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S1 2019 Jan Q4

A Level / Edexcel / S1

IAL 2019 Jan Paper · Question 4

题目

Problem

A group of 100 adults recorded the amount of time, tt minutes, they spent exercising each day. Their results are summarised in the table below.

Time (tt minutes)Frequency (ff)Time midpoint (xx)
0≤t<150\leq t<15257.5
15≤t<3015\leq t<301722.5
30≤t<6030\leq t<602845
60≤t<12060\leq t<1202490
120≤t≤240120\leq t\leq2406180

[You may use ∑fx2=455 512.5\sum fx^2=455\,512.5]

A histogram is drawn to represent these data.

The bar representing the time 0≤t<150\leq t<15 has width 0.5 cm and height 6 cm.

(a) Calculate the width and height of the bar representing a time of 60≤t<12060\leq t<120

(3)

(b) Use linear interpolation to estimate the median time spent exercising by these adults each day.

(2)

(c) Find an estimate of the mean time spent exercising by these adults each day.

(2)

(d) Calculate an estimate for the standard deviation of these times.

(2)

(e) Describe, giving a reason, the skewness of these data.

(1)

Further analysis of the above data revealed that 18 of the 25 adults in the 0≤t<150\leq t<15 group took no exercise each day.

(f) State, giving a reason, what effect, if any, this new information would have on your answers to

(i) the estimate of the median in part (b),

(ii) the estimate of the mean in part (c),

(iii) the estimate of the standard deviation in part (d).

(3)
题目中文翻译

100 名成年人记录了每天运动的时间 tt（分钟），结果如表所示。可使用 ∑fx2=455 512.5\sum fx^2=455\,512.5。

现用直方图表示这些数据。区间 0≤t<150\leq t<15 的长条宽 0.5 cm、高 6 cm。

(a) 计算区间 60≤t<12060\leq t<120 的长条宽度和高度。

(b) 用线性插值估计这些成年人每天运动时间的中位数。

(c) 估计每天运动时间的平均数。

(d) 估计这些时间的标准差。

(e) 描述数据的偏态，并说明理由。

进一步分析发现，0≤t<150\leq t<15 组的 25 人中有 18 人每天完全没有运动。

(f) 说明这一新资料会使以下估计发生什么变化（若没有变化也须说明），并给出理由：(i) (b) 的中位数；(ii) (c) 的平均数；(iii) (d) 的标准差。

解答

(a)

解法一

思路

展开

直方图面积代表频数。第一组宽 15 分钟画成 0.5 cm，所以 60 分钟画成 2 cm。

答题过程

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Since 1515 minutes is represented by 0.50.5 cm, 6060 minutes is represented by

4(0.5)=2 cm.\begin{align*} 4(0.5)=2\text{ cm}. \end{align*}

The first bar has area

0.5(6)=3,\begin{align*} 0.5(6)=3, \end{align*}

representing frequency 2525. For frequency 2424, the area is

3⋅2425=2.88.\begin{align*} 3\cdot\frac{24}{25}=2.88. \end{align*}

Hence the height is

2.882=1.44 cm.\begin{align*} \frac{2.88}{2}=1.44\text{ cm}. \end{align*}

(b)

解法一

思路

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中位数是第 50 个数据，落在 30≤t<6030\leq t<60 组。

答题过程

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The cumulative frequency before 30≤t<6030\leq t<60 is

25+17=42.\begin{align*} 25+17=42. \end{align*}

So

median=30+50−4228×30=38.571….\begin{align*} \text{median}=30+\frac{50-42}{28}\times30=38.571\ldots. \end{align*}

The estimated median is

38.6 minutes.\begin{align*} 38.6\text{ minutes}. \end{align*}

(c)

解法一

思路

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用组中值估计平均数。

答题过程

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Using the midpoints,

∑fx=5070.\begin{align*} \sum fx=5070. \end{align*}

Therefore

xˉ=5070100=50.7.\begin{align*} \bar x=\frac{5070}{100}=50.7. \end{align*}

The estimated mean is

50.7 minutes.\begin{align*} 50.7\text{ minutes}. \end{align*}

(d)

解法一

思路

展开

题目给了 ∑fx2\sum fx^2，直接套标准差公式。

答题过程

展开 σ= 455512.5100−(50.7)2= 44.549….\begin{align*} \sigma =&\,\sqrt{\frac{455512.5}{100}-(50.7)^2}\\ =&\,44.549\ldots. \end{align*}

So the estimated standard deviation is

44.5 minutes.\begin{align*} 44.5\text{ minutes}. \end{align*}

(e)

解法一

思路

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平均数大于中位数，说明右尾较长，是正偏态。

答题过程

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The data are positively skewed, since the mean 50.750.7 is greater than the median 38.638.6.

(f)

(i)

思路

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这 18 个数仍然位于中位数组之前，因此不会改变中位数所在位置及插值。

答题过程

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The median will not change, because the 18 values are still below the median.

(ii)

思路

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原先以组中值 7.5 估计这 18 个数；知道它们实际为 0 后，数据总和会减小。

答题过程

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The mean will be smaller, because replacing some midpoint estimates by 0 reduces the total.

(iii)

思路

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新增的 18 个零使数据相对平均数更加分散。

答题过程

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The standard deviation will be greater, because the data are more spread out.