题目

Problem

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

Time ( $t$ minutes)	Frequency ( $f$ )	Time midpoint ( $x$ )
$0\leq t<15$	25	7.5
$15\leq t<30$	17	22.5
$30\leq t<60$	28	45
$60\leq t<120$	24	90
$120\leq t\leq240$	6	180

A histogram is drawn to represent these data.

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

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

(3)

(b) Use linear interpolation to estimate the median 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 $0\leq t<15$ 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)

解答

(a)

解法一

思路

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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 $15$ minutes is represented by $0.5$ cm, $60$ minutes is represented by

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

The first bar has area

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

representing frequency $25$ . For frequency $24$ , the area is

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

Hence the height is

\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 $30\leq t<60$ 组。

答题过程

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The cumulative frequency before $30\leq t<60$ is

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

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

The estimated median is

\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,

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

Therefore

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

The estimated mean is

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

(d)

解法一

思路

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题目给了 $\sum fx^2$ ，直接套标准差公式。

答题过程

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\begin{aligned} \sigma =&\,\sqrt{\frac{455512.5}{100}-(50.7)^2}\\ =&\,44.549\ldots. \end{aligned}

So the estimated standard deviation is

\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 $50.7$ is greater than the median $38.6$ .

(f)

解法一

思路

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这 18 个零仍然在中位数以下，所以中位数不变。把原本用 7.5 代表的部分改成 0，会降低平均数，并且让数据更分散，所以标准差增大。

答题过程

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

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

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

IAL 2019 Jan Q4

题目

解答

(a)

解法一

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(b)

解法一

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(c)

解法一

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(d)

解法一

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(e)

解法一

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(f)

解法一

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