CalcGospel 國際數學圖譜
返回

IAL 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)

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\left(Z>\frac{640-700}{50}\right)\\[3mm] =&\,P(Z>-1.2)\\[3mm] =&\,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*}

Therefore

P(675<S<725)= P(675−70050<Z<725−70050)= P(−0.5<Z<0.5).\begin{align*} P(675<S<725) =&\,P\left(\frac{675-700}{50}<Z<\frac{725-700}{50}\right)\\[3mm] =&\,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)\\[3mm] =&\,0.6915-(1-0.6915)\\[3mm] =&\,0.3830. \end{align*}

So the probability is

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

解法二

思路

展开

正态分布对称性。由于区间 [675,725][675, 725] 的中点恰好是均值 μ=700\mu = 700，即区间关于均值完美对称。在标准化后，对应的区间为 [−0.5,0.5][-0.5, 0.5]，也关于 00 对称。因此可以使用对称公式直接求概率：

P(−a<Z<a)=2Φ(a)−1\begin{align*} P(-a < Z < a) = 2\Phi(a) - 1 \end{align*}

代入 a=0.5a = 0.5 即可。这种方法大大简化了查表和计算对立事件的繁琐步骤。

答题过程

展开

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

Standardising the boundaries:

P(675<S<725)=  P(675−70050<Z<725−70050)=  P(−0.5<Z<0.5)\begin{align*} P(675 < S < 725) =&\,\, P\left(\frac{675-700}{50} < Z < \frac{725-700}{50}\right)\\[3mm] =&\,\, 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\\[3mm] =&\,\, 2(0.6915) - 1\\[3mm] =&\,\, 1.3830 - 1\\[3mm] =&\,\, 0.383. \end{align*}

So the probability is 0.3830.383.

(b)(i)

解法一

思路

展开

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,\qquad 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)\\[3mm] =&\,2.0244\sigma. \end{align*}

Hence

σ= 812.0244= 40.011….\begin{align*} \sigma =&\,\frac{81}{2.0244}\\[3mm] =&\,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)\\[3mm] =&\,619.98\ldots. \end{align*}

Therefore

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