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IAL 2019 June Q4

A Level / Edexcel / S1

IAL 2019 June Paper · Question 4

题目

Problem

The weights of packages delivered to Susie are normally distributed with a mean of 510 grams and a standard deviation of 45 grams.

(a) Find the probability that a randomly selected package delivered to Susie weighs less than 450 grams.

(3)

The heaviest 5% of packages delivered to Susie are delivered by Rav in his van, the others are delivered by Taruni on foot.

(b) Find the weight of the lightest package that Rav would deliver to Susie.

(3)

Susie randomly selects a package from those delivered by Taruni.

(c) Find the probability that this package weighs more than 450 grams.

(4)

On Tuesday there are 5 packages delivered to Susie.

(d) Find the probability that 4 are delivered by Taruni and 1 is delivered by Rav.

(3)

解答

(a)

解法一

思路

展开

标准化 450，再查左尾概率。

答题过程

展开

Let WW be the weight of a package.

W∼N(510,452).\begin{align*} W\sim N(510,45^2). \end{align*}

Then

P(W<450)=P(Z<450−51045)=P(Z<−1.333…).P(W<450)=P\left(Z<\frac{450-510}{45}\right) =P(Z<-1.333\ldots).

Using symmetry,

P(W<450)=1−0.9082=0.0918.\begin{align*} P(W<450)=1-0.9082=0.0918. \end{align*}

(b)

解法一

思路

展开

最重的 5% 是右尾 0.05，对应 z=1.6449z=1.6449。

答题过程

展开

Let dd be the required weight.

Since P(W>d)=0.05P(W>d)=0.05,

d−51045=1.6449.\begin{align*} \frac{d-510}{45}=1.6449. \end{align*}

Therefore

d=510+1.6449(45)=584.02….\begin{align*} d=510+1.6449(45)=584.02\ldots. \end{align*}

The lightest package Rav would deliver weighs approximately

584 grams.\begin{align*} 584\text{ grams}. \end{align*}

(c)

解法一

思路

展开

Taruni 送的是不属于最重 5% 的包裹，所以条件是 W<584W<584。在这个条件下求 W>450W>450。

答题过程

展开

We need

P(W>450∣W<584.02…).\begin{align*} P(W>450\mid W<584.02\ldots). \end{align*}

So

P(W>450∣W<584.02…)=P(450<W<584.02…)P(W<584.02…).P(W>450\mid W<584.02\ldots) =\frac{P(450<W<584.02\ldots)}{P(W<584.02\ldots)}.

Now

P(W<584.02…)=0.95\begin{align*} P(W<584.02\ldots)=0.95 \end{align*}

and

P(450<W<584.02…)=0.95−0.0918=0.8582.\begin{align*} P(450<W<584.02\ldots)=0.95-0.0918=0.8582. \end{align*}

Therefore

0.85820.95=0.903….\begin{align*} \frac{0.8582}{0.95}=0.903\ldots. \end{align*}

So the probability is approximately

0.903.\begin{align*} 0.903. \end{align*}

(d)

解法一

思路

展开

每个包裹由 Taruni 送的概率是 0.95，由 Rav 送的概率是 0.05。5 个中恰好 4 个 Taruni、1 个 Rav，用排列数乘上概率。

答题过程

展开 P=(51)(0.95)4(0.05).\begin{align*} P=\binom{5}{1}(0.95)^4(0.05). \end{align*}

Therefore

P=5(0.95)4(0.05)=0.2036….\begin{align*} P=5(0.95)^4(0.05)=0.2036\ldots. \end{align*}

So the probability is

0.204.\begin{align*} 0.204. \end{align*}