题目

Problem

Jim records the length, $l$ mm, of $81$ salmon. The data are coded using $x=l-600$ and the following summary statistics are obtained.

n=81\qquad \sum x=3711\qquad \sum x^2=475181

(a) Find the mean length of these salmon.

(3)

(b) Find the variance of the lengths of these salmon.

(2)

The weight, $w$ grams, of each of the $81$ salmon is recorded to the nearest gram. The recorded results for the $81$ salmon are summarised in the box plot below.

Box plot

4600<w\leqslant7700

(1)

Raj says that the box plot is incorrect as Jim has not included outliers. For these data an outlier is defined as a value that is more than

1.5\times\operatorname{IQR}

above the upper quartile or $1.5\times\operatorname{IQR}$ below the lower quartile

(d) Show that there are no outliers.

(3)

解答

(a)

解法一

思路

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由 $x=l-600$ 得 $l=x+600$ ，所以平均长度等于 $\bar{x}+600$ 。

答题过程

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\begin{align*} \bar{x}=\frac{3711}{81}=45.814\ldots \end{align*}

Since $l=x+600$ ,

\begin{align*} \bar{l} =&\,\bar{x}+600\\[3mm] =&\,45.814\ldots+600\\[3mm] =&\,645.814\ldots \end{align*}

The mean length is

\begin{align*} 646\text{ mm} \end{align*}

to $3$ significant figures.

(b)

解法一

思路

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加上 $600$ 不会改变 variance，所以 $l$ 的 variance 与 $x$ 的 variance 相同。

答题过程

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\begin{align*} \operatorname{Var}(X) =&\,\frac{\sum x^2}{n}-\left(\frac{\sum x}{n}\right)^2\\[3mm] =&\,\frac{475181}{81}-\left(\frac{3711}{81}\right)^2\\[3mm] =&\,3767.43\ldots \end{align*}

Since $l=x+600$ , adding $600$ does not change the variance.

Therefore

\begin{align*} \operatorname{Var}(L)=3770 \end{align*}

to $3$ significant figures.

(c)

解法一

思路

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从 box plot 读数， $4600$ 到 $7700$ 最多可包含上半部分，也就是约 $50\%$ 的 $81$ 条鱼。人数必须是整数，所以最大为 $40$ 。

答题过程

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The interval $4600<w\leqslant7700$ can contain at most the upper half of the data.

So the maximum number is

\begin{align*} 40. \end{align*}

(d)

解法一

思路

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由 box plot 读出 $Q_1=3800$ ， $Q_3=5400$ ，最小值 $1600$ ，最大值 $7700$ 。算出 outlier boundaries，再检查最小值和最大值是否越界。

答题过程

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From the box plot,

\begin{align*} Q_1=3800,\qquad Q_3=5400. \end{align*}

\begin{align*} \operatorname{IQR}=5400-3800=1600. \end{align*}

The upper boundary is

\begin{align*} 5400+1.5(1600)=7800. \end{align*}

The lower boundary is

\begin{align*} 3800-1.5(1600)=1400. \end{align*}

The maximum value is $7700$ , and

\begin{align*} 7700<7800. \end{align*}

The minimum value is $1600$ , and

\begin{align*} 1600>1400. \end{align*}

Therefore there are no outliers.

IAL 2023 June Q3

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程

(c)

解法一

思路

答题过程

(d)

解法一

思路

答题过程