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S1 2022 June Q6

A Level / Edexcel / S1

IAL 2022 June Paper · Question 6

题目

Problem

A manufacturer fills bottles with oil. The volume of oil in a bottle, VV ml, is normally distributed with V∼N(100,2.52)V\sim N(100,2.5^2)

(a) Find P(V>104.9)P(V>104.9)

(3)

(b) In a pack of 150150 bottles, find the expected number of bottles containing more than 104.9104.9 ml

(2)

(c) Find the value of vv, to 22 decimal places, such that P(V>v∣V<104.9)=0.2801P(V>v\mid V<104.9)=0.2801

(6)
题目中文翻译

一家制造商向瓶中灌装食用油。每瓶油的体积 VV ml 服从正态分布

V∼N(100,2.52).V\sim N(100,2.5^2).

(a) 求 P(V>104.9)P(V>104.9)。

(b) 一箱有 150150 瓶油，求其中装有超过 104.9104.9 ml 油的瓶数的期望值。

(c) 求 vv 的值，答案保留 22 位小数，使得

P(V>v∣V<104.9)=0.2801.P(V>v\mid V<104.9)=0.2801.

解答

(a)

解法一

思路

展开

先标准化 104.9104.9，再求右尾概率。

答题过程

展开 P(V>104.9)= P(Z>104.9−1002.5)= P(Z>1.96)= 1−0.9750= 0.0250.\begin{align*} P(V>104.9) =&\,P\left(Z>\frac{104.9-100}{2.5}\right)\\ =&\,P(Z>1.96)\\ =&\,1-0.9750\\ =&\,0.0250. \end{align*}

(b)

解法一

思路

展开

Expected number = number of bottles ×\times probability.

答题过程

展开 150(0.0250)=3.75.\begin{align*} 150(0.0250)=3.75. \end{align*}

The expected number of bottles is

3.75.\begin{align*} 3.75. \end{align*}

(c)

解法一

思路

展开

条件概率的分母是 P(V<104.9)=0.9750P(V<104.9)=0.9750。条件 V>vV>v 且 V<104.9V<104.9 表示区间 v<V<104.9v<V<104.9。

答题过程

展开

We need

P(V>v∣V<104.9)=0.2801.\begin{align*} P(V>v\mid V<104.9)=0.2801. \end{align*}

So

P(v<V<104.9)P(V<104.9)=0.2801.\begin{align*} \frac{P(v<V<104.9)}{P(V<104.9)}=0.2801. \end{align*}

Since

P(V<104.9)=0.9750,\begin{align*} P(V<104.9)=0.9750, \end{align*}

we get

P(v<V<104.9)=0.2801(0.9750).\begin{align*} P(v<V<104.9)=0.2801(0.9750). \end{align*}

Therefore

P(V<v)= 0.9750−0.2801(0.9750)= 0.7019…\begin{align*} P(V<v) =&\,0.9750-0.2801(0.9750)\\ =&\,0.7019\ldots \end{align*}

The standard normal value for 0.7019…0.7019\ldots is approximately

z=0.53.\begin{align*} z=0.53. \end{align*}

Thus

v−1002.5=0.53.\begin{align*} \frac{v-100}{2.5}=0.53. \end{align*}

So

v=100+0.53(2.5)=101.325.\begin{align*} v=100+0.53(2.5)=101.325. \end{align*}

Hence

v=101.33\begin{align*} v=101.33 \end{align*}

to 22 decimal places.

解法二

思路

展开

先在条件 V<104.9V<104.9 下取补事件，直接求 P(V<v)P(V<v)，再标准化查表。

答题过程

展开

Using the complement rule,

P(V<v∣V<104.9)= 1−0.2801= 0.7199.\begin{align*} P(V<v\mid V<104.9) =&\,1-0.2801\\ =&\,0.7199. \end{align*}

Since v<104.9v<104.9, the event V<vV<v is a subset of V<104.9V<104.9. Hence

P(V<v)P(V<104.9)= 0.7199,P(V<v)= 0.7199(0.9750)= 0.7019…\begin{align*} \frac{P(V<v)}{P(V<104.9)}=&\,0.7199,\\ P(V<v)=&\,0.7199(0.9750)\\ =&\,0.7019\ldots \end{align*}

Therefore z≈0.53z\approx0.53, so

v−1002.5= 0.53,v= 101.325= 101.33to 2 d.p.\begin{align*} \frac{v-100}{2.5}=&\,0.53,\\ v=&\,101.325\\ =&\,101.33\quad\text{to 2 d.p.} \end{align*}