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IAL 2019 Oct Q2

A Level / Edexcel / S1

IAL 2019 Oct Paper · Question 2

题目

Problem

The histogram shows the times taken, in seconds, by each of 260 people to complete a puzzle.

(a) Use the histogram to complete the frequency table for the times taken to complete the puzzle.

(3)
Time taken (seconds)15-3030-4545-5555-6060-7575-90
Frequency (ff)20145105
Time midpoint (tt seconds)22.537.55057.567.582.5

Given that ∑ft=11087.5\sum ft=11087.5 and ∑ft2=505718.75\sum ft^2=505718.75

(b) find an estimate for

(i) the mean time taken to complete the puzzle,

(1)

(ii) the standard deviation of the times taken to complete the puzzle.

(2)

(c) Use linear interpolation to estimate the median time taken to complete the puzzle.

(2)

(d) Describe the skewness of these data. Give a reason for your answer.

(1)

Three of the 260 people are chosen at random.

(e) Estimate the probability that all 3 of their times are less than 36 seconds.

(4)

解答

(a)

解法一

思路

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直方图面积代表频数。根据图中相对高度可补出 4545-5555 和 5555-6060 两组，再用总频数 260 检查。

答题过程

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From the histogram,

45≤t<55\begin{align*} 45\leq t<55 \end{align*}

has frequency 6565, and

55≤t<60\begin{align*} 55\leq t<60 \end{align*}

has frequency 1515.

The completed missing frequencies are therefore

65and15.\begin{align*} 65\quad\text{and}\quad15. \end{align*}

(b)

解法一

思路

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题目已给 ∑ft\sum ft 与 ∑ft2\sum ft^2，直接套分组数据公式。

答题过程

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The estimated mean is

tˉ=11087.5260=42.644….\begin{align*} \bar t=\frac{11087.5}{260}=42.644\ldots. \end{align*}

So

tˉ=42.6 seconds.\begin{align*} \bar t=42.6\text{ seconds}. \end{align*}

The estimated standard deviation is

σ= 505718.75260−(11087.5260)2= 11.249….\begin{aligned} \sigma =&\,\sqrt{\frac{505718.75}{260} -\left(\frac{11087.5}{260}\right)^2}\\ =&\,11.249\ldots. \end{aligned}

So

σ=11.2 seconds.\begin{align*} \sigma=11.2\text{ seconds}. \end{align*}

(c)

解法一

思路

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总数 260，所以中位数是第 130 个数据附近。累计频数到 30 秒前为 20，因此中位数落在 3030-4545 组。

答题过程

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The median is the 130th value.

The cumulative frequency before 30≤t<4530\leq t<45 is 2020.

So

median=30+130−20145×15.\begin{align*} \text{median}=30+\frac{130-20}{145}\times15. \end{align*}

Hence

median=41.379….\begin{align*} \text{median}=41.379\ldots. \end{align*}

Therefore the estimated median is

41.4 seconds.\begin{align*} 41.4\text{ seconds}. \end{align*}

(d)

解法一

思路

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平均数大于中位数，说明右尾较长，所以是正偏态。

答题过程

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The data are positively skewed, since the mean 42.642.6 is greater than the median 41.441.4.

(e)

解法一

思路

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先估计小于 36 秒的人数。15-30 秒有 20 人；30-45 秒这一组中，30 到 36 占 6/156/15。

答题过程

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The estimated number of people taking less than 3636 seconds is

20+615(145)=78.\begin{align*} 20+\frac{6}{15}(145)=78. \end{align*}

So the probability for one selected person is

78260=0.3.\begin{align*} \frac{78}{260}=0.3. \end{align*}

Choosing 3 people without replacement,

P=78260⋅77259⋅76258.\begin{align*} P=\frac{78}{260}\cdot\frac{77}{259}\cdot\frac{76}{258}. \end{align*}

Therefore

P=0.02627….\begin{align*} P=0.02627\ldots. \end{align*}

So the required probability is approximately

0.0263.\begin{align*} 0.0263. \end{align*}