Descriptive statistics formulas

Raw data. For grouped data with frequencies $f_i$ and class midpoints $x_i$, use $\bar{x}=\sum f_i{x}_i/\sum f_i$. This treats every value as sitting at its class midpoint, so it is an approximation unless the values really are the midpoints.

Here $x_{(k)}$ is the $k$-th smallest value; sort first.

The median class is the first class whose cumulative frequency is at least $N/2$. $L$ = its lower class boundary, $N=\sum f_i$, $F$ = cumulative frequency of the classes before it, $f$ = its frequency, $h$ = its width. Classes must be continuous: if the class limits leave gaps (10-19, 20-29), use the boundaries (9.5-19.5, 19.5-29.5).

$L$ = lower boundary of the modal class (highest frequency $f_1$), $f_0$ and $f_2$ = frequencies of the classes just before and after it, $h$ = class width. The modal class and its two neighbours must have the same width. If the modal class is the first or last class, take the missing neighbour's frequency as $0$.

$p$ is the pmf, $f$ the pdf and $F$ the CDF. The median condition $F(m)=\frac{1}{2}$ is for a continuous $X$ (in general a median satisfies $P(X\le m)\ge \frac{1}{2}$ and $P(X\ge m)\ge \frac{1}{2}$). The mean exists only when the sum or integral of $|x|$ converges. Example: for the exponential distribution with rate $\lambda$, $1-e^{-\lambda m}=\frac{1}{2}$ gives median $=\ln 2/\lambda$, while the mean is $1/\lambda$ and the mode is $0$.

For any constant $c$, the sum of squared deviations is therefore smallest when $c=\bar{x}$. The sum of absolute deviations $\sum |x_i-c|$ is smallest when $c$ is the median (for even $n$, any $c$ between the two middle values).

$\mu$ is the population mean. The second form is the shortcut $E[X^2]-(E[X]{)}^2$.

$s^2$ is an unbiased estimator of the population variance. For the same data it relates to the divide-by-$n$ value by $s^2=\frac{n}{n-1}{\sigma }^2$.

$\mathrm{SD}=\sqrt{\mathrm{Var}(X)}$. If $X$ picks one of the $n$ data values uniformly at random, this equals the population (divide-by-$n$) variance of the data.

Defined for data with a positive mean. It has no units, so it compares the spread of data sets measured on different scales: the one with the larger CV is relatively more variable.

This holds for the population and the sample versions alike. The median and the mode also transform as $a\cdot (\text{value})+b$.

Weighted by group size. It equals the plain average $(m_1+m_2)/2$ only when $n_1=n_2$.

Both group variances must be the divide-by-$n$ kind. If sample variances are given, first convert them with ${\sigma }_i^2=\frac{n_i-1}{n_i}{s}_i^2$.

$\mathrm{Cov}(X,X)=\mathrm{Var}(X)$, and $\mathrm{Cov}(X,c)=0$ for a constant $c$.

For the population version, divide by $n$ instead. State the convention you use.

Both variances must be positive and finite. The $n$ or $n-1$ factors cancel in $r$, so the sample value does not depend on the convention.

For $a\ne 0$ and $c\ne 0$. A positive-slope linear change of either variable leaves the correlation unchanged; a negative multiplier on exactly one variable flips its sign.

Covariance is linear in each argument separately. Note the $+\mathrm{Var}(Y)$ in $\mathrm{Var}(X-Y)$: variances never subtract.

The cross term vanishes when the covariance is zero (for example, when $X$ and $Y$ are independent).

If the $X_i$ are pairwise uncorrelated, the variance of the sum is the sum of the variances.

For $X_1,\dots ,X_n$ independent (pairwise uncorrelated is enough) with common variance ${\sigma }^2$; here $\bar{X}=\frac{1}{n}\sum X_i$.

$b_{yx}$ is the least-squares slope of the line of $y$ on $x$, and $b_{xy}$ that of $x$ on $y$. Both lines pass through $(\bar{x},\bar{y})$. The two slopes and $r$ share one sign, so $r$ is the square root of $b_{yx}{b}_{xy}$ taken with that sign. Use the same divisor in the covariance and the variances.

$d_i$ is the difference between the ranks of $x_i$ and $y_i$. This shortcut is exact only when there are no tied ranks. With ties, compute Pearson's $r$ on the ranks instead.