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IAL 2025 Oct S2 A Q1

A Level / Edexcel / S2

IAL 2025 Oct A Paper · Question 1

题目

Problem

A continuous random variable XX has cumulative distribution function

F(x)={0x<2120(x24)2x415(2x5)4<x51x>5F(x) = \begin{cases} 0 & x < 2 \\ \dfrac{1}{20}(x^2 - 4) & 2 \leqslant x \leqslant 4 \\ \dfrac{1}{5}(2x - 5) & 4 < x \leqslant 5 \\ 1 & x > 5 \end{cases}

(a) Calculate P(X>4)P(X > 4)

(2)

(b) Find the value of aa such that P(3<X<a)=0.642P(3 < X < a) = 0.642

(4)

(c) Find the probability density function of XX, specifying it for all values of xx.

(4)

(Total for Question 1 is 10 marks)

题目中文翻译

连续随机变量 XX 的累积分布函数为

F(x)={0x<2120(x24)2x415(2x5)4<x51x>5F(x) = \begin{cases} 0 & x < 2 \\ \dfrac{1}{20}(x^2 - 4) & 2 \leqslant x \leqslant 4 \\ \dfrac{1}{5}(2x - 5) & 4 < x \leqslant 5 \\ 1 & x > 5 \end{cases}

(a) 计算 P(X>4)P(X > 4)

(b) 求 aa 的值使得 P(3<X<a)=0.642P(3 < X < a) = 0.642

(c) 求 XX 的概率密度函数,对所有 xx 值进行说明。

(第 1 题共 10 分)

解答