데이터 분석: 분포, 가중 평균 및 조건부 확률
GRE General Test 주제 14 2:33 영어 내레이션 · 영어 + 중국어 자막 burned-in
장
전사본
This unit distinguishes mean, median, spread and weighted averages, interprets distributions and overlapping or conditional counts, and uses the correct reference total in table-based inference.
本单元区分平均数、中位数、离散程度和加权平均，解释分布以及重叠或条件计数，并在表格推断中使用正确的参照总数。
Mean is sum over count; median is the middle ordered value.
平均数是总和除以个数；中位数是排序后居中的值。
An extreme value can move the mean and range without changing the median.
一个极端值可以改变平均数和极差，而不改变中位数。
Adding a constant shifts the centre but not the spread, while multiplying every value by c multiplies the standard deviation by the absolute value of c.
每个值加上常数会平移中心但不改变离散程度；每个值乘以 c，标准差乘以 |c|。
Combine groups by their totals: 10 with mean 6 and 30 with mean 10 give 360 over 40, which is 9.
合并各组要用总和：10 个均值 6 和 30 个均值 10，得 360/40 = 9。
A condition restricts the denominator to the given group.
条件把分母限制在给定的组。
Draws without replacement change later counts, overlapping events need the overlap subtracted, and independence must be justified.
不放回抽取会改变后续计数，重叠事件求并时要减去重叠部分，独立性必须有依据。
Four counters are marked, three of them red, so red given marked is 3 over 4.
有 4 个标记筹码，其中 3 个是红色，所以已知有标记时是红色的概率是 3/4。
Two reds without replacement is 4 tenths times 3 ninths, 2 fifteenths.
不放回连取两个红色是 4/10 × 3/9 = 2/15。
For data 1, 2, 3, 4, 20, mean equals 30 over 5 equals 6, but median equals 3: the outlier pulls the mean.
数据 1、2、3、4、20 的平均数是 30/5 = 6，但中位数是 3：离群值把平均数拉高。
A condition changes the reference population.
条件改变参照总体。
Group means need weights, and draws without replacement need changing denominators.
组均值需要加权，不放回抽取需要改变分母。
The combined total is 140 over twenty observations, so the mean is seven; averaging 5 and 10 would ignore group sizes.
合并后总和为 140，共 20 个观测，所以平均数是 7；直接平均 5 和 10 会忽略组的大小。
The data have mean six, median three and range nineteen.
数据的平均数 6，中位数 3，极差 19。
Adding seven shifts every value and the mean together, so the standard deviation is unchanged.
每个值都加 7，平均数一起平移，所以标准差不变。
Of 40 students, 24 study art, 18 study music and 10 study both.
40 名学生中，24 人学美术，18 人学音乐，10 人两者都学。
Music given art divides by 24: 5 twelfths.
已知学美术时学音乐的概率要除以 24：5/12。
The union count is 32, so at least one subject has probability four fifths.
并集人数是 32，所以至少学一门的概率是 4/5。
Independence would need a quarter to equal 27 hundredths: it does not.
独立需要 1/4 等于 27/100：并不相等。
Multiplying data by minus 3 triples the standard deviation.
数据乘以 −3，标准差变为 3 倍。
Check yourself.
自我检测。
The list has mean 6 and median 3.
这组数据平均数 6，中位数 3。
Adding 7 leaves the standard deviation unchanged.
都加 7，标准差不变。
The combined mean is 7.
合并平均数是 7。
Music given art is 5 twelfths.
已知美术时音乐的概率是 5/12。
And multiplying by minus 3 triples the standard deviation.
乘以 −3，标准差变为 3 倍。