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IAL 2022 Jan Q3

A Level / Edexcel / S1

IAL 2022 Jan Paper · Question 3

题目

Problem

The stem and leaf diagram shows the number of deliveries made by Pat each day for 2424 days

Stem and leaf diagram

where aa, bb and cc are positive integers with a<b<ca<b<c

An outlier is defined as any value greater than 1.5×1.5\times interquartile range above the upper quartile.

Given that there is only one outlier for these data,

(a) show that c=9c=9

(3)

The number of deliveries made by Pat each day is represented by dd

The data in the stem and leaf diagram are coded using

x=d−125x=d-125

and the following summary statistics are obtained

∑x=−96and∑(x−xˉ)2=1306\sum x=-96\quad\text{and}\quad \sum (x-\bar{x})^2=1306

(b) Find the mean number of deliveries.

(3)

(c) Find the standard deviation of the number of deliveries.

(2)

One of these 2424 days is selected at random. The random variable DD represents the number of deliveries made by Pat on this day.

The random variable X=D−125X=D-125

(d) Find P(D>118∣X<0)P(D>118\mid X<0)

(2)

解答

(a)

解法一

思路

展开

从有序数据读出 Q1=116Q_1=116，Q3=125Q_3=125。按题意，上界是 Q3+1.5IQR⁡Q_3+1.5\operatorname{IQR}。只有一个 outlier，说明只有 139139 超过上界，因此 cc 必须是 99。

答题过程

展开

From the stem and leaf diagram,

Q1=116,Q3=125.\begin{align*} Q_1=116,\qquad Q_3=125. \end{align*}

So

IQR⁡=125−116=9.\begin{align*} \operatorname{IQR}=125-116=9. \end{align*}

The outlier boundary is

125+1.5(9)=138.5.\begin{align*} 125+1.5(9)=138.5. \end{align*}

Since there is only one outlier, the only value greater than 138.5138.5 is 139139. Therefore the largest leaf in the 1313 stem must be 99, so

c=9.\begin{align*} c=9. \end{align*}

(b)

解法一

思路

展开

xˉ=∑x24\bar{x}=\dfrac{\sum x}{24}，且 d=x+125d=x+125，所以 dˉ=xˉ+125\bar{d}=\bar{x}+125。

答题过程

展开 xˉ=−9624=−4.\begin{align*} \bar{x}=\frac{-96}{24}=-4. \end{align*}

Since d=x+125d=x+125,

dˉ=−4+125=121.\begin{align*} \bar{d}=-4+125=121. \end{align*}

The mean number of deliveries is

121.\begin{align*} 121. \end{align*}

(c)

解法一

思路

展开

加上 125125 不改变 standard deviation，所以 dd 的 standard deviation 与 xx 相同。

答题过程

展开 σd= σx= 130624= 7.376…\begin{align*} \sigma_d =&\,\sigma_x\\[3mm] =&\,\sqrt{\frac{1306}{24}}\\[3mm] =&\,7.376\ldots \end{align*}

So the standard deviation is

7.38.\begin{align*} 7.38. \end{align*}

(d)

解法一

思路

展开

X<0X<0 等价于 D<125D<125。在这些天中，再要求 D>118D>118，就是 118<D<125118<D<125。

答题过程

展开

Since X=D−125X=D-125,

X<0  ⟺  D<125.\begin{align*} X<0\iff D<125. \end{align*}

From the stem and leaf diagram,

P(D>118∣X<0)=number with 118<D<125number with D<125.P(D>118\mid X<0) =\frac{\text{number with }118<D<125} {\text{number with }D<125}.

There are 55 values with 118<D<125118<D<125, and 1414 values with D<125D<125.

Therefore

P(D>118∣X<0)=514.\begin{align*} P(D>118\mid X<0)=\frac{5}{14}. \end{align*}