概率与统计1

Choose a diagram that suits the data. You should be able to draw and read:

• a stem-and-leaf diagram 茎叶图 (keeps the original values and shows the shape); two data sets are compared with a back-to-back 背靠背 version – a shared central stem, one set's leaves increasing to the left and the other's to the right, so you can compare their medians and spreads at a glance;
• a box-and-whisker plot 箱线图 (shows the lowest value, the three quartiles, and the highest value);
• a histogram 直方图 (for grouped data, where the area of each bar shows the frequency);
• a cumulative frequency 累积频数 graph (running totals, used to estimate the median and quartiles).

Averages and spread

A measure of central tendency 集中趋势 is a single "middle" value:

• the mean 平均数 $\bar{x} = \dfrac{\sum x}{n}$ (the average);
• the median 中位数 (the middle ordered value, or the mean of the middle pair for an even number of values);
• the mode 众数 (the most common value).

A measure of variation 离散程度 shows how spread out the data is:

• the range (of data) 极差 (highest $-$ lowest);
• the interquartile range 四分位距 (upper quartile $-$ lower quartile);
• the standard deviation 标准差 $\sigma = \sqrt{\dfrac{\sum x^2}{n} - \bar{x}^2}$.

You often work from the totals $\sum x$ and $\sum x^2$. The square of the standard deviation is the variance 方差.

Worked example. For $10$ values, $\sum x = 50$ and $\sum x^2 = 300$. Find the mean and standard deviation.

Coding 编码 makes big numbers easier. Replace each value by $t=x-a$ for a convenient assumed mean 假定平均数 $a$. Then $\bar{x}=a+\bar{t}$, while the standard deviation is unchanged (shifting every value does not spread the data). So from the coded totals $\sum(x-a)$ and $\sum(x-a)^2$ you get $\bar x$ and $\sigma$ directly, and two data sets can be compared through their coded totals.

Worked example: coded totals. For $8$ values, $\sum(x - 50) = 24$ and $\sum(x - 50)^2 = 200$. Find the mean and standard deviation of $x$.

Work with $t = x - 50$ first: $\bar t = \dfrac{24}{8} = 3$ and $\sigma_t = \sqrt{\dfrac{200}{8} - 3^2} = \sqrt{16} = 4$. Shift back for the mean only: $\bar x = 50 + 3 = 53$, and $\sigma_x = 4$, unchanged. Adding $50$ to the standard deviation is the usual error.

A permutation 排列 is an arrangement where order matters; a combination 组合 is a selection where order does not matter. The numbers are

Worked example. How many different arrangements are there of the letters of the word NEEDLESS?

There are $8$ letters, with E repeated $3$ times and S repeated $2$ times:

Worked example: together and apart. Seven friends, including Amy and Bo, stand in a row. In how many ways can they stand if (a) Amy and Bo are together, (b) Amy and Bo are not together?

(a) Tie Amy and Bo into one block. Now arrange $6$ items, $6! = 720$ ways, and the block itself can be AB or BA: $720 \times 2 = 1440$.

(b) Take the total minus the "together" cases: $7! - 1440 = 5040 - 1440 = 3600$.

Worked example: "at least one". A committee of $4$ is chosen from $5$ men and $4$ women. How many committees contain at least one woman?

Count the opposite case: committees with no women use only men, ${}^5C_4 = 5$ ways. All committees: ${}^9C_4 = 126$. So $126 - 5 = 121$.

Common mistakes. Using ${}^nP_r$ for a committee (order does not matter in a committee), and counting "at least one" case by case, which double-counts easily; the complement is safer.

Find a probability by counting equally likely outcomes, or by using permutations and combinations. Combine probabilities with these rules:

• addition for "or": $P(A \cup B) = P(A) + P(B) - P(A \cap B)$;
• multiplication for "and" when events are independent: $P(A \cap B) = P(A)\,P(B)$.

Two events are mutually exclusive events 互斥事件 if they cannot both happen, and independent events 独立事件 if one happening does not change the chance of the other. To test independence, check whether $P(A \cap B) = P(A)\times P(B)$. A conditional probability 条件概率 is the chance of $A$ given that $B$ has happened: $P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}$, provided $P(B)>0$.

