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S3 2022 Jan UNUSED Q7

A Level / Edexcel / S3

IAL 2022 Jan Unused Paper · Question 7

题目

Problem

7.

A market stall sells vegetables. Two of the vegetables sold are broccoli heads and cabbages.

The weights of these broccoli heads, BB kilograms, follow a normal distribution

B∼N(0.588,0.0842)B \sim N(0.588, 0.084^2)

The weights of these cabbages, CC kilograms, follow a normal distribution

C∼N(0.908,0.0392)C \sim N(0.908, 0.039^2)

(a) Find the probability that the total weight of two randomly chosen broccoli heads is less than the weight of a randomly chosen cabbage.

(4)

Broccoli heads cost £2.50 per kg and cabbages cost £3.00 per kg.

Jaymini buys 1 broccoli head and 2 cabbages, chosen randomly.

(b) Find the probability that she pays more than £7

(4)

The market stall offers a discount for buying 5 or more broccoli heads. The price with the discount is £ww per kg.

Let £DD be the price with the discount of 5 broccoli heads.

(c) Find, in terms of ww, the mean and standard deviation of DD

(2)

Given that P(D<6)<0.1P(D<6)<0.1

(d) find the smallest possible value of ww, giving your answer to 2 decimal places.

(4)
(Total 14 marks)
题目中文翻译

一个菜摊出售蔬菜。其中两种蔬菜分别是西兰花和卷心菜。

这些西兰花的重量 BB（单位：千克）服从正态分布

B∼N(0.588,0.0842)B \sim N(0.588, 0.084^2)

这些卷心菜的重量 CC（单位：千克）服从正态分布

C∼N(0.908,0.0392)C \sim N(0.908, 0.039^2)

(a) 求两颗随机选取的西兰花总重量小于一颗随机选取的卷心菜重量的概率。

(4)

西兰花每千克 £2.50，卷心菜每千克 £3.00。

Jaymini 随机购买 1 颗西兰花和 2 颗卷心菜。

(b) 求她支付超过 £7 的概率。

(4)

购买 5 颗或以上西兰花可享受折扣，折扣后的单价为每千克 ww 英镑。 设 £DD 为 5 颗西兰花的折扣价格。

(c) 求 DD 的均值和标准差，用 ww 表示。

(2)

已知 P(D<6)<0.1P(D<6)<0.1。

(d) 求 ww 的最小可能值，答案保留 2 位小数。

(4)

解答

(a)

解法一

思路

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设两颗西兰花总重量减去一颗卷心菜重量为 XX。独立正态变量的线性组合仍服从正态分布；相减时均值相减，但方差仍相加。所求事件就是 X<0X<0。

答题过程

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Let

X=B1+B2−C.X=B_1+B_2-C.

Assuming that the randomly chosen vegetables have independent weights,

E⁡(X)= 2(0.588)−0.908= 0.268,\begin{align*} \operatorname{E}(X) =&\,2(0.588)-0.908\\ =&\,0.268, \end{align*}

and

Var⁡(X)= 2(0.0842)+0.0392= 0.015633.\begin{align*} \operatorname{Var}(X) =&\,2(0.084^2)+0.039^2\\ =&\,0.015633. \end{align*}

Therefore,

X∼N(0.268,0.015633).X\sim N(0.268,0.015633).

Hence

P⁡(B1+B2<C)= P⁡(X<0)= P⁡(Z<0−0.2680.015633)= P⁡(Z<−2.143…)≈ 0.0160.\begin{align*} \operatorname{P}(B_1+B_2<C) =&\,\operatorname{P}(X<0)\\ =&\,\operatorname{P}\bigg( Z<\frac{0-0.268}{\sqrt{0.015633}} \bigg)\\ =&\,\operatorname{P}(Z<-2.143\ldots)\\ \approx&\,\boxed{0.0160}. \end{align*}

Using rounded normal tables gives 0.01620.0162, which is also acceptable.

(b)

解法一

思路

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把总价写成各重量的线性组合：西兰花重量乘 2.52.5，两颗卷心菜重量分别乘 33。计算线性组合的方差时，各系数需要平方。

答题过程

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Let YY be the total cost. Then

Y=2.5B+3C1+3C2.Y=2.5B+3C_1+3C_2.

Its mean is

E⁡(Y)= 2.5(0.588)+2(3)(0.908)= 6.918,\begin{align*} \operatorname{E}(Y) =&\,2.5(0.588)+2(3)(0.908)\\ =&\,6.918, \end{align*}

and its variance is

Var⁡(Y)= 2.52(0.0842)+2(32)(0.0392)= 0.071478.\begin{align*} \operatorname{Var}(Y) =&\,2.5^2(0.084^2)+2(3^2)(0.039^2)\\ =&\,0.071478. \end{align*}

Thus

Y∼N(6.918,0.071478),Y\sim N(6.918,0.071478),

so

P⁡(Y>7)= P⁡(Z>7−6.9180.071478)= P⁡(Z>0.3067…)≈ 0.3795.\begin{align*} \operatorname{P}(Y>7) =&\,\operatorname{P}\bigg( Z>\frac{7-6.918}{\sqrt{0.071478}} \bigg)\\ =&\,\operatorname{P}(Z>0.3067\ldots)\\ \approx&\,\boxed{0.3795}. \end{align*}

(c)

解法一

思路

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五颗西兰花的总重量均值和方差分别是单颗的 55 倍；总价 DD 是总重量乘以每千克价格 ww，所以均值乘 ww，标准差也乘 ww。

答题过程

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The total cost is

D=w(B1+B2+B3+B4+B5).D=w(B_1+B_2+B_3+B_4+B_5).

Therefore,

E⁡(D)= 5w(0.588)= 2.94w,\begin{align*} \operatorname{E}(D) =&\,5w(0.588)\\ =&\,\boxed{2.94w}, \end{align*}

and

SD⁡(D)= w5(0.0842)= 0.0845 w.\begin{align*} \operatorname{SD}(D) =&\,w\sqrt{5(0.084^2)}\\ =&\,\boxed{0.084\sqrt{5}\,w}. \end{align*}

(d)

解法一

思路

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由 (c) 标准化 DD。因为标准正态分布的第 1010 百分位数为 −1.2816-1.2816，要使 P⁡(D<6)<0.1\operatorname{P}(D<6)<0.1，对应的标准化界值必须小于 −1.2816-1.2816。解出 ww 的下界，再按货币的两位小数选择最小可行值。

答题过程

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From part (c),

D∼N(2.94w,(0.0845 w)2).D\sim N\bigg(2.94w,\bigg(0.084\sqrt{5}\,w\bigg)^2\bigg).

Since P⁡(Z<−1.2816)=0.1\operatorname{P}(Z<-1.2816)=0.1, the condition P⁡(D<6)<0.1\operatorname{P}(D<6)<0.1 requires

6−2.94w0.0845 w<−1.2816.\frac{6-2.94w}{0.084\sqrt{5}\,w}<-1.2816.

Since w>0w>0 and 2.94−1.2816(0.0845)=2.699277…>02.94-1.2816(0.084\sqrt5)=2.699277\ldots>0,

6−2.94w< −1.2816(0.0845 w)6< (2.94−1.2816(0.0845))ww> 2.222…\begin{align*} 6-2.94w <&\,-1.2816(0.084\sqrt{5}\,w)\\ 6 <&\,\bigg(2.94-1.2816(0.084\sqrt{5})\bigg)w\\ w >&\,2.222\ldots \end{align*}

Therefore, the smallest possible price per kilogram, to the nearest penny, is

w=2.23.\boxed{w=2.23}.