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IAL 2025 Oct S2 Q4

A Level / Edexcel / S2

IAL 2025 Oct Paper · Question 4

题目

Problem

The continuous random variable XX has cumulative distribution function F(x)F(x) given by

F(x)={0x<1124(x1)21x316(x2)3<x6k(x220x+52)6<x101x>10F(x) = \begin{cases} 0 & x < 1 \\ \dfrac{1}{24}(x - 1)^2 & 1 \leqslant x \leqslant 3 \\ \dfrac{1}{6}(x - 2) & 3 < x \leqslant 6 \\ -k(x^2 - 20x + 52) & 6 < x \leqslant 10 \\ 1 & x > 10 \end{cases}

where kk is a positive constant.

(a) Show that k=148k = \dfrac{1}{48}

(2)

(b) Write down the value of P(X=4)P(X = 4)

(1)

(c) Find the value of P(2.5<X<4.5)+P(5.5<X<8.5)P(2.5 < X < 4.5) + P(5.5 < X < 8.5)

(3)

The probability density function of XX is given by

f(x)={r(x)1x3163<x6s(x)6<x100otherwisef(x) = \begin{cases} r(x) & 1 \leqslant x \leqslant 3 \\ \dfrac{1}{6} & 3 < x \leqslant 6 \\ s(x) & 6 < x \leqslant 10 \\ 0 & \text{otherwise} \end{cases}

(d) Find r(x)r(x) and s(x)s(x)

(3)

(Total for Question 4 is 9 marks)

题目中文翻译

连续随机变量 XX 的累积分布函数 F(x)F(x)

F(x)={0x<1124(x1)21x316(x2)3<x6k(x220x+52)6<x101x>10F(x) = \begin{cases} 0 & x < 1 \\ \dfrac{1}{24}(x - 1)^2 & 1 \leqslant x \leqslant 3 \\ \dfrac{1}{6}(x - 2) & 3 < x \leqslant 6 \\ -k(x^2 - 20x + 52) & 6 < x \leqslant 10 \\ 1 & x > 10 \end{cases}

其中 kk 为正常数。

(a) 证明 k=148k = \dfrac{1}{48}

(b) 写出 P(X=4)P(X = 4) 的值。

(c) 求 P(2.5<X<4.5)+P(5.5<X<8.5)P(2.5 < X < 4.5) + P(5.5 < X < 8.5) 的值。

XX 的概率密度函数为

f(x)={r(x)1x3163<x6s(x)6<x100otherwisef(x) = \begin{cases} r(x) & 1 \leqslant x \leqslant 3 \\ \dfrac{1}{6} & 3 < x \leqslant 6 \\ s(x) & 6 < x \leqslant 10 \\ 0 & \text{otherwise} \end{cases}

(d) 求 r(x)r(x)s(x)s(x)

(第 4 题共 9 分)

解答