Worked example. Events have $P(A) = 0.5$, $P(B) = 0.4$ and $P(A \cap B) = 0.2$. Are $A$ and $B$ independent?

Test: $P(A)\times P(B) = 0.5 \times 0.4 = 0.2 = P(A \cap B)$. The values are equal, so $A$ and $B$ are independent.

Worked example: a tree diagram. The probability that it rains on a given day is $0.3$. If it rains, the probability that Kim is late is $0.4$; if it does not rain, it is $0.1$. Find the probability that Kim is late, and the probability that it rained given that Kim was late.

Multiply along each branch that ends in "late", then add the branches:

Given that Kim was late, only those two branches count:

The common error is to answer $0.4$, which is $P(L \mid R)$, the other way round.

A discrete random variable 离散型随机变量 $X$ takes separate values, each with a probability. List them in a probability distribution table 概率分布表; the probabilities must add to $1$. Then the expectation 期望 (mean) and variance are

Two special models:

• the binomial distribution 二项分布 $X \sim B(n, p)$, for the number of successes in $n$ independent trials, each with the same success probability $p$: $P(X = r) = \binom{n}{r}p^r(1-p)^{n-r}$, with $E(X) = np$ and $\mathrm{Var}(X) = np(1-p)$;
• the geometric distribution 几何分布, for the trial on which the first success happens in independent trials with constant success probability $p$: $P(X = r) = (1-p)^{r-1}p$, with $E(X) = \dfrac{1}{p}$.

Worked example: a probability table. The variable $X$ has this distribution. Find $k$, $E(X)$ and $\mathrm{Var}(X)$.

The probabilities add to $1$, so $k = 1 - 0.6 = 0.4$. Then

Remember to subtract $\big(E(X)\big)^2$: stopping at $8.1$ is a frequent slip.

Worked example. $X \sim B(10, 0.3)$. Find $P(X = 2)$ and $E(X)$.

Worked example: geometric. A fair die is thrown until a six appears. Find the probability that the first six comes on the third throw, and that more than $4$ throws are needed.

$P(X = 3) = \left(\tfrac56\right)^2\tfrac16 = 0.116$. "More than $4$ throws" means the first four throws are all not six: $P(X > 4) = \left(\tfrac56\right)^4 = 0.482$.

The normal distribution 正态分布 models a continuous random variable 连续型随机变量 with a symmetric bell shape. Write $X \sim N(\mu, \sigma^2)$, where $\mu$ is the mean and $\sigma$ is the standard deviation. To use the tables, by standardisation 标准化 convert to the variable $Z \sim N(0, 1)$:

Then $P(X < x) = P\!\left(Z < \dfrac{x - \mu}{\sigma}\right)$, which you read from the normal table $\Phi$.

The normal distribution is also a good approximation 近似 to the binomial when both $np$ and $n(1-p)$ are sufficiently large. Because you replace a discrete variable by a continuous one, apply a continuity correction 连续性校正 (adjust by $0.5$).

Worked example. Bags of rice have mass $X \sim N(\mu, 0.14^2)$. Given that $P(X < 1.48) = 0.22$, find $\mu$.

From the table, $P(Z < z) = 0.22$ gives $z = -0.772$. So

Worked example: approximating a binomial. $40\%$ of a town's residents cycle to work. In a random sample of $100$ residents, find the probability that at least $45$ cycle to work.

$X \sim B(100, 0.4)$. Here $np = 40$ and $n(1-p) = 60$ are both more than $5$, so use $N(np, np(1-p)) = N(40, 24)$. "At least $45$" includes $45$, so the continuity correction starts at $44.5$:

For the boundary, picture the bar for $45$: it runs from $44.5$ to $45.5$, and "at least $45$" keeps the whole bar. So $P(X > 45)$ would start at $45.5$, and $P(X \leqslant 45)$ would end at $45.5$.

• Decide whether order matters: permutations ($^nP_r$) when it does, combinations ($^nC_r$) when it does not.
• For a discrete random variable, check the probabilities sum to $1$ and use $E(X) = \sum x\,P(X=x)$.
• For the normal distribution, standardise with $z = (x - \mu)/\sigma$, sketch and shade, then read the table.
• Apply a continuity correction when approximating a discrete variable by the normal.