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IAL 2025 Jun S2 Q7

A Level / Edexcel / S2

IAL 2025 June Paper · Question 7

题目

Problem

When purchasing a pair of glasses from an optician, customers are offered insurance to cover accidental damage.

Past records from the optician show that 30% of customers buy insurance when they purchase a pair of glasses.

In a random sample of 40 customers who have purchased a pair of glasses, XX represents the number who buy insurance.

(a) State a suitable model for the distribution of XX

(1)

(b) State a necessary assumption for the model in part (a) to be valid.

(1)

The probability that fewer than rr customers who have purchased a pair of glasses buy insurance is less than 0.05

(c) Find the largest possible value of rr

(2)

A second random sample of 200 customers who have purchased a pair of glasses is taken.

Using a normal approximation, the probability that at least tt of these 200 customers buy insurance is 0.9474, correct to 4 decimal places.

(d) Find the value of tt. You must show your working.

(5)

The optician decided to introduce a special offer in which customers who purchase a pair of glasses get the first three months of insurance free if they buy insurance.

Following this, a random sample of 25 customers who purchase a pair of glasses is taken and 11 of them buy insurance.

(e) Test, at the 10% level of significance, whether or not there is evidence that the proportion of customers who buy insurance when purchasing a pair of glasses has increased. State your hypotheses clearly.

(5)

(Total for Question 7 is 14 marks)

题目中文翻译

从眼科医生处购买眼镜时,顾客可获得意外损坏保险。

眼科医生的过往记录显示,30% 的顾客在购买眼镜时会购买保险。

在随机抽取的 40 名已购买眼镜的顾客中,XX 表示购买保险的人数。

(a) 说明 XX 分布的合适模型。

(b) 说明 (a) 中模型有效所需的一个必要假设。

购买眼镜的顾客中少于 rr 人购买保险的概率小于 0.05

(c) 求 rr 的最大可能值。

又随机抽取了 200 名已购买眼镜的顾客样本。

使用正态近似,这 200 名顾客中至少 tt 人购买保险的概率为 0.9474(精确到 4 位小数)。

(d) 求 tt 的值。你必须展示你的计算过程。

眼科医生决定推出一项特别优惠:购买眼镜的顾客如果购买保险,前三个月免费。

此后,随机抽取了 25 名购买眼镜的顾客样本,其中 11 人购买了保险。

(e) 在 10% 的显著性水平下,检验是否有证据表明购买眼镜时购买保险的顾客比例已经增加。清楚地陈述你的假设。

(第 7 题共 14 分)

解答

(a)

解法一

思路

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样本量固定为 40,每名顾客只有“购买保险”或“不购买保险”两种结果,购买保险的概率为 0.3。因此可使用二项分布模型。

答题过程

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A suitable model is

XB(40,0.3).\boxed{X\sim\operatorname{B}(40,0.3)}.

(b)

解法一

思路

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二项模型要求各次试验相互独立,并且成功概率保持不变。题目只要求一个必要假设,可说明不同顾客是否购买保险彼此独立。

答题过程

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A necessary assumption is that

customers make their decisions about buying insuranceindependently of one another.\boxed{ \begin{gathered} \text{customers make their decisions about buying insurance}\\ \text{independently of one another} \end{gathered} }.

(c)

解法一

思路

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因为 XX 只能取整数,“少于 rr 人”表示 P(X<r)=P(Xr1)P(X<r)=P(X\leq r-1)。检查相邻整数边界:P(X6)P(X\leq6) 小于 0.05,而 P(X7)P(X\leq7) 已大于 0.05,因此最大整数 rr 为 7。

答题过程

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Using XB(40,0.3)X\sim\operatorname{B}(40,0.3),

P(X<7)=P(X6)=0.0238<0.05P(X<7)=P(X\leq6)=0.0238<0.05

but

P(X<8)=P(X7)=0.0553>0.05.P(X<8)=P(X\leq7)=0.0553>0.05.

Therefore, the largest possible integer value is

r=7.\boxed{r=7}.

(d)

解法一

思路

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200 人中购买保险的人数服从 B(200,0.3)B(200,0.3),其均值为 60、方差为 42。用 N(60,42)N(60,42) 近似时,“至少 tt 人”经连续性修正成为正态变量大于 t0.5t-0.5。上尾概率 0.9474 对应标准正态分位数约为 1.620-1.620,由此解出整数 tt

答题过程

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Let YY be the number, out of 200 customers, who buy insurance. Then

YB(200,0.3).Y\sim\operatorname{B}(200,0.3).

Its mean and variance are

μ=200(0.3)=60\mu=200(0.3)=60

and

σ2=200(0.3)(0.7)=42.\sigma^2=200(0.3)(0.7)=42.

Therefore,

Y˙N(60,42).Y\mathrel{\dot\sim}N(60,42).

Using a continuity correction,

P(Yt)P(Y>t0.5)=0.9474.P(Y\geq t)\approx P(Y>t-0.5)=0.9474.

Since P(Z>1.62015)=0.9474P(Z>-1.62015\ldots)=0.9474,

t0.56042=1.62015.\frac{t-0.5-60}{\sqrt{42}} =-1.62015\ldots.

Hence,

t=60.51.6201542=50.001.\begin{align*} t =&\,60.5-1.62015\ldots\sqrt{42}\\ =&\,50.001\ldots. \end{align*}

As tt is an integer,

t=50.\boxed{t=50}.

(e)

解法一

思路

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pp 为推出优惠后购买保险的顾客比例。题目检验比例是否增加,因此使用右尾二项检验;在原假设下样本人数服从 B(25,0.3)B(25,0.3)。计算观察到至少 11 人的概率,并与显著性水平 0.10 比较。

答题过程

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Let pp be the proportion of customers who buy insurance after the special offer. The hypotheses are

H0:p=0.3,H1:p>0.3.H_0:p=0.3, \qquad H_1:p>0.3.

Let WW be the number who buy insurance in the sample of 25. Under H0H_0,

WB(25,0.3).W\sim\operatorname{B}(25,0.3).

For the observed value W=11W=11,

P(W11)=1P(W10)=0.0978.\begin{align*} P(W\geq11) =&\,1-P(W\leq10)\\ =&\,0.0978. \end{align*}

Since

0.0978<0.10,0.0978<0.10,

H0H_0 is rejected. There is sufficient evidence at the 10%10\% significance level to suggest that the proportion of customers who buy insurance has increased.

解法二

思路

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官方评分资料也接受临界区域法。在 B(25,0.3)B(25,0.3) 下,右尾概率首次不超过 0.10 的边界为 11,因此临界区域是 W11W\geq11;观察值正好落入临界区域。

答题过程

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Under H0H_0, WB(25,0.3)W\sim\operatorname{B}(25,0.3). The upper-tail probabilities at adjacent boundaries are

P(W10)=0.1894>0.10P(W\geq10)=0.1894>0.10

and

P(W11)=0.0978<0.10.P(W\geq11)=0.0978<0.10.

Hence, the critical region is

W11.\boxed{W\geq11}.

The observed value is W=11W=11, which lies in the critical region. Therefore, H0H_0 is rejected, and there is sufficient evidence at the 10%10\% significance level to suggest that the proportion of customers who buy insurance has increased.