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IAL 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)

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)\\[3mm] =&\,P(Z>1.96)\\[3mm] =&\,1-0.9750\\[3mm] =&\,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)\\[3mm] =&\,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

条件概率补集法。 已知 P(V>v∣V<104.9)=0.2801P(V > v \mid V < 104.9) = 0.2801。我们要求出 vv 的值，可以先对其应用对立事件公式（补集法则）：

P(V⩽v∣V<104.9)=1−P(V>v∣V<104.9)=1−0.2801=0.7199\begin{align*} P(V \leqslant v \mid V < 104.9) = 1 - P(V > v \mid V < 104.9) = 1 - 0.2801 = 0.7199 \end{align*}

根据条件概率定义，有：

P(V⩽v∩V<104.9)P(V<104.9)=0.7199\begin{align*} \frac{P(V \leqslant v \cap V < 104.9)}{P(V < 104.9)} = 0.7199 \end{align*}

由于 v<104.9v < 104.9（因为 V>vV > v 的条件概率非零），事件 V⩽vV \leqslant v 必然完全包含在 V<104.9V < 104.9 之中。 因此交集部分可以直接化简为 P(V⩽v)P(V \leqslant v)：

P(V⩽v)P(V<104.9)=0.7199⟹P(V⩽v)=0.7199×P(V<104.9)\begin{align*} \frac{P(V \leqslant v)}{P(V < 104.9)} = 0.7199 \Longrightarrow P(V \leqslant v) = 0.7199 \times P(V < 104.9) \end{align*}

代入 P(V<104.9)=0.9750P(V < 104.9) = 0.9750 即可算得 P(V⩽v)P(V \leqslant v)，最后标准化查表。这种方法通过先求補集，直接避免了分子中区间概率 P(v<V<104.9)P(v < V < 104.9) 的拆分，使代数方程极为清爽。

答题过程

展开

Use the complement rule for conditional probability:

P(V<v∣V<104.9)=  1−P(V>v∣V<104.9)=  1−0.2801=  0.7199.\begin{align*} P(V < v \mid V < 104.9) =&\,\, 1 - P(V > v \mid V < 104.9)\\[3mm] =&\,\, 1 - 0.2801\\[3mm] =&\,\, 0.7199. \end{align*}

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

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

By definition of conditional probability:

P(V<v∣V<104.9)=  P(V<v)P(V<104.9)=  0.7199.\begin{align*} P(V < v \mid V < 104.9) =&\,\, \frac{P(V < v)}{P(V < 104.9)}\\[3mm] =&\,\, 0.7199. \end{align*}

From part (a), we have P(V<104.9)=1−0.0250=0.9750P(V < 104.9) = 1 - 0.0250 = 0.9750. So:

P(V<v)=  0.7199×0.9750=  0.7019…\begin{align*} P(V < v) =&\,\, 0.7199 \times 0.9750\\[3mm] =&\,\, 0.7019\ldots \end{align*}

Find the corresponding standard normal zz-score for a cumulative probability of 0.70190.7019:

z≈0.53.\begin{align*} z \approx 0.53. \end{align*}

Standardise vv:

v−1002.5=  0.53v=  100+0.53(2.5)=  101.325.\begin{align*} \frac{v - 100}{2.5} =&\,\, 0.53\\[3mm] v =&\,\, 100 + 0.53(2.5)\\[3mm] =&\,\, 101.325. \end{align*}

So

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

to 22 decimal places.