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S1 2023 June Q7

A Level / Edexcel / S1

IAL 2023 June Paper · Question 7

题目

Problem

A machine squeezes apples to extract their juice. The volume of juice, JJ ml, extracted from 11 kg of apples is modelled by a normal distribution with mean μ\mu and standard deviation σ\sigma

Given that μ=500\mu=500 and σ=25\sigma=25 use standardisation to

(a) (i) show that P(J>510)=0.3446P(J>510)=0.3446

(2)

(ii) calculate the value of dd such that P(J>d)=0.9192P(J>d)=0.9192

(3)

Zen randomly selects 55 bags each containing 11 kg of apples and records the volume of juice extracted from each bag of apples.

(b) Calculate the probability that each of the 55 bags of apples produce less than 510510 ml of juice.

(2)

Following adjustments to the machine, the volume of juice, RR ml, extracted from 11 kg of apples is such that μ=520\mu=520 and σ=k\sigma=k

Given that P(R<r)=0.15P(R<r)=0.15 and P(R>3r−800)=0.005P(R>3r-800)=0.005

(c) find the value of rr and the value of kk

(7)
题目中文翻译

一台机器压榨苹果以提取果汁。每 11 kg 苹果所提取的果汁体积 JJ ml 用平均数为 μ\mu、标准差为 σ\sigma 的正态分布建模。

已知 μ=500\mu=500、σ=25\sigma=25，使用标准化：

(a) (i) 证明 P(J>510)=0.3446P(J>510)=0.3446；

(ii) 求满足 P(J>d)=0.9192P(J>d)=0.9192 的 dd 值。

Zen 随机选择 55 袋苹果，每袋 11 kg，并记录每袋所提取的果汁体积。

(b) 计算这 55 袋苹果各自产生少于 510510 ml 果汁的概率。

机器调整后，每 11 kg 苹果所提取的果汁体积 RR ml 满足 μ=520\mu=520、σ=k\sigma=k。

已知 P(R<r)=0.15P(R<r)=0.15 且 P(R>3r−800)=0.005P(R>3r-800)=0.005。

(c) 求 rr 和 kk 的值。

解答

(a)(i)

解法一

思路

展开

标准化 510510，然后用标准正态表找右尾概率。

答题过程

展开 P(J>510)= P(Z>510−50025)= P(Z>0.4)= 1−0.6554= 0.3446.\begin{align*} P(J>510) =&\,P\left(Z>\frac{510-500}{25}\right)\\ =&\,P(Z>0.4)\\ =&\,1-0.6554\\ =&\,0.3446. \end{align*}

(a)(ii)

解法一

思路

展开

P(J>d)=0.9192P(J>d)=0.9192，所以左侧概率是 0.08080.0808，对应 z≈−1.4z\approx -1.4。

答题过程

展开

Since

P(J>d)=0.9192,\begin{align*} P(J>d)=0.9192, \end{align*}

we have

P(J<d)=0.0808.\begin{align*} P(J<d)=0.0808. \end{align*}

So

d−50025=−1.4.\begin{align*} \frac{d-500}{25}=-1.4. \end{align*}

Hence

d=500−1.4(25)=465.\begin{align*} d=500-1.4(25)=465. \end{align*}

(b)

解法一

思路

展开

每袋少于 510510 ml 的概率是 1−0.34461-0.3446。55 袋都少于 510510 ml，要把这个概率乘 55 次。

答题过程

展开 P(all 5 bags produce less than 510)= (1−0.3446)5= 0.1209…\begin{align*} P(\text{all 5 bags produce less than }510) =&\,(1-0.3446)^5\\ =&\,0.1209\ldots \end{align*}

Therefore the probability is

0.121\begin{align*} 0.121 \end{align*}

to 33 significant figures.

(c)

解法一

思路

展开

把两个概率条件分别转成标准正态方程。P(R<r)=0.15P(R<r)=0.15 给出 z=−1.0364z=-1.0364；P(R>3r−800)=0.005P(R>3r-800)=0.005 表示左侧概率为 0.9950.995，给出 z=2.5758z=2.5758。

答题过程

展开

From

P(R<r)=0.15,\begin{align*} P(R<r)=0.15, \end{align*}

we get

r−520k=−1.0364.\begin{align*} \frac{r-520}{k}=-1.0364. \end{align*}

So

r=520−1.0364k.\begin{align*} r=520-1.0364k. \end{align*}

Also,

P(R>3r−800)=0.005\begin{align*} P(R>3r-800)=0.005 \end{align*}

means

P(R<3r−800)=0.995.\begin{align*} P(R<3r-800)=0.995. \end{align*}

Thus

3r−800−520k=2.5758.\begin{align*} \frac{3r-800-520}{k}=2.5758. \end{align*}

Substitute r=520−1.0364kr=520-1.0364k:

3(520−1.0364k)−1320k= 2.5758240−3.1092kk= 2.5758240−3.1092k= 2.5758k240= 5.6850kk= 42.216…\begin{align*} \frac{3(520-1.0364k)-1320}{k}=&\,2.5758\\ \frac{240-3.1092k}{k}=&\,2.5758\\ 240-3.1092k=&\,2.5758k\\ 240=&\,5.6850k\\ k=&\,42.216\ldots \end{align*}

Then

r= 520−1.0364(42.216…)= 476.246…\begin{align*} r =&\,520-1.0364(42.216\ldots)\\ =&\,476.246\ldots \end{align*}

Therefore

k=42.2,r=476\begin{align*} k=42.2,\qquad r=476 \end{align*}

to 33 significant figures.