Skip to content
CalcGospel 國際數學圖譜
返回

IAL 2024 Oct S2 Q5

A Level / Edexcel / S2

IAL 2024 Oct Paper · Question 5

题目

Problem

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

f(x)={14(3x)1x2142<x314(x2)3<x40otherwisef(x) = \begin{cases} \dfrac{1}{4}(3 - x) & 1 \leqslant x \leqslant 2 \\[6pt] \dfrac{1}{4} & 2 < x \leqslant 3 \\[6pt] \dfrac{1}{4}(x - 2) & 3 < x \leqslant 4 \\[6pt] 0 & \text{otherwise} \end{cases}

The cumulative distribution function of XX is F(x)F(x).

(a) Show that F(x)=14(x22+3x)58F(x) = \dfrac{1}{4}\left(-\dfrac{x^2}{2} + 3x\right) - \dfrac{5}{8} for 1x21 \leqslant x \leqslant 2

(2)

(b) Find F(x)F(x) for all values of xx

(5)

(c) Find P(1.2<X<3.1)P(1.2 < X < 3.1)

(2)

(Total for Question 5 is 9 marks)

题目中文翻译

连续随机变量 XX 的概率密度函数为

f(x)={14(3x)1x2142<x314(x2)3<x40otherwisef(x) = \begin{cases} \dfrac{1}{4}(3 - x) & 1 \leqslant x \leqslant 2 \\[6pt] \dfrac{1}{4} & 2 < x \leqslant 3 \\[6pt] \dfrac{1}{4}(x - 2) & 3 < x \leqslant 4 \\[6pt] 0 & \text{otherwise} \end{cases}

XX 的累积分布函数为 F(x)F(x)

(a) 证明对于 1x21 \leqslant x \leqslant 2F(x)=14(x22+3x)58F(x) = \dfrac{1}{4}\left(-\dfrac{x^2}{2} + 3x\right) - \dfrac{5}{8}

(b) 求对所有 xx 值的 F(x)F(x)

(c) 求 P(1.2<X<3.1)P(1.2 < X < 3.1)

(第 5 题共 9 分)

解答