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CIE 9231 2024 June Paper 41 Q7

A Level / CIE / FS

CIE 9231 2024 June Paper 41 Paper · Question 7

题目

Problem

The continuous random variable XX has probability density function ff given by

f(x)={x4(4−x2),0≤x≤2,0,otherwise.f(x) = \begin{cases} \dfrac{x}{4}(4 - x^2), & 0 \leq x \leq 2, \\ 0, & \text{otherwise.} \end{cases}

(a) Find Var⁡(X)\operatorname{Var}(\sqrt{X}).

[4]

The continuous random variable YY is defined by Y=X2Y = X^2.

(b) Find the probability density function of YY.

[4]

(c) Find the exact value of the median of YY.

[2]
题目中文翻译

连续随机变量 XX 的概率密度函数 ff 为

f(x)={x4(4−x2),0≤x≤2,0,otherwise.f(x) = \begin{cases} \dfrac{x}{4}(4 - x^2), & 0 \leq x \leq 2, \\ 0, & \text{otherwise.} \end{cases}

(a) 求 Var⁡(X)\operatorname{Var}(\sqrt{X})。

(4)

连续随机变量 YY 定义为 Y=X2Y = X^2。

(b) 求 YY 的概率密度函数。

(4)

(c) 求 YY 的中位数的精确值。

(2)
原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

解答

(a)

解法一

思路

展开

X\sqrt X 的二阶矩是 XX 的均值。分别积分求 E⁡[X]\operatorname E[X] 和 E⁡[X]\operatorname E[\sqrt X]。

答题过程

展开

On [0,2][0,2], the density is x−x3/4x-x^3/4. Thus

 E⁡[X]=∫02(x2−x44) dx= [x33−x520]02= 1615.\begin{align*} &\,\operatorname E[X]=\int_0^2\Big(x^2-\frac{x^4}{4}\Big)\,\mathrm{d}x\\ =&\,\Big[\frac{x^3}{3}-\frac{x^5}{20}\Big]_0^2\\ =&\,\frac{16}{15}. \end{align*}

Also,

 E⁡[X]=∫02(x3/2−x7/24) dx= [25x5/2−118x9/2]02= 32245.\begin{align*} &\,\operatorname E[\sqrt X]=\int_0^2\Big(x^{3/2}-\frac{x^{7/2}}4\Big)\,\mathrm{d}x\\ =&\,\Big[\frac25x^{5/2}-\frac1{18}x^{9/2}\Big]_0^2\\ =&\,\frac{32\sqrt2}{45}. \end{align*}

Therefore

 Var⁡(X)= 1615−(32245)2= 1122025≈0.0553.\begin{align*} &\,\operatorname{Var}(\sqrt X)\\ =&\,\frac{16}{15}-\Big(\frac{32\sqrt2}{45}\Big)^2\\ =&\,\frac{112}{2025}\approx0.0553. \end{align*}

(b)

解法一

思路

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在 X≥0X\ge0 的支撑区间内，平方是递增变换。先求新变量的累积分布函数，再分段求导。

答题过程

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For 0≤y≤40\le y\le4,

 FY(y)=Pr⁡(X≤y)= ∫0y(x−x34) dx= y2−y216.\begin{align*} &\,F_Y(y)=\Pr(X\le\sqrt y)\\ =&\,\int_0^{\sqrt y}\Big(x-\frac{x^3}{4}\Big)\,\mathrm{d}x\\ =&\,\frac y2-\frac{y^2}{16}. \end{align*}

The CDF is constant outside [0,4][0,4]. A density is therefore

fY(y)={(4−y)/8,0≤y≤4,0,otherwise.f_Y(y)= \begin{cases} (4-y)/8, & 0\le y\le4,\\ 0, & \text{otherwise}. \end{cases}

(c)

解法一

思路

展开

将 (b) 的累积概率设为 1/21/2。解二次方程后，只保留支撑区间内的根。

答题过程

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The median mm satisfies m/2−m2/16=1/2m/2-m^2/16=1/2, giving m2−8m+8=0m^2-8m+8=0. The roots are 4±224\pm2\sqrt2. Only the minus choice is in [0,4][0,4], so the exact median is 4−224-2\sqrt2.