Skip to content
CalcGospel 國際數學圖譜
返回

IAL 2026 Jan S2 Q4

A Level / Edexcel / S2

IAL 2026 Jan Paper · Question 4

题目

Problem

An optician is testing patients for a specific eye condition.

It is known from past records that 5% of patients have this eye condition.

(a) Find the probability that from a random sample of 30 patients

(i) exactly one patient has the eye condition,

(ii) no more than 2 patients have the eye condition.

(2)

The optician claims that the proportion of patients with the eye condition has changed.

A random sample of 150 patients is taken and 13 have the eye condition.

(b) Using a suitable approximation, carry out an appropriate test to investigate the optician’s claim. Use a 5% level of significance and state your hypotheses clearly.

(6)

(Total for Question 4 is 8 marks)

题目中文翻译

一位眼科医生正在检测患者是否患有特定眼疾。

根据过去的记录,已知 5% 的患者患有此眼疾。

(a) 求在随机抽取的 30 名患者中

(i) 恰好有一名患者患有此眼疾的概率,

(ii) 不超过 2 名患者患有此眼疾的概率。

眼科医生声称患有此眼疾的患者比例已经改变。

随机抽取了 150 名患者样本,其中 13 人患有此眼疾。

(b) 使用适当的近似方法,进行适当的检验来调查眼科医生的声明。使用 5% 的显著性水平并清楚地陈述你的假设。

(第 4 题共 8 分)

解答

(a)(i)

解法一

思路

展开

XX 表示 30 名患者中患有该眼疾的人数,则 XB(30,0.05)X\sim B(30,0.05)。把 X=1X=1 代入二项分布的单点概率公式。

答题过程

展开

Let XX be the number of patients, out of 30, who have the eye condition. Then

XB(30,0.05).X\sim\operatorname{B}(30,0.05).

Therefore,

P(X=1)=(301)(0.05)(0.95)29=0.338903=0.339(3 s.f.).\begin{align*} P(X=1) =&\,\binom{30}{1}(0.05)(0.95)^{29}\\ =&\,0.338903\ldots\\ =&\,\boxed{0.339}\quad\text{(3 s.f.)}. \end{align*}

(a)(ii)

解法一

思路

展开

“不超过 2 人”对应 X2X\leq2,因此把 X=0,1,2X=0,1,2 的二项概率相加,或直接使用二项分布的累积概率。

答题过程

展开

Using the same binomial distribution,

P(X2)=x=02(30x)(0.05)x(0.95)30x=0.812178=0.812(3 s.f.).\begin{align*} P(X\leq2) =&\,\sum_{x=0}^{2} \binom{30}{x}(0.05)^x(0.95)^{30-x}\\ =&\,0.812178\ldots\\ =&\,\boxed{0.812}\quad\text{(3 s.f.)}. \end{align*}

(b)

解法一

思路

展开

“比例已经改变”表示双尾检验。原假设下样本人数服从 B(150,0.05)B(150,0.05);由于 nn 大而 pp 小,用均值 np=7.5np=7.5 的泊松分布近似。观察值 1313 在均值上方,因此计算上尾概率,并与双尾检验单侧的 0.0250.025 比较。

答题过程

展开

Let pp be the proportion of patients who have the eye condition. The hypotheses are

H0:p=0.05,H1:p0.05.H_0:p=0.05, \qquad H_1:p\neq0.05.

Under H0H_0, the number YY of affected patients in the sample has distribution B(150,0.05)B(150,0.05). Since nn is large and pp is small,

Y˙Po(150×0.05)=Po(7.5).Y\mathrel{\dot\sim}\operatorname{Po}(150\times0.05) =\operatorname{Po}(7.5).

For the observed value Y=13Y=13,

P(Y13)=1P(Y12)=0.0426658=0.0427(4 s.f.).\begin{align*} P(Y\geq13) =&\,1-P(Y\leq12)\\ =&\,0.0426658\ldots\\ =&\,0.0427\quad\text{(4 s.f.)}. \end{align*}

This is a two-tailed test, and

0.0427>0.052=0.025.0.0427>\frac{0.05}{2}=0.025.

Therefore, H0H_0 is not rejected. There is insufficient evidence at the 5%5\% significance level to suggest that the proportion of patients with the eye condition has changed.

解法二

思路

展开

也可直接找出泊松近似下的双尾临界区域。每一尾的概率不能超过 0.0250.025:下尾边界为 Y2Y\leq2,上尾边界为 Y14Y\geq14。观察值 1313 不在临界区域,因此结论相同。

答题过程

展开

Using Y˙Po(7.5)Y\mathrel{\dot\sim}\operatorname{Po}(7.5),

P(Y2)=0.02026<0.025P(Y\leq2)=0.02026<0.025

and

P(Y14)=0.02156<0.025.P(Y\geq14)=0.02156<0.025.

The adjacent outcomes would make the corresponding tail probability exceed 0.0250.025, so the critical region is

Y2orY14.Y\leq2 \quad\text{or}\quad Y\geq14.

The observed value Y=13Y=13 is not in the critical region. Therefore, H0H_0 is not rejected, and there is insufficient evidence that the proportion of patients with the eye condition has changed.