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IAL 2023 Jan Q5

A Level / Edexcel / S1

IAL 2023 Jan Paper · Question 5

题目

Problem

The lengths, LL mm, of housefly wings are normally distributed with L∼N(4.5,0.42)L\sim N(4.5,0.4^2)

(a) Find the probability that a randomly selected housefly has a wing length of less than 3.863.86 mm.

(3)

(b) Find

(i) the upper quartile (Q3Q_3) of LL

(ii) the lower quartile (Q1Q_1) of LL

(4)

A value that is greater than Q3+1.5×(Q3−Q1)Q_3+1.5\times(Q_3-Q_1) or smaller than Q1−1.5×(Q3−Q1)Q_1-1.5\times(Q_3-Q_1) is defined as an outlier.

(c) Find these two outlier limits.

(3)

A housefly is selected at random.

(d) Using standardisation, show that the probability that this housefly is not an outlier is 0.9930.993 to 33 decimal places.

(3)

Given that this housefly is not an outlier,

(e) showing your working, find the probability that the wing length of this housefly is greater than 55 mm.

(4)

解答

(a)

解法一

思路

展开

先标准化 3.863.86，再查左尾概率。

答题过程

展开 P(L<3.86)= P(Z<3.86−4.50.4)= P(Z<−1.6)= 1−0.9452= 0.0548.\begin{align*} P(L<3.86) =&\,P\left(Z<\frac{3.86-4.5}{0.4}\right)\\[3mm] =&\,P(Z<-1.6)\\[3mm] =&\,1-0.9452\\[3mm] =&\,0.0548. \end{align*}

(b)

解法一

思路

展开

上四分位数对应左侧概率 0.750.75，下四分位数对应左侧概率 0.250.25。标准正态值约为 ±0.67\pm0.67。

答题过程

展开

For Q3Q_3,

Q3−4.50.4=0.67.\begin{align*} \frac{Q_3-4.5}{0.4}=0.67. \end{align*}

So

Q3=4.5+0.67(0.4)=4.768.\begin{align*} Q_3=4.5+0.67(0.4)=4.768. \end{align*}

For Q1Q_1,

Q1−4.50.4=−0.67.\begin{align*} \frac{Q_1-4.5}{0.4}=-0.67. \end{align*}

So

Q1=4.5−0.67(0.4)=4.232.\begin{align*} Q_1=4.5-0.67(0.4)=4.232. \end{align*}

Therefore

Q3=4.77,Q1=4.23\begin{align*} Q_3=4.77,\qquad Q_1=4.23 \end{align*}

to 33 significant figures.

(c)

解法一

思路

展开

先求 IQR，再用上下 outlier limits 的公式。

答题过程

展开 IQR⁡= 4.768−4.232= 0.536.\begin{align*} \operatorname{IQR} =&\,4.768-4.232\\[3mm] =&\,0.536. \end{align*}

So

1.5IQR⁡=1.5(0.536)=0.804.\begin{align*} 1.5\operatorname{IQR}=1.5(0.536)=0.804. \end{align*}

The lower limit is

4.232−0.804=3.428.\begin{align*} 4.232-0.804=3.428. \end{align*}

The upper limit is

4.768+0.804=5.572.\begin{align*} 4.768+0.804=5.572. \end{align*}

(d)

解法一

思路

展开

“不是 outlier” 即 wing length 在两个 limits 之间。把两个 limits 标准化。

答题过程

展开 P(not an outlier)= P(3.428<L<5.572)= P(3.428−4.50.4<Z<5.572−4.50.4)= P(−2.68<Z<2.68).\begin{align*} P(\text{not an outlier}) =&\,P(3.428<L<5.572)\\[3mm] =&\,P\left(\frac{3.428-4.5}{0.4}<Z <\frac{5.572-4.5}{0.4}\right)\\[3mm] =&\,P(-2.68<Z<2.68). \end{align*}

Using the standard normal distribution,

P(−2.68<Z<2.68)=0.9926…\begin{align*} P(-2.68<Z<2.68)=0.9926\ldots \end{align*}

Therefore the probability is

0.993\begin{align*} 0.993 \end{align*}

to 33 decimal places.

(e)

解法一

思路

展开

这是条件概率。已知不是 outlier，所以分母是 P(3.428<L<5.572)P(3.428<L<5.572)；又要求 L>5L>5，所以分子是 P(5<L<5.572)P(5<L<5.572)。

答题过程

展开

First,

P(not an outlier)=0.9926…\begin{align*} P(\text{not an outlier})=0.9926\ldots \end{align*}

Also,

P(5<L<5.572)= P(5−4.50.4<Z<5.572−4.50.4)= P(1.25<Z<2.68)= 0.1021…\begin{align*} P(5<L<5.572) =&\,P\left(\frac{5-4.5}{0.4}<Z <\frac{5.572-4.5}{0.4}\right)\\[3mm] =&\,P(1.25<Z<2.68)\\[3mm] =&\,0.1021\ldots \end{align*}

Therefore

P(L>5∣not an outlier)= P(5<L<5.572)P(3.428<L<5.572)= 0.1021…0.9926…= 0.1028…\begin{align*} P(L>5\mid \text{not an outlier}) =&\,\frac{P(5<L<5.572)}{P(3.428<L<5.572)}\\[3mm] =&\,\frac{0.1021\ldots}{0.9926\ldots}\\[3mm] =&\,0.1028\ldots \end{align*}

So the probability is

0.103\begin{align*} 0.103 \end{align*}

to 33 significant figures.