Skip to content
CalcGospel 國際數學圖譜
返回

S1 2018 Oct Q6

A Level / Edexcel / S1

IAL 2018 Oct Paper · Question 6

题目

Problem

A machine makes bolts such that the length, LL cm, of a bolt has distribution

L∼N(4.1,0.1252)L\sim N(4.1,0.125^2)

A bolt is selected at random.

(a) Find the probability that the length of this bolt is more than 4.3 cm.

(3)

(b) Show that P(3.9<L<4.3)P(3.9<L<4.3) is 0.890 correct to 3 decimal places.

(1)

The machine makes 500 bolts.

The cost to make each bolt is 5 pence.

Only bolts with length between 3.9 cm and 4.3 cm can be used.

These are sold for 9 pence each.

All the bolts that cannot be used are recycled with a scrap value of 1 pence each.

(c) Calculate an estimate for the profit made on these 500 bolts.

(4)

Following adjustments to the machine, the length of a bolt, BB cm, made by the machine is such that B∼N(μ,σ2)B\sim N(\mu,\sigma^2)

Given that P(B>4.198)=0.025P(B>4.198)=0.025 and P(B<4.065)=0.242P(B<4.065)=0.242

(d) find the value of μ\mu and the value of σ\sigma

(6)

(e) State, giving a reason, whether the adjustments to the machine will result in a decrease or an increase in the profit made on 500 bolts.

(2)

中文翻译

某机器生产螺栓,螺栓长度 LL cm 服从 L∼N(4.1,0.1252)L\sim N(4.1,0.125^2)。随机选取一枚螺栓。

(a) 求该螺栓长度超过 4.3 cm 的概率。

(b) 证明 P(3.9<L<4.3)P(3.9<L<4.3) 为 0.890(精确至 3 位小数)。

机器生产 500 枚螺栓。每枚制造成本为 5 便士;只有长度介于 3.9 cm 与 4.3 cm 之间的螺栓可用,并以每枚 9 便士出售;不可用的螺栓回收后,每枚废料价值为 1 便士。

(c) 估计这 500 枚螺栓带来的利润。

调整机器后,螺栓长度 BB cm 服从 B∼N(μ,σ2)B\sim N(\mu,\sigma^2)。已知 P(B>4.198)=0.025P(B>4.198)=0.025 且 P(B<4.065)=0.242P(B<4.065)=0.242。

(d) 求 μ\mu 与 σ\sigma。

(e) 说明机器调整后,生产 500 枚螺栓的利润会减少还是增加,并给出理由。

解答

(a)

思路

展开

把 4.3 标准化,再求标准正态分布的右尾概率。

答题过程

展开 P(L>4.3)=P(Z>4.3−4.10.125)=P(Z>1.6)=1−0.9452=0.0548.\begin{aligned} P(L>4.3) &=P\left(Z>\frac{4.3-4.1}{0.125}\right)\\ &=P(Z>1.6)\\ &=1-0.9452\\ &=0.0548. \end{aligned}

(b)

思路

展开

3.9 与 4.3 关于均值 4.1 对称,标准化后分别为 −1.6-1.6 与 1.61.6。

答题过程

展开 P(3.9<L<4.3)=P(−1.6<Z<1.6)=0.9452−0.0548=0.8904=0.890(3 d.p.).\begin{aligned} P(3.9<L<4.3) &=P(-1.6<Z<1.6)\\ &=0.9452-0.0548\\ &=0.8904\\ &=0.890\quad\text{(3 d.p.)}. \end{aligned}

(c)

解法一

思路

展开

分别估计可用与不可用螺栓的数量,再计算销售及废料收入,最后减去总制造成本。

答题过程

展开

The estimated numbers of usable and unusable bolts are

500(0.8904)=445.2,500−445.2=54.8.500(0.8904)=445.2,\qquad 500-445.2=54.8.

Hence the estimated profit, in pence, is

445.2(9)+54.8(1)−500(5)=1561.6.445.2(9)+54.8(1)-500(5)=1561.6. Estimated profit≈1560 pence=£15.60.\text{Estimated profit}\approx1560\text{ pence}=\pounds15.60.

解法二

思路

展开

先计算每枚螺栓的期望收入,再减去每枚 5 便士的成本并乘以 500。

答题过程

展开 E(revenue per bolt)=9(0.8904)+1(1−0.8904)=8.1232 pence.\begin{aligned} E(\text{revenue per bolt}) &=9(0.8904)+1(1-0.8904)\\ &=8.1232\text{ pence}. \end{aligned}

Therefore

500(8.1232−5)=1561.6 pence≈£15.60.500(8.1232-5)=1561.6\text{ pence} \approx\pounds15.60.

(d)

思路

展开

把两个概率转换成标准正态分位点,建立关于 μ\mu、σ\sigma 的联立方程。

答题过程

展开

From P(B>4.198)=0.025P(B>4.198)=0.025,

4.198−μσ=1.960.\frac{4.198-\mu}{\sigma}=1.960.

From P(B<4.065)=0.242P(B<4.065)=0.242,

4.065−μσ≈−0.70.\frac{4.065-\mu}{\sigma}\approx-0.70.

Thus

μ+1.960σ=4.198,μ−0.70σ=4.065.\mu+1.960\sigma=4.198,\qquad \mu-0.70\sigma=4.065.

Subtracting gives

2.660σ=0.133,2.660\sigma=0.133,

so

σ=0.0500,μ=4.065+0.70(0.0500)=4.10.\sigma=0.0500,\qquad \mu=4.065+0.70(0.0500)=4.10.

(e)

思路

展开

比较调整前后的均值与标准差,判断落入可用长度范围的比例如何变化。

答题过程

展开

The mean remains approximately 4.104.10 cm, while the standard deviation decreases from 0.1250.125 cm to 0.05000.0500 cm. A greater proportion of bolts will therefore lie between 3.9 cm and 4.3 cm, so the profit will increase.