Statistics and probability
Pearson Edexcel · International GCSE · Mathematics A Topic 6 3:48 English narration · English + 中文 subtitles burned in
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Four families carry topic 6.
第 6 主题有四类要点。
A grouped mean uses midpoints.
分组平均数用组中值。
A histogram bar's area is its frequency, so its height is frequency density.
直方图长条的面积是频数，所以高度是频数密度。
Cumulative frequency plots at upper boundaries.
累积频数在上组界处描点。
And probabilities use the complement and the addition rule.
概率用对立事件和加法法则。
In a histogram with unequal classes, the area of a bar is the frequency, so the height is the frequency density: frequency divided by class width.
在组距不等的直方图中，长条的面积代表频数，所以高度是频数密度：频数除以组距。
Widths of 2, 3, 5, 10 and 15 with frequencies 5, 12, 18, 14 and 9 give densities of 2.5, 4, 3.6, 1.4 and 0.6.
组距为 2、3、5、10 和 15，频数为 5、12、18、14 和 9， 得到的密度是 2.5、4、3.6、1.4 和 0.6。
The last bar is the lowest, yet it still holds 9 people, because its area is 0.6 times 15.
最后一个长条最矮，但仍然有 9 人，因为它的面积是 0.6 乘 15。
25 students' times fall in four classes.
25 名学生的用时分为四组。
Treat everyone as sitting at their class midpoint: 5, 15, 25 and 35.
把每个人都看作在组中值：5、15、25、35。
Multiply by the frequencies and add: 525.
乘以频数再相加：525。
Divide by 25, the number of students, not by 4: about 21 minutes.
除以学生人数 25，而不是除以 4：约 21 分钟。
The modal class is 20 to 30.
众数组是 20 到 30。
Order the eight values.
把八个数排好。
The median sits between 9 and 10: 9.5.
中位数在 9 和 10 之间：9.5。
The lower half, 4, 6, 7 and 9, has median 6.5, which is Q1; the upper half has median 13, which is Q3.
下半部分 4、6、7、9 的中位数是 6.5，即 Q1；上半部分的中位数是 13，即 Q3。
The box spans the middle half, so the interquartile range is 6.5.
箱子覆盖中间一半，所以四分位距是 6.5。
An outlier would stretch the range, but not the box.
异常值会拉长极差，但不会拉长箱子。
Plot cumulative frequencies at the upper class boundaries and join them with a smooth increasing curve.
在每组的上组界处描出累积频数，再用一条平滑上升的曲线连起来。
For 60 workers the running totals are 6, 17, 27, 42 and 60 at 10, 15, 20, 25 and 30.
对 60 名工人，累计总数在 10、15、20、25 和 30 处分别是 6、17、27、42 和 60。
The median is the 30th worker: read across from 30 to the curve, then down, to about 21.
中位数是第 30 名工人：从 30 横向读到曲线，再向下读，约为 21。
The lower quartile at the 15th is about 14.1 and the upper quartile at the 45th about 25.8, so the interquartile range is about 11.7.
下四分位数在第 15 名，约为 14.1；上四分位数在第 45 名，约为 25.8， 所以四分位距约为 11.7。
Draw the reading lines, because the marks are for the reads.
要画出读数线，因为分数就给在读数上。
30 students: 18 like cricket, 15 like tennis and 7 like both.
30 名学生：18 人喜欢板球，15 人喜欢网球，7 人两样都喜欢。
Fill the overlap first: 7 like both, so cricket only is 11 and tennis only is 8.
先填重叠部分：7 人两样都喜欢，所以只喜欢板球的 11 人，只喜欢网球的 8 人。
Either one is 18 plus 15 minus 7, which is 26 out of 30, because the 7 were counted twice.
至少喜欢一样的是 18 + 15 − 7 = 26，共 30 人，因为那 7 人被数了两次。
So neither is 1 minus 26 thirtieths, which is 2 fifteenths.
所以两样都不喜欢的概率是 1 − 26/30 = 2/15。
Multiply along a path and add between paths.
沿一条路径相乘，不同路径之间相加。
A bag holds 5 red, 6 blue and 9 green, and two are taken without replacement, so the second denominator drops to 19.
袋中有 5 个红球、6 个蓝球、9 个绿球，不放回地取两个，所以第二次的分母变成 19。
Red then blue is 5 over 20 times 6 over 19.
先红后蓝是二十分之五 乘十九分之六。
Blue then red is 6 over 20 times 5 over 19.
先蓝后红是二十分之六乘十九分之五。
Adding the two paths gives 60 over 380, which is 3 over 19.
把两条路径相加得三百八十分之六十，即十九分之三。
Check yourself.
自我检测。
Frequency 9 over width 15 gives a height of 0.6.
频数 9 除以组距 15，高度 0.6。
The mean of 3, 7, 8, 10 and 12 is 8.
3、7、8、10、12 的平均数是 8。
On an ogive for 60, the median is the 30th value.
60 个数据的累积频数图上，中位数是第 30 个。
If P of A is 0.35, P of not A is 0.65.
若 P(A) = 0.35，则 P(非 A) = 0.65。
And two reds without replacement from 3 red and 2 blue is 3 fifths times 2 quarters: 0.3.
3 红 2 蓝不放回取两次都是红：3/5 × 2/4 = 0.3。