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IAL 2026 Jan S2 Q7

A Level / Edexcel / S2

IAL 2026 Jan Paper · Question 7

题目

Problem

A train travels daily between two cities.

The probability that the train arrives no more than 2 minutes late is 0.45.

Bobby randomly selects 30 of these train journeys between the two cities. The random variable XX represents the number of times the train arrives no more than 2 minutes late.

(a) Find P(X>16)P(X>16)

(2)

Hallie takes a random sample of 180 of these train journeys between the two cities.

Using a normal approximation, the probability that more than kk of these 180 train journeys arrive no more than 2 minutes late is less than 0.0427 to 4 decimal places.

(b) Using standardisation, find the smallest possible value of kk.

You must show all stages of your working.

(8)
(Total for Question 7 is 10 marks)
题目中文翻译

一列火车每天往返于两座城市之间。

火车到达时晚点不超过 2 分钟的概率为 0.45。

Bobby 从这些往返行程中随机选择 30 次。随机变量 XX 表示火车到达时晚点不超过 2 分钟的次数。

(a) 求 P(X>16)P(X>16)。

Hallie 从这些往返行程中随机抽取 180 次。

使用正态近似，在精确到 4 位小数时，这 180 次行程中晚点不超过 2 分钟的次数多于 kk 的概率小于 0.0427。

(b) 使用标准化，求 kk 的最小可能值。

你必须展示计算过程的所有步骤。

（第 7 题共 10 分）

解答

(a)

解法一

思路

展开

每次行程“晚点不超过 2 分钟”的概率固定为 0.45，随机抽取 30 次，因此使用二项分布；“大于 16”用补事件计算。

答题过程

展开 X∼B⁡(30,0.45).X\sim\operatorname{B}(30,0.45).

Therefore,

P(X>16)= 1−P(X≤16)= 1−0.8644= 0.136 (approximately).\begin{align*} P(X>16) =&\,1-P(X\leq16)\\[4mm] =&\,1-0.8644\\[4mm] =&\,\boxed{0.136}\text{ (approximately).} \end{align*}

(b)

解法一

思路

展开

先用二项分布的均值与方差建立正态近似。事件“多于 kk”用 k+0.5k+0.5 作连续性修正；再以右尾概率 0.0427 对应的标准正态临界值约 1.721.72 建立不等式。

答题过程

展开

Let YY be the number of the 180 journeys that arrive no more than 2 minutes late.

Y∼B⁡(180,0.45),E(Y)=180(0.45)=81,Var⁡(Y)=180(0.45)(0.55)=44.55.\begin{align*} Y&\sim\operatorname{B}(180,0.45),\\[4mm] E(Y)&=180(0.45)=81,\\[4mm] \operatorname{Var}(Y)&=180(0.45)(0.55)=44.55. \end{align*}

Thus

Y≈N(81,44.55).Y\approx N(81,44.55).

Using a continuity correction,

P(Y>k)≈P(Z>k+0.5−8144.55).P(Y>k)\approx P\left(Z>\frac{k+0.5-81}{\sqrt{44.55}}\right).

For this probability to be less than 0.04270.0427, the standardised value must exceed 1.721.72:

k+0.5−8144.55>1.72k>91.98…\begin{align*} \frac{k+0.5-81}{\sqrt{44.55}}&>1.72\\[4mm] k&>91.98\ldots \end{align*}

Therefore, the smallest possible integer value is

k=92.\boxed{k=92}.