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IAL 2023 Jan S2 Q5

A Level / Edexcel / S2

IAL 2023 Jan Paper · Question 5

题目

Problem

A company produces steel cable.

Defects in the steel cable produced by this company occur at random, at a constant rate of 1 defect per 16 metres.

On one day the company produces a piece of steel cable 80 metres long.

(a) Find the probability that there are at most 5 defects in this piece of steel cable.

(2)

The company produces a piece of steel cable 80 metres long on each of the next 4 days.

(b) Find the probability that fewer than 2 of these 4 pieces of steel cable contain at most 5 defects.

(4)

The following week the company produces a piece of steel cable xx metres long.

Using a normal approximation, the probability that this piece of steel cable has fewer than 26 defects is 0.5398

(c) Find the value of xx

(8)
题目中文翻译

一家公司生产钢缆。

该公司生产的钢缆会随机出现缺陷，缺陷以每 16 米平均 1 个的恒定速率出现。

某天，该公司生产了一段 80 米长的钢缆。

(a) 求这段钢缆中缺陷数不超过 5 的概率。

在接下来的 4 天中，该公司每天都生产一段 80 米长的钢缆。

(b) 求这 4 段钢缆中少于 2 段含有不超过 5 个缺陷的概率。

下一周，该公司生产了一段 xx 米长的钢缆。

使用正态近似时，这段钢缆少于 26 个缺陷的概率为 0.5398。

(c) 求 xx 的值。

解答

(a)

解法一

思路

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缺陷随机且以恒定速率出现，因此使用泊松分布。80 米钢缆的期望缺陷数为 80/16=580/16=5，“不超过 5 个”对应累积概率 P(D⩽5)P(D\leqslant5)。

答题过程

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Let DD be the number of defects in an 80-metre piece of cable. Then

D∼Po⁡(8016)=Po⁡(5).D\sim\operatorname{Po}\bigg(\frac{80}{16}\bigg) =\operatorname{Po}(5).

Therefore,

P(D⩽5)= 0.615960…≈ 0.616.\begin{align*} P(D\leqslant5)=&\,0.615960\ldots \\ \approx&\,\boxed{0.616}. \end{align*}

(b)

解法一

思路

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由 (a)，每段 80 米钢缆“缺陷数不超过 5”的概率约为 0.6160.616。把这件事视为成功，4 天中满足条件的钢缆段数服从二项分布；“少于 2 段”即 0 段或 1 段。

答题过程

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Let YY be the number of the four pieces that contain at most 5 defects. Using the result from part (a),

Y∼B(4,0.616).Y\sim B(4,0.616).

Hence,

P(Y<2)= P(Y=0)+P(Y=1)= (0.384)4 +4(0.616)(0.384)3= 0.1613…≈ 0.161.\begin{align*} P(Y<2) =&\,P(Y=0)+P(Y=1) \\ =&\,(0.384)^4 \\ &\,+4(0.616)(0.384)^3 \\ =&\,0.1613\ldots \\ \approx&\,\boxed{0.161}. \end{align*}

(c)

解法一

思路

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xx 米钢缆的缺陷数原本服从均值为 x/16x/16 的泊松分布，正态近似的均值与方差也都是 x/16x/16。“少于 26 个”是缺陷数不超过 25，因此使用连续性修正 25.5。由给定概率反查标准正态分位数约为 0.10.1，再令 t=xt=\sqrt{x} 解方程，并舍去负根。

答题过程

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Let NN be the number of defects in xx metres of cable. Using a normal approximation,

N≈N⁡(x16,x16).N\approx\operatorname{N}\bigg(\frac{x}{16},\frac{x}{16}\bigg).

Since P(N<26)=P(N⩽25)P(N<26)=P(N\leqslant25), applying a continuity correction gives

P(Z<25.5−x/16x/16)=0.5398.P\bigg( Z<\frac{25.5-x/16}{\sqrt{x/16}} \bigg)=0.5398.

From the standard normal distribution,

Φ(0.1)=0.5398\Phi(0.1)=0.5398

to 4 decimal places. Therefore,

25.5−x/16x/4=0.1.\frac{25.5-x/16}{\sqrt{x}/4}=0.1.

Let t=xt=\sqrt{x}, where t⩾0t\geqslant0. Then

25.5−t216= t405t2+2t−2040= 0(t−20)(5t+102)= 0.\begin{align*} 25.5-\frac{t^2}{16}=&\,\frac{t}{40} \\ 5t^2+2t-2040=&\,0 \\ (t-20)(5t+102)=&\,0. \end{align*}

Thus t=20t=20 or t=−20.4t=-20.4. Since t=x⩾0t=\sqrt{x}\geqslant0, t=20t=20, so

x=202=400.\boxed{x=20^2=400}.