Elementary statistics and probability

Uses every observation, so one extreme value can move it a lot. Total of all values $=n\bar{x}$, and many mean questions become simple once they are worked through totals.

Weights $w_i\ge 0$, not all zero. For a frequency table put $w_i=f_i$, the frequency of the value $x_i$. For grouped data (class intervals), take $x_i$ as the mid-point of each class.

The group sizes act as weights. The combined mean always lies between the two group means (it equals both when they are equal). When the group sizes differ, it lies nearer the mean of the larger group; when the sizes are equal, it is exactly midway.

$N$ is the total frequency and $h$ the common class width. Median: the median class is the first class whose cumulative frequency reaches $\frac{N}{2}$; $L_m$ is its lower boundary, $f_m$ its frequency and $F$ the cumulative frequency of all the classes before it. Mode: the modal class is the class with the highest frequency; $L_o$ is its lower boundary, $f_1$ its frequency, and $f_0$, $f_2$ are the frequencies of the classes just before and just after it (take 0 if there is no such class). Both formulas are estimates that assume the values are spread evenly within each class.

Approximate, and only for moderately skewed data with a single peak. Use it when a question asks for it, or gives two of the three averages and asks for the third. Equivalent form: $\text{Mean}-\text{Mode}\approx 3(\text{Mean}-\text{Median})$.

For positive values only. Equality holds throughout exactly when all the values are equal. For two positive numbers $a$ and $b$: $\text{AM}=\frac{a+b}{2}$, $\text{GM}=\sqrt{ab}$, $\text{HM}=\frac{2ab}{a+b}$, and ${\text{GM}}^2=\text{AM}\times \text{HM}$.

Depends only on the two extreme values, so a single outlier changes it.

The mean of the squares minus the square of the mean. Convention in these notes: divide by $n$ unless the question says 'sample'. The variance is always $\ge 0$ and is zero only when all values are equal.

Use it only when a question explicitly asks for the sample variance or sample standard deviation ($n\ge 2$).

The SD is in the same units as the data and is never negative. The CV is used for data with a positive mean. It compares the consistency of data sets with different means: the set with the smaller CV is the more consistent.

For $n$ terms in AP with common difference $d$. For the first $n$ natural numbers this is $\frac{n^2-1}{12}$. The mean and the median of an AP both equal $\frac{\text{first}+\text{last}}{2}$.

${\sigma }_1^2$ and ${\sigma }_2^2$ are the population variances of the two groups, and $\bar{x}$ is the combined mean.

Valid only when $S$ is finite and its outcomes are equally likely. Always $0\le P(A)\le 1$, $P(S)=1$ and $P(\varnothing )=0$.

${}^n{C}_r$ counts selections of $r$ items from $n$ distinct items when order does not matter; ${}^n{P}_r$ counts arrangements, where order matters. Here $0\le r\le n$ and $0!=1$. When $r>n$ no such selection exists, and ${}^n{C}_r$ is taken as $0$.

$A^{\prime }$ ('not $A$') contains every outcome of $S$ that is outside $A$. Use this rule whenever 'not $A$' is easier to count than $A$, above all for 'at least one' questions.

Always true. For mutually exclusive events $P(A\cap B)=0$, so it reduces to $P(A)+P(B)$.

Inclusion and exclusion: add the single events, subtract the pairs, then add back the triple.

'$A$ only' has probability $P(A)-P(A\cap B)$. 'Neither' means not $A$ and not $B$, that is $P(A^{\prime }\cap B^{\prime })$.

The odds against $A$ are then $b:a$. Odds compare the chance of $A$ with the chance of not $A$ (favourable with unfavourable outcomes, when they are equally likely), not with the total.

For a point chosen uniformly at random from a line segment or a plane region, so that parts of equal length (or area) are equally likely. For example, a point chosen at random inside a square of side 2 lies within distance 1 of the centre of the square with probability $\frac{\pi \times 1^2}{2^2}=\frac{\pi }{4}$.

With equally likely outcomes this is $P(A\mid B)=\frac{n(A\cap B)}{n(B)}$: count inside $B$ only.

The first form needs $P(A)>0$ and the second $P(B)>0$, so that the conditional probability is defined. For successive draws: $P(A\cap B\cap C)=P(A)P(B\mid A)P(C\mid A\cap B)$.

The second form needs $P(B)>0$. If $A$ and $B$ are independent, so are $A$ and $B^{\prime }$, $A^{\prime }$ and $B$, and $A^{\prime }$ and $B^{\prime }$. Successive tosses, rolls, and draws WITH replacement are independent.

For INDEPENDENT events with $P(A_i)=p_i$. For $n$ independent trials, each with success probability $p$, it becomes $1-(1-p{)}^n$. 'At least one' covers exactly one, exactly two, and so on up to all of them.

For $n$ independent trials, each with the same success probability $p$ (tosses of a coin, rolls of a die, draws WITH replacement). For a fair coin it is ${}^n{C}_k/2^n$.

$B_1,\dots ,B_k$ must be mutually exclusive and exhaustive (exactly one of them happens), with each $P(B_i)>0$. The second formula also needs $P(A)>0$. Use it when a question gives $P(A\mid B)$ but asks for $P(B\mid A)$.

Counts ordered pairs, giving 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1 ways for totals 2 to 12 (36 in all). Totals $s$ and $14-s$ are equally likely.

For $k$ items drawn together (or one by one without replacement) from $N$ items, of which $a$ are of type 1. The probability is $0$ unless $r\le a$ and $k-r\le N-a$. WITH replacement, use the binomial formula with $p=a/N$ instead.

$p_i=P(X=x_i)$, summed over ALL possible values, including zero and negative values such as a loss.

Holds for any random variables; independence is NOT needed. So the expected total over $n$ repeated games is $n$ times the expected value of one game.

The same 'mean of the squares minus the square of the mean' idea as for data. For one fair die, $E(X)=\frac{7}{2}$, $E\left(X^2\right)=\frac{91}{6}$ and $\mathrm{Var}(X)=\frac{91}{6}-\frac{49}{4}=\frac{35}{12}$.

For independent trials with a constant success probability $p$ ($0<p\le 1$). The number of successes in $n$ trials also has variance $np(1-p)$. For example, a fair die takes 6 rolls on average to show its first 6.