統計学と確率
国際バカロレア · IB Diploma · Mathematics: Applications and Interpretation · SL トピック 4 4:24 英語ナレーション・英語＋中文字幕 burning-in
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Transcript
These are selected topics, not the whole syllabus, and every example is fictional.
这些是精选主题，不是完整大纲，所有例子都是虚构的。
Compare an average and a spread, and remember a histogram shows frequency as area.
比较要同时用平均数和离散程度；直方图用面积表示频数。
On a tree, multiply along branches and add between them; a condition restricts the denominator.
树状图上沿枝相乘、枝间相加；条件会限制分母。
A binomial count needs fixed n, independent trials and constant p; a normal value is read through z.
二项分布要求 n 固定、试验独立、p 不变；正态分布的值用 z 来读。
A test asks how likely data this extreme would be if the null hypothesis were true.
假设检验问：若原假设为真，出现这样极端数据的可能性有多大。
Three classes: 0 to 10 with frequency 20, 10 to 20 with 30, and 20 to 40 with 20.
三组：0–10 频数 20，10–20 频数 30，20–40 频数 20。
Divide each frequency by its width: densities 2, 3 and 1.
每组频数除以组距：密度 2、3、1。
The wide last bar is the lowest, yet its area, 20 times 1, still holds 20 observations.
最宽的最后一条最矮，但面积 20 × 1 仍是 20 个数据。
With unequal widths, height alone misleads.
组距不等时，只看高度会被误导。
Five values: 2, 4, 4, 6 and 9.
五个数：2、4、4、6、9。
The median is the middle one, 4.
中位数是中间那个，4。
The mean adds them, 25, and divides by 5: 5.
平均数是总和 25 除以 5：5。
The 9 pulls the mean above the median, but not the median.
9 把平均数拉到中位数之上，但拉不动中位数。
Two groups with the same median can still differ in spread, so compare an average and a spread.
中位数相同的两组仍可能离散程度不同，所以要同时比较平均数和离散程度。
A bag holds 3 red and 2 blue counters, and two are taken without replacement.
袋中有 3 个红、2 个蓝，不放回取两个。
After a red, 2 reds remain out of 4, so the second branch changes.
取出红后剩 4 个中有 2 个红，所以第二层的概率变了。
Multiply along: two reds is 3 fifths times 2 quarters, 3 tenths.
沿枝相乘：两红是 3/5 × 2/4 = 3/10。
One of each can happen two ways, so add them: 3 fifths.
一红一蓝有两条路径，相加：3/5。
Suppose P of A and B is 0.12 and P of B is 0.3.
设 P(A∩B) = 0.12，P(B) = 0.3。
Given B, only the B circle counts: the overlap is 0.12 out of 0.3, so P of A given B is 0.4.
已知 B 时只看 B 这个圆：重叠部分 0.12 占 0.3，所以 P(A|B) = 0.4。
Mutually exclusive events have no overlap at all, so if one happens the other cannot: they are dependent, not independent.
互斥事件完全没有重叠，一个发生另一个就不能发生：它们是相关的，不是独立的。
An inspector checks 5 items, each defective with probability 0.2, independently.
检验员检查 5 件产品，每件次品概率 0.2，相互独立。
The count X is binomial, 5 and 0.2.
次品数 X 服从 B(5, 0.2)。
No defects has probability 0.8 to the 5, about 0.33; one defect is the tallest bar, about 0.41.
没有次品的概率是 0.8⁵，约 0.33；一件次品的柱最高，约 0.41。
The mean is n p, 1, and the variance n p times 1 minus p, 0.8.
均值 np = 1，方差 np(1 − p) = 0.8。
A quantity is normal with mean 100 and standard deviation 15.
某量服从正态分布，均值 100，标准差 15。
How unusual is 130?
130 有多不寻常？
Its z-score is 130 minus 100 over 15: 2 standard deviations above the mean, out in the shaded upper tail.
z = (130 − 100)/15 = 2：比均值高 2 个标准差，位于阴影的上尾。
For a continuous variable, ask about ranges: one exact value has probability zero.
连续变量要问区间：单个确切值的概率为零。
And a Poisson rate of 3 per hour means 6 in two hours.
泊松分布每小时均值 3，两小时就是 6。
A factory claims 10 percent of items are defective.
工厂声称次品率 10%。
Test H nought, p equals 0.1, against H one, p greater than 0.1.
检验 H₀: p = 0.1，对 H₁: p > 0.1。
A sample of 10 has 3 defects.
10 件样本有 3 件次品。
Under H nought the count is binomial 10, 0.1, and the upper tail, 3 or more, has probability about 0.070.
在 H₀ 下次品数服从 B(10, 0.1)，上尾“3 件及以上”的概率约 0.070。
That is above 5 percent, so there is insufficient evidence that the defect rate exceeds 10 percent.
大于 5%，所以没有足够证据说明次品率超过 10%。
Every test follows the same chain.
每个检验都是同一条链。
State the hypotheses about the population before looking.
看数据前先写出关于总体的假设。
Compute the tail probability under H nought, in the direction H one points.
按 H₁ 的方向，计算 H₀ 下的尾概率。
Compare it with the significance level.
与显著性水平比较。
Then write the conclusion in context: failing to reject H nought is not proof that it is true.
最后结合情境写结论：不拒绝 H₀ 并不证明它为真。
Check yourself.
自我检测。
Frequency 30 over width 10 is a density of 3.
频数 30 ÷ 组距 10 = 密度 3。
P of A given B is 0.12 over 0.3: 0.4.
P(A|B) = 0.12 ÷ 0.3 = 0.4。
Five trials at 0.2 give a mean of 1.
5 次试验、p = 0.2，均值是 1。
A p-value of 0.07 at 5 percent is not enough evidence, which does not prove H nought.
5% 水平下 p 值 0.07 证据不足，但这并不证明原假设成立。