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S1 2025 Jan Q5

A Level / Edexcel / S1

IAL 2025 Jan Paper · Question 5

题目

Problem

A recycling centre measures the weight of glass deposited by the public each day. The weight of glass, SS kg, deposited at the recycling centre in a day during the summer can be modelled by S∼N(700,502)S\sim N(700,50^2).

(a) Using standardisation and showing your working, find the probability that, in one randomly selected day during the summer,

(i) more than 640640 kg of glass is deposited at the recycling centre,

(2)

(ii) 700700 kg of glass, correct to the nearest 5050 kg, is deposited at the recycling centre.

(5)

The weight of glass, WW kg, deposited at the recycling centre in a day during the winter can be modelled by W∼N(μ,σ2)W\sim N(\mu,\sigma^2).

(b) Given that P(W>680)=0.0668P(W>680)=0.0668 and P(W<599)=0.3P(W<599)=0.3

(i) find two equations in terms of μ\mu and σ\sigma

(3)

(ii) Hence, showing your working, find the value of μ\mu and the value of σ\sigma

(3)
题目中文翻译

一家回收中心测量公众每天送来的玻璃重量。夏季某一天送来的玻璃重量 SS kg 可用 S∼N(700,502)S\sim N(700,50^2) 建模。

(a) 使用标准化并写出计算过程，求夏季随机选取的一天中：

(i) 送来的玻璃超过 640640 kg 的概率。（2 分）

(ii) 送来的玻璃重量四舍五入到最接近的 5050 kg 后为 700700 kg 的概率。（5 分）

冬季某一天送来的玻璃重量 WW kg 可用 W∼N(μ,σ2)W\sim N(\mu,\sigma^2) 建模。

(b) 已知 P(W>680)=0.0668P(W>680)=0.0668，P(W<599)=0.3P(W<599)=0.3。

(i) 求两个包含 μ\mu 和 σ\sigma 的方程。（3 分）

(ii) 由此写出计算过程，求 μ\mu 和 σ\sigma 的值。（3 分）

解答

(a)(i)

解法一

思路

展开

标准化：

Z=S−70050.\begin{align*} Z=\frac{S-700}{50}. \end{align*}

640640 比平均数低 6060，所以对应 z=−1.2z=-1.2。要求 P(S>640)P(S>640)，也就是 P(Z>−1.2)P(Z>-1.2)。

答题过程

展开 P(S>640)= P(Z>640−70050)= P(Z>−1.2)= 0.8849….\begin{align*} P(S>640) =&\,P\Big(Z>\frac{640-700}{50}\Big)\\ =&\,P(Z>-1.2)\\ =&\,0.8849\ldots. \end{align*}

Therefore

P(S>640)=0.885\begin{align*} P(S>640)=0.885 \end{align*}

to 33 significant figures.

(a)(ii)

解法一

思路

展开

“700700 kg correct to the nearest 5050 kg” 表示实际值在 675675 到 725725 之间。然后标准化两个端点。

答题过程

展开

700700 kg correct to the nearest 5050 kg means

675<S<725.\begin{align*} 675<S<725. \end{align*}

The standardised boundaries are

zlower= 675−70050=−0.5,zupper= 725−70050=0.5.\begin{align*} z_{\text{lower}}=&\,\frac{675-700}{50}=-0.5,\\ z_{\text{upper}}=&\,\frac{725-700}{50}=0.5. \end{align*}

Therefore

P(675<S<725)= P(−2550<Z<2550)= P(−0.5<Z<0.5).\begin{align*} P(675<S<725) =&\,P\Big(-\frac{25}{50}<Z<\frac{25}{50}\Big)\\ =&\,P(-0.5<Z<0.5). \end{align*}

Using the standard normal distribution,

P(−0.5<Z<0.5)= P(Z<0.5)−P(Z<−0.5)= 0.6915−(1−0.6915)= 0.3830.\begin{align*} P(-0.5<Z<0.5) =&\,P(Z<0.5)-P(Z<-0.5)\\ =&\,0.6915-(1-0.6915)\\ =&\,0.3830. \end{align*}

So the probability is

0.383.\begin{align*} 0.383. \end{align*}

解法二

思路

展开

标准化后区间关于 00 对称，可取均值右侧面积的两倍，也可从 11 减去两端尾部面积。

答题过程

展开

700700 kg correct to the nearest 5050 kg means 675<S<725675 < S < 725.

