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IAL 2024 Oct Q4

A Level / Edexcel / S1

IAL 2024 Oct Paper · Question 4

题目

Problem

In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.

The distances, mm miles, a motorbike travels on a full tank of petrol can be modelled by a normal distribution with mean 170170 miles and standard deviation 1616 miles.

(a) Find the probability that, on a randomly selected journey, the motorbike could travel at least 190190 miles on a full tank of petrol.

(2)

The probability that, on a randomly selected journey, the motorbike could travel at least dd miles on a full tank of petrol is 0.90.9.

(b) Find the value of dd.

(3)

解答

(a)

解法一

思路

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设 XX 表示满油可行驶距离，则 X∼N(170,162)X\sim N(170,16^2)。先标准化 190190，再查标准正态表。

答题过程

展开

Let

X∼N(170,162).\begin{align*} X\sim N(170,16^2). \end{align*}

Then

P(X⩾190)= P(Z⩾190−17016)= P(Z⩾1.25).\begin{align*} P(X\geqslant190) =&\,P\left(Z\geqslant\frac{190-170}{16}\right)\\[3mm] =&\,P(Z\geqslant1.25). \end{align*}

Using the standard normal distribution table,

P(Z⩾1.25)= 1−P(Z<1.25)= 1−0.8944= 0.1056.\begin{align*} P(Z\geqslant1.25) =&\,1-P(Z<1.25)\\[3mm] =&\,1-0.8944\\[3mm] =&\,0.1056. \end{align*}

Therefore

P(X⩾190)=0.106\begin{align*} P(X\geqslant190)=0.106 \end{align*}

to 33 significant figures.

(b)

解法一

思路

展开

P(X⩾d)=0.9P(X\geqslant d)=0.9 表示 dd 在平均数左边，而且左侧概率是 0.10.1。查标准正态表可得对应的 zz 值大约是 −1.2816-1.2816。

答题过程

展开

Since

P(X⩾d)=0.9,\begin{align*} P(X\geqslant d)=0.9, \end{align*}

we have

P(X<d)=0.1.\begin{align*} P(X<d)=0.1. \end{align*}

The corresponding standard normal value is

z=−1.2816.\begin{align*} z=-1.2816. \end{align*}

So

d−17016=−1.2816.\begin{align*} \frac{d-170}{16}=-1.2816. \end{align*}

Hence

d= 170−1.2816(16)= 149.4944.\begin{align*} d =&\,170-1.2816(16)\\[3mm] =&\,149.4944. \end{align*}

Therefore

d=149\begin{align*} d=149 \end{align*}

to 33 significant figures.