题目
The parking times, hours, for cars in a car park are summarised below.
| Time ( hours) | Frequency () | Time midpoint () |
|---|---|---|
| 10 | 0.5 | |
| 18 | 1.5 | |
| 15 | 3 | |
| 12 | 5 | |
| 5 | 9 |
(You may use and )
A histogram is drawn to represent these data.
The bar representing the time has a width of 1.5 cm and a height of 6 cm.
(a) Calculate the width and the height of the bar representing the time
(b) Use linear interpolation to estimate the median parking time for the cars in the car park.
(c) Estimate the mean and the standard deviation of the parking time for the cars in the car park.
(d) Describe, giving a reason, the skewness of the data.
One of these cars is selected at random.
(e) Estimate the probability that this car is parked for more than 75 minutes.
中文翻译
某停车场内汽车的停车时间 (小时)如表所示。(可使用 和 。)
现根据这些数据绘制直方图。区间 的长条宽 1.5 cm、高 6 cm。
(a) 计算区间 的长条宽度及高度。
(b) 用线性插值估计停车时间的中位数。
(c) 估计停车时间的平均数及标准差。
(d) 描述数据的偏态,并说明理由。
随机选取其中一辆汽车。
(e) 估计该车停车超过 75 分钟的概率。
解答
(a)
思路
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横向比例由组距决定;直方图面积与频数成正比。
答题过程
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The class width doubles from hour to hours, so the drawn width is
The first bar has area for frequency 18. Hence the required area is
and its height is
(b)
思路
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总频数为 60,中位数位置是第 30 个数据,落在 。
答题过程
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(c)
思路
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利用题目给出的 与 估计平均数及标准差。
答题过程
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The estimated mean is hours and the estimated standard deviation is hours.
(d)
思路
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比较平均数与中位数。
答题过程
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The data are positively skewed because the mean is greater than the median.
(e)
解法一
思路
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75 分钟是 1.25 小时。假设组内均匀分布,估计 中超过 1.25 小时的数量,再加上其后各组。
答题过程
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The later classes contain cars, so
解法二
思路
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也可利用补集,只计算不超过 1.25 小时的汽车数量。
答题过程
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Therefore