Standardising each boundary:

zlower= 675−70050=−0.5,zupper= 725−70050=0.5.\begin{align*} z_{\text{lower}}=&\,\frac{675-700}{50}=-0.5,\\ z_{\text{upper}}=&\,\frac{725-700}{50}=0.5. \end{align*}

Therefore

P(675<S<725)= P(−0.5<Z<0.5).\begin{align*} P(675 < S < 725) =&\,P(-0.5 < Z < 0.5). \end{align*}

By symmetry of the standard normal distribution:

P(−0.5<Z<0.5)= 2P(Z<0.5)−1= 2(0.6915)−1= 1.3830−1= 0.383.\begin{align*} P(-0.5 < Z < 0.5) =&\, 2P(Z < 0.5) - 1\\ =&\, 2(0.6915) - 1\\ =&\, 1.3830 - 1\\ =&\, 0.383. \end{align*}

So the probability is 0.3830.383.

Equivalently, subtract the two equal tails:

P(−0.5<Z<0.5)= 1−2P(Z<−0.5)= 1−2(0.3085)= 0.383.\begin{align*} P(-0.5<Z<0.5)=&\,1-2P(Z<-0.5)\\ =&\,1-2(0.3085)\\ =&\,0.383. \end{align*}

(b)(i)

解法一

思路

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把两个概率都转成标准正态边界。P(W>680)=0.0668P(W>680)=0.0668 表示左侧概率 0.93320.9332，对应 z≈1.5z\approx1.5。P(W<599)=0.3P(W<599)=0.3 对应 z≈−0.5244z\approx-0.5244。

答题过程

展开

From

P(W>680)=0.0668,\begin{align*} P(W>680)=0.0668, \end{align*}

we have

P(W<680)=0.9332.\begin{align*} P(W<680)=0.9332. \end{align*}

The corresponding zz-value is 1.51.5, so

680−μσ=1.5.\begin{align*} \frac{680-\mu}{\sigma}=1.5. \end{align*}

Also,

P(W<599)=0.3.\begin{align*} P(W<599)=0.3. \end{align*}

The corresponding zz-value is −0.5244-0.5244, so

599−μσ=−0.5244.\begin{align*} \frac{599-\mu}{\sigma}=-0.5244. \end{align*}

(b)(ii)

解法一

思路

展开

由两个方程：

680−μ= 1.5σ,599−μ= −0.5244σ.\begin{align*} 680-\mu=&\,1.5\sigma,\\ 599-\mu=&\,-0.5244\sigma. \end{align*}

两式相减可以消去 μ\mu。

答题过程

展开

From part (b)(i),

680−μ=1.5σ\begin{align*} 680-\mu=1.5\sigma \end{align*}

and

599−μ=−0.5244σ.\begin{align*} 599-\mu=-0.5244\sigma. \end{align*}

Subtract the second equation from the first:

81= 1.5σ−(−0.5244σ)= 2.0244σ.\begin{align*} 81 =&\,1.5\sigma-(-0.5244\sigma)\\ =&\,2.0244\sigma. \end{align*}

Hence

σ= 812.0244= 40.011….\begin{align*} \sigma =&\,\frac{81}{2.0244}\\ =&\,40.011\ldots. \end{align*}

Using 680−μ=1.5σ680-\mu=1.5\sigma,

μ= 680−1.5(40.011…)= 619.98….\begin{align*} \mu =&\,680-1.5(40.011\ldots)\\ =&\,619.98\ldots. \end{align*}

Therefore

μ=620,σ=40.0.\begin{align*} \mu=620,\qquad \sigma=40.0. \end{align*}