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IAL 2024 Jan S2 Q4

A Level / Edexcel / S2

IAL 2024 Jan Paper · Question 4

题目

Problem

The continuous random variable G has probability density function f(g) given by

f(g) = { (1/15)(g + 3), -1 < g ≤ 2

{ 3/20, 2 < g ≤ 4

{ 0, otherwise

(a) Sketch the graph of f(g)

(2)

(b) Find P(1 ≤ 2G ≤ 6 | G ≤ 2)

(4)

The continuous random variable H is such that E(H) = 12 and Var(H) = 2.4

(c) Find E(2H² + 3G + 3)

Show your working clearly.

(Solutions relying on calculator technology are not acceptable.)

(6)

(Total for Question 4 is 12 marks)

题目中文翻译

连续随机变量 G 的概率密度函数 f(g) 为

f(g) = { (1/15)(g + 3), -1 < g ≤ 2

{ 3/20, 2 < g ≤ 4

{ 0, otherwise

(a) 画出 f(g) 的图像

(b) 求 P(1 ≤ 2G ≤ 6 | G ≤ 2)

连续随机变量 H 满足 E(H) = 12 且 Var(H) = 2.4

(c) 求 E(2H² + 3G + 3)

清楚地展示你的计算过程。 （依赖计算器技术的解答不可接受。）

（第 4 题共 12 分）

解答

(a)

解法一

思路

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在 −1<g⩽2-1<g\leqslant2 上，密度是斜率为 1/151/15 的上升直线，端点高度分别趋近 2/152/15 和等于 1/31/3；在 2<g⩽42<g\leqslant4 上，密度是高度 3/203/20 的水平线。注意 g=2g=2 处有向下跳跃，两段不能连接，区间外密度为 0，并标清横轴端点及关键高度。

答题过程

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For −1<g⩽2-1<g\leqslant2, draw the increasing line segment from the open endpoint (−1,2/15)(-1,2/15) to the closed endpoint (2,1/3)(2,1/3).

For 2<g⩽42<g\leqslant4, draw the horizontal segment f(g)=3/20f(g)=3/20, open at g=2g=2 and closed at g=4g=4. Also show f(g)=0f(g)=0 outside these intervals.

(b)

解法一

思路

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先把 1⩽2G⩽61\leqslant2G\leqslant6 化成 0.5⩽G⩽30.5\leqslant G\leqslant3。再结合条件 G⩽2G\leqslant2，分子事件缩成 0.5⩽G⩽20.5\leqslant G\leqslant2。分别对密度积分求分子、分母，再套用条件概率公式。

答题过程

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The condition 1⩽2G⩽61\leqslant2G\leqslant6 is equivalent to

0.5⩽G⩽3.0.5\leqslant G\leqslant3.

Also,

P(G⩽2)= ∫−12g+315 dg= [g230+g5]−12= 710.\begin{align*} P(G\leqslant2)=&\,\int_{-1}^{2}\frac{g+3}{15}\,\mathrm{d}g\\ =&\,\left[\frac{g^2}{30}+\frac g5\right]_{-1}^{2}\\ =&\,\frac7{10}. \end{align*}

Under the condition G⩽2G\leqslant2, the required numerator is

P(0.5⩽G⩽2)= ∫0.52g+315 dg= 1740.\begin{align*} P(0.5\leqslant G\leqslant2) =&\,\int_{0.5}^{2}\frac{g+3}{15}\,\mathrm{d}g\\ =&\,\frac{17}{40}. \end{align*}

Therefore,

P(1⩽2G⩽6∣G⩽2)= 17/407/10= 1728.\begin{align*} P(1\leqslant2G\leqslant6\mid G\leqslant2) =&\,\frac{17/40}{7/10}\\ =&\,\boxed{\frac{17}{28}}. \end{align*}

解法二

思路

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在条件 G⩽2G\leqslant2 下，目标事件的补集是 G<0.5G<0.5。先求 P(G<0.5)P(G<0.5)，除以 P(G⩽2)P(G\leqslant2) 得到条件补事件概率，再从 1 中减去。

答题过程

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From Method 1,

P(G⩽2)=710.P(G\leqslant2)=\frac7{10}.

Also,

P(G<0.5)= ∫−10.5g+315 dg= 1140.\begin{align*} P(G<0.5)=&\,\int_{-1}^{0.5}\frac{g+3}{15}\,\mathrm{d}g\\ =&\,\frac{11}{40}. \end{align*}

Hence

P(1⩽2G⩽6∣G⩽2)= 1−11/407/10= 1728.\begin{align*} P(1\leqslant2G\leqslant6\mid G\leqslant2) =&\,1-\frac{11/40}{7/10}\\ =&\,\boxed{\frac{17}{28}}. \end{align*}

(c)

解法一

思路

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先由 Var⁡(H)=E⁡(H2)−[E⁡(H)]2\operatorname{Var}(H)=\operatorname{E}(H^2)-[\operatorname{E}(H)]^2 求 E⁡(H2)\operatorname{E}(H^2)。再按 GG 的分段密度积分求 E⁡(G)\operatorname{E}(G)。最后使用期望的线性性质；这里不需要假设 HH 与 GG 独立。

答题过程

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Using

Var⁡(H)=E⁡(H2)−[E⁡(H)]2,\operatorname{Var}(H) =\operatorname{E}(H^2)-[\operatorname{E}(H)]^2,

we have

E⁡(H2)= 2.4+122= 146.4.\begin{align*} \operatorname{E}(H^2)=&\,2.4+12^2\\ =&\,146.4. \end{align*}

Also,

E⁡(G)= ∫−12g(g+315) dg +∫24g(320) dg= 115[g33+3g22]−12 +340[g2]24= 0.5+0.9= 1.4.\begin{align*} \operatorname{E}(G) =&\,\int_{-1}^{2}g\left(\frac{g+3}{15}\right)\,\mathrm{d}g\\ &\,\hspace{2pt}+\int_{2}^{4}g\left(\frac3{20}\right)\,\mathrm{d}g\\ =&\,\frac1{15}\left[\frac{g^3}{3}+\frac{3g^2}{2}\right]_{-1}^{2}\\ &\,\hspace{2pt}+\frac3{40}\left[g^2\right]_{2}^{4}\\ =&\,0.5+0.9\\ =&\,1.4. \end{align*}

By linearity of expectation,

E⁡(2H2+3G+3)= 2E⁡(H2)+3E⁡(G)+3= 2(146.4)+3(1.4)+3= 300.\begin{align*} \operatorname{E}(2H^2+3G+3) =&\,2\operatorname{E}(H^2)+3\operatorname{E}(G)+3\\ =&\,2(146.4)+3(1.4)+3\\ =&\,\boxed{300}. \end{align*}