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CIE 9231 2023 June Paper 41 Q3

A Level / CIE / FS

CIE 9231 2023 June Paper 41 Paper · Question 3

题目

Problem

A random sample of 50 values of the continuous random variable XX was taken. These values are summarised in the following table.

Interval1x<1.51 \leq x < 1.51.5x<21.5 \leq x < 22x<2.52 \leq x < 2.52.5x<32.5 \leq x < 33x<3.53 \leq x < 3.53.5x43.5 \leq x \leq 4
Observed frequency338111312

It is required to test the goodness of fit of the distribution with probability density function f given by

f(x)={124(4x2+x2),1x4,0,otherwise.f(x) = \begin{cases} \dfrac{1}{24}\bigg(\dfrac{4}{x^2} + x^2\bigg), & 1 \leq x \leq 4,\\ 0, & \text{otherwise.} \end{cases}

The expected frequencies, correct to 4 decimal places, are given in the following table.

Interval1x<1.51 \leq x < 1.51.5x<21.5 \leq x < 22x<2.52 \leq x < 2.52.5x<32.5 \leq x < 33x<3.53 \leq x < 3.53.5x43.5 \leq x \leq 4
Expected frequency4.4271aa6.12858.4549bb14.9678

(a) Show that a=4.6007a = 4.6007 and find the value of bb.

[3]

(b) Carry out a goodness of fit test, at the 10% significance level, to test whether f is a satisfactory model for the data.

[6]
题目中文翻译

从连续随机变量 XX 中随机抽取 50 个值。这些值汇总在下表中。

区间1x<1.51 \leq x < 1.51.5x<21.5 \leq x < 22x<2.52 \leq x < 2.52.5x<32.5 \leq x < 33x<3.53 \leq x < 3.53.5x43.5 \leq x \leq 4
观察频数338111312

现在要检验具有以下概率密度函数 f 的分布的拟合优度:

f(x)={124(4x2+x2),1x4,0,otherwise.f(x) = \begin{cases} \dfrac{1}{24}\bigg(\dfrac{4}{x^2} + x^2\bigg), & 1 \leq x \leq 4,\\ 0, & \text{otherwise.} \end{cases}

下表给出了期望频数,结果精确到 4 位小数。

区间1x<1.51 \leq x < 1.51.5x<21.5 \leq x < 22x<2.52 \leq x < 2.52.5x<32.5 \leq x < 33x<3.53 \leq x < 3.53.5x43.5 \leq x \leq 4
期望频数4.4271aa6.12858.4549bb14.9678

(a) 证明 a=4.6007a = 4.6007,并求 bb 的值。

(b) 在 10% 显著性水平下进行拟合优度检验,检验 f 是否是该数据的一个令人满意的模型。

解答