>12)= ∫1/2118(x+1)2 dx +∫111/314 dx= [(x+1)324]1/21 +[x4]111/3= 37192+23= 5564.\begin{align*} &\,P\big(X>\tfrac12\big)\\[4mm] =&\,\int_{1/2}^{1}\frac18(x+1)^2\,\mathrm{d}x\\[4mm] &\,\hspace{2pt}+\int_{1}^{11/3}\frac14\,\mathrm{d}x\\[4mm] =&\,\Big[\frac{(x+1)^3}{24}\Big]_{1/2}^{1}\\[4mm] &\,\hspace{2pt}+\Big[\frac{x}{4}\Big]_{1}^{11/3}\\[4mm] =&\,\frac{37}{192}+\frac23\\[4mm] =&\,\frac{55}{64}. \end{align*}

Thus,

 P(X<−12) +P(X>12)= 1192+5564= 8396= 0.865(3 s.f.).\begin{align*} &\,P\big(X<-\tfrac12\big)\\[4mm] &\,\quad+P\big(X>\tfrac12\big)\\[4mm] =&\,\frac{1}{192}+\frac{55}{64}\\[4mm] =&\,\frac{83}{96}\\[4mm] =&\,\boxed{0.865}\quad\text{(3 s.f.)}. \end{align*}

解法二

思路

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两个尾部事件的补集是 −12⩽X⩽12-\tfrac12\leqslant X\leqslant\tfrac12。这个区间完全落在概率密度函数的第一段内，因此只需计算一个积分，再用 1 减去它。

答题过程

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Using the complementary event,

 P(X<−12) +P(X>12)= 1−P(−12⩽X⩽12)= 1−∫−1/21/218(x+1)2 dx= 1−[(x+1)324]−1/21/2= 1−1396= 8396= 0.865(3 s.f.).\begin{align*} &\,P\big(X<-\tfrac12\big)\\[4mm] &\,\quad+P\big(X>\tfrac12\big)\\[4mm] =&\,1-P\big(-\tfrac12\leqslant X\leqslant\tfrac12\big)\\[4mm] =&\,1-\int_{-1/2}^{1/2}\frac18(x+1)^2\,\mathrm{d}x\\[4mm] =&\,1-\Big[\frac{(x+1)^3}{24}\Big]_{-1/2}^{1/2}\\[4mm] =&\,1-\frac{13}{96}\\[4mm] =&\,\frac{83}{96}\\[4mm] =&\,\boxed{0.865}\quad\text{(3 s.f.)}. \end{align*}

解法三

思路

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先求两个端点所在区间的累积分布函数。左尾是 F(−12)F(-\tfrac12)，右尾是 1−F(12)1-F(\tfrac12)，相加即可；不需要求出所有区间的累积分布函数。

答题过程

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For −1⩽x<1-1\leqslant x<1,

F(x)= ∫−1x18(t+1)2 dt= [(t+1)324]−1x= (x+1)324.\begin{align*} F(x)=&\,\int_{-1}^{x}\frac18(t+1)^2\,\mathrm{d}t\\[4mm] =&\,\Big[\frac{(t+1)^3}{24}\Big]_{-1}^{x}\\[4mm] =&\,\frac{(x+1)^3}{24}. \end{align*}

Hence

F(−12)= (1/2)324= 1192,F(12)= (3/2)324= 964.\begin{align*} F\big(-\tfrac12\big)=&\,\frac{(1/2)^3}{24}\\[4mm] =&\,\frac1{192},\\[4mm] F\big(\tfrac12\big)=&\,\frac{(3/2)^3}{24}\\[4mm] =&\,\frac9{64}. \end{align*}

Therefore,

 P(X<−12) +P(X>12)= F(−12)+1−F(12)= 1192+1−964= 8396= 0.865(3 s.f.).\begin{align*} &\,P\big(X<-\tfrac12\big)\\[4mm] &\,\quad+P\big(X>\tfrac12\big)\\[4mm] =&\,F\big(-\tfrac12\big)+1-F\big(\tfrac12\big)\\[4mm] =&\,\frac1{192}+1-\frac9{64}\\[4mm] =&\,\frac{83}{96}\\[4mm] =&\,\boxed{0.865}\quad\text{(3 s.f.)}. \end{align*}