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CIE 9231 2026 June Paper 43 Q4

A Level / CIE / FS

CIE 9231 2026 June Paper 43 Paper · Question 4

题目

Problem

A random sample of 200200 observations of the continuous random variable XX was taken. The values are summarised in the following table.

interval1≤x<1.51\le x<1.51.5≤x<21.5\le x<22≤x<2.52\le x<2.52.5≤x<32.5\le x<33≤x<3.53\le x<3.53.5≤x<43.5\le x<4x≥4x\ge4
observed frequency9735231510614

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

f(x)={2x3,x≥1,0,otherwise.\mathrm{f}(x)=\begin{cases} \frac{2}{x^3},&x\ge1,\\[2mm] 0,&\text{otherwise.} \end{cases}

Most of the relevant expected frequencies, correct to 22 decimal places, are given in the following table.

interval1≤x<1.51\le x<1.51.5≤x<21.5\le x<22≤x<2.52\le x<2.52.5≤x<32.5\le x<33≤x<3.53\le x<3.53.5≤x<43.5\le x<4x≥4x\ge4
expected frequency111.11aa18.009.78bb3.8312.50

(a) Show that a=38.89a=38.89 and find the value of bb correct to 22 decimal places.

(3)

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

(6)
题目中文翻译

对连续随机变量 XX 随机抽取 200200 个观测值，结果汇总如下表。

区间1≤x<1.51\le x<1.51.5≤x<21.5\le x<22≤x<2.52\le x<2.52.5≤x<32.5\le x<33≤x<3.53\le x<3.53.5≤x<43.5\le x<4x≥4x\ge4
观测频数9735231510614

现需检验以下概率密度函数 f\mathrm{f} 所对应分布的拟合优度：

f(x)={2x3,x≥1,0,其他情况。\mathrm{f}(x)=\begin{cases} \frac{2}{x^3},&x\ge1,\\[2mm] 0,&\text{其他情况。} \end{cases}

大部分相应的期望频数如下表所示，均精确到小数点后 22 位。

区间1≤x<1.51\le x<1.51.5≤x<21.5\le x<22≤x<2.52\le x<2.52.5≤x<32.5\le x<33≤x<3.53\le x<3.53.5≤x<43.5\le x<4x≥4x\ge4
期望频数111.11aa18.009.78bb3.8312.50

(a) 证明 a=38.89a=38.89，并求 bb 的值，精确到小数点后 22 位。

(3)

(b) 在 10%10\% 的显著性水平下进行拟合优度检验，检验 f\mathrm{f} 是否是这些数据的合适模型。

(6)
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

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期望频数等于区间概率乘 200200。先积分 2x−32x^{-3}，再代入两组端点。

答题过程

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For [u,v][u,v] within the support, the expected frequency is

200∫uv2x−3 dx= 200[−x−2]uv= 200(u−2−v−2).\begin{align*} 200\int_u^v2x^{-3}\,\mathrm{d}x =&\,200\big[-x^{-2}\big]_u^v\\ =&\,200(u^{-2}-v^{-2}). \end{align*}

Thus a=200(1.5−2−2−2)=38.88888…a=200(1.5^{-2}-2^{-2})=38.88888\ldots, which rounds to 38.8938.89. Also b=200(3−2−3.5−2)=5.89569…b=200(3^{-2}-3.5^{-2})=5.89569\ldots, which rounds to 5.905.90.

解法二

思路

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先积分求 aa，再利用总期望频数为 200200 求 bb。表中数据已经舍入，因此与逐项积分的末位可能不同。

答题过程

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First,

a= 200∫1.522x−3 dx= 200[−x−2]1.52= 3509=38.88888….\begin{align*} a=&\,200\int_{1.5}^{2}2x^{-3}\,\mathrm{d}x\\ =&\,200\big[-x^{-2}\big]_{1.5}^{2}\\ =&\,\frac{350}{9}=38.88888\ldots. \end{align*}

Hence a=38.89a=38.89 to 22 decimal places. Since the expected frequencies sum to 200200, using the rounded table entries gives

b≈ 200−(111.11+38.89+18.00 +9.78+3.83+12.50)= 5.89.\begin{align*} b\approx&\,200-(111.11+38.89+18.00\\ &\,\hspace{2pt}+9.78+3.83+12.50)\\ =&\,5.89. \end{align*}

The difference from the integrated value 5.905.90 arises from rounding the other expected frequencies.

(b)

解法一

思路

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把期望频数小于 55 的 [3.5,4)[3.5,4) 与前一格合并。密度参数已知，按合并后格数确定自由度。

答题过程

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Test H0H_0 that the stated density fits the observations against H1H_1 that it does not. Combine [3,3.5)[3,3.5) with [3.5,4)[3.5,4). The combined observed count is 1616, and the expected count is 200(3−2−4−2)=9.72222…200(3^{-2}-4^{-2})=9.72222\ldots. All six expected frequencies then exceed 55. Using unrounded frequencies,

 χ2=∑(O−E)2E= (97−111.111…)2111.111…+ ⋯+(16−9.72222…)29.72222…+ (14−12.5)212.5= 10.59268….\begin{align*} &\,\chi^2=\sum\frac{(O-E)^2}{E}\\ =&\,\frac{(97-111.111\ldots)^2}{111.111\ldots}\\ +&\,\cdots+\frac{(16-9.72222\ldots)^2}{9.72222\ldots}\\ +&\,\frac{(14-12.5)^2}{12.5}\\ =&\,10.59268\ldots. \end{align*}

There are 6−1=56-1=5 degrees of freedom. The upper 10%10\% critical value is 9.2369.236. Since the statistic exceeds it, we reject H0H_0. There is sufficient evidence at the 10%10\% level that the density is unsuitable for the data.

解法二

思路

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也可以将小频数格并入右侧的尾格。重新计算统计量后用相同的自由度判断。

答题过程

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Combining [3.5,4)[3.5,4) with [4,∞)[4,\infty) gives observed count 2020 and expected count 200(3.5−2)=16.32653…200(3.5^{-2})=16.32653\ldots. The other five groups are unchanged. In particular, [3,3.5)[3,3.5) contributes (10−b)2/b(10-b)^2/b with unrounded bb. Summing the six contributions now gives χ2=10.04279…\chi^2=10.04279\ldots. The degrees of freedom are still 55. Since 10.04279…>9.23610.04279\ldots>9.236, we again reject the proposed model at the 10%10\% level.