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IAL 2023 Jan S2 Q1

A Level / Edexcel / S2

IAL 2023 Jan Paper · Question 1

题目

Problem

A shop sells shoes at a mean rate of 4 pairs of shoes per hour on a weekday.

(a) Suggest a suitable distribution for modelling the number of sales of pairs of shoes made per hour on a weekday.

(1)

(b) State one assumption necessary for this distribution to be a suitable model of this situation.

(1)

(c) Find the probability that on a weekday the shop sells

(i) more than 4 pairs of shoes in a one-hour period,

(ii) more than 4 pairs of shoes in each of 3 consecutive one-hour periods.

(4)

The area manager visits the shop on a weekday, the day after an advert for the shop appears in a local paper.

In a one-hour period during the manager’s visit, the shop sells 7 pairs of shoes. This leads the manager to believe that the advert has increased the shop’s sales of pairs of shoes.

(d) Stating your hypotheses clearly, test at the 5% level of significance whether or not there is evidence of an increase in sales of pairs of shoes following the appearance of the advert.

(5)
题目中文翻译

某鞋店在工作日的平均销售速率为每小时 4 双鞋。

(a) 建议一个适合建模工作日每小时售出鞋双数的分布。

(b) 写出该分布作为这个情境的合适模型所必需的一个假设。

(c) 求下列概率:

(i) 在一个小时内售出超过 4 双鞋;

(ii) 在连续 3 个一小时区间内,每个区间都售出超过 4 双鞋。

区域经理在某个工作日到店拜访,恰好是在一则广告刊登在当地报纸后的第二天。

在经理来访期间的一小时内,店里卖出了 7 双鞋。 这使经理相信该广告提高了鞋店的销售量。

(d) 清楚写出假设,在 5% 显著性水平下检验:在广告刊登后,是否有证据表明鞋双销量增加。

解答

(a)

解法一

思路

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每小时售出的鞋双数是固定时间区间内的事件次数,且题目给出平均发生率,因此适合用泊松分布建模。

答题过程

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A suitable model is

XPo(4),\boxed{X\sim\operatorname{Po}(4)},

where XX is the number of pairs of shoes sold in one hour.

(b)

解法一

思路

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泊松模型要求事件以恒定平均速率独立发生。结合本题语境,可陈述为各双鞋的售出相互独立。

答题过程

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The sales of pairs of shoes occur independently of one another.

(c)(i)

解法一

思路

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沿用 (a) 的泊松模型。“超过 4 双”是 X>4X>4,使用补事件把它化为 1 减去累积概率 P(X4)P(X\leqslant4)

答题过程

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For XPo(4)X\sim\operatorname{Po}(4),

P(X>4)=1P(X4)=0.3712.\begin{align*} P(X>4) =&\,1-P(X\leqslant4) \\ =&\,\boxed{0.3712}. \end{align*}

(c)(ii)

解法一

思路

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三个连续一小时区间内都要售出超过 4 双。由泊松过程在互不重叠区间内的事件次数相互独立,所求概率是 (c)(i) 未取整概率的三次方。

答题过程

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The numbers of sales in the three one-hour periods are independent. Therefore,

P(more than 4 in each period)=[P(X>4)]3=(0.371163)3=0.0511.\begin{align*} P(\text{more than 4 in each period}) =&\,\big[P(X>4)\big]^3 \\ =&\,(0.371163\ldots)^3 \\ =&\,\boxed{0.0511}. \end{align*}

(d)

解法一

思路

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经理认为广告使平均销量增加,所以进行右尾检验。原假设下每小时销量仍服从均值为 4 的泊松分布;计算观察到 7 双或更多的概率,并与 5% 显著性水平比较。

答题过程

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Let λ\lambda be the mean number of pairs of shoes sold per hour after the advert.

H0:λ=4,H1:λ>4.H_0:\lambda=4, \qquad H_1:\lambda>4.

Under H0H_0, XPo(4)X\sim\operatorname{Po}(4). The p-value is

P(X7)=1P(X6)=0.1107.\begin{align*} P(X\geqslant7) =&\,1-P(X\leqslant6) \\ =&\,0.1107. \end{align*}

Since 0.1107>0.050.1107>0.05, we do not reject H0H_0. There is insufficient evidence at the 5% significance level of an increase in sales of pairs of shoes following the appearance of the advert.

解法二

思路

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官方评分资料也接受临界域法。寻找最小整数 rr,使右尾概率 P(Xr)P(X\geqslant r) 不超过 0.05,再判断观测值 7 是否落入临界域。

答题过程

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Using the same hypotheses and XPo(4)X\sim\operatorname{Po}(4),

P(X8)=1P(X7)=0.0511>0.05,P(X9)=1P(X8)=0.0214<0.05.\begin{align*} P(X\geqslant8) =&\,1-P(X\leqslant7) \\ =&\,0.0511>0.05, \\ P(X\geqslant9) =&\,1-P(X\leqslant8) \\ =&\,0.0214<0.05. \end{align*}

Hence the critical region is X9X\geqslant9. Since the observed value 77 is not in the critical region, we do not reject H0H_0. There is insufficient evidence at the 5% significance level of an increase in sales of pairs of shoes following the appearance of the advert.