Foundations + Measures of Central Tendency

Data is the set of observed values of a variable measured on a collection of items. The full collection is the POPULATION; a subset actually observed is a SAMPLE. Statistics builds summary measures from samples to draw conclusions about the population.

• Qualitative (categorical): takes labels, not numbers — operations like mean don't apply.
• Quantitative — discrete: numerical, jumping in integer-sized steps (kids per family, integer marks).
• Quantitative — continuous: numerical, filling an interval smoothly (height, weight, time).

A frequency distribution lists each distinct value (or class interval) alongside its frequency. The TOTAL FREQUENCY equals the number of observations: $N = \sum f_i$. Every chapter formula that involves grouped or repeated data uses this $N$, not the count of distinct values.

For a class interval with lower bound $L$ and upper bound $U$: the CLASS MARK is $x = (L + U)/2$ and the CLASS WIDTH is $h = U - L$. All grouped-data formulas (mean, median, mode, variance) use the class mark as the representative value of the interval.

For a sequence $x_1, x_2, \ldots, x_n$, $\sum_{i=1}^{n} x_i = x_1 + x_2 + \cdots + x_n$. Two identities are load-bearing throughout this chapter: $\sum_{i=1}^{n} c = nc$ (a constant summed $n$ times) and $\sum (a x_i + b) = a \sum x_i + nb$ (linearity).

For raw data, the total is $\sum x_i$ and the count is $n$. For frequency-tabulated data with distinct values $x_i$ of frequency $f_i$, the total is $\sum f_i x_i$ and the count is $N = \sum f_i$. Every measure has a "raw" form (unweighted) and a "grouped" form (weighted) — they're the same idea with frequencies multiplied in.

For $n$ observations $x_1, x_2, \ldots, x_n$, the arithmetic mean is the total sum divided by the number of observations.

If the value $x_i$ occurs with frequency $f_i$, the mean is the frequency-weighted sum divided by the total frequency.

If a new variable $y_i = a\,x_i + b$ is formed from each observation, the new mean is $\bar{y} = a\,\bar{x} + b$.

If the mean of $n$ observations is $M$ and a single value $x$ is replaced by $y$, the new mean is $M_{\text{new}} = M + \dfrac{y - x}{n}$. For a wrong-value correction, $x$ is what was recorded and $y$ is the correct value. For discards or additions, $n$ itself changes — reason about the new total $(n \pm k)\,M_{\text{new}}$ directly.

Three shortcuts are load-bearing: (a) Mean of consecutive integers from $a$ to $b$ is $(a+b)/2$. (b) Mean of squares $1^2, 2^2, \ldots, n^2$ is $\dfrac{(n+1)(2n+1)}{6}$. (c) For an AP, the mean equals the average of the first and last terms (or equivalently the middle term). For binomial-weighted means, the denominator is $\sum \binom{n}{k} = 2^n$, not the number of terms.

For group 1 of size $n_1$ with mean $M_1$ and group 2 of size $n_2$ with mean $M_2$, the combined mean of the pooled dataset is the frequency-weighted average. Generalises to $k$ groups as a weighted average of the group means, with each weight equal to the group's size. Variance does NOT pool the same way. The combined variance carries a between-group term: $\sigma^2_{12} = \dfrac{n_1(\sigma_1^2 + d_1^2) + n_2(\sigma_2^2 + d_2^2)}{n_1 + n_2}$ with $d_i = M_i - M_{12}$. Only when both group means are equal does it reduce to the weighted average of the variances — so "the combined variance is the mean of the two variances" is false in general, and an option list built on that assumption has "none of the above" as its key.

For raw data with $n$ sorted observations, the median is the middle value if $n$ is odd, and the average of the two middle values if $n$ is even. For grouped data, use the class-interval formula below. Discrete series (values with frequencies, no class intervals): build the cumulative frequency and take the value of the $\tfrac{N+1}{2}$-th item — the first value whose cumulative frequency reaches $\tfrac{N+1}{2}$. This is the raw-data rule applied to the expanded list, NOT the grouped $\tfrac{N}{2}$ formula; using $N/2$ here is the planted error.

For raw data, the mode is the value with the highest frequency. If multiple values tie for highest, the dataset is multimodal. For grouped data, use the class-interval formula below.

For $n$ positive observations $x_1, x_2, \ldots, x_n$, the geometric mean is the $n$-th root of their product.

• Two numbers: $\text{GM} = \sqrt{mn}$, and $\text{AM} - \text{GM} = \dfrac{m+n}{2} - \sqrt{mn} = \dfrac{(\sqrt m - \sqrt n)^2}{2} \ge 0$ — the one-line proof of $\text{AM} \ge \text{GM}$, with equality iff $m = n$.
• Multiplicative rules: $\text{GM}(cx_i) = c\,\text{GM}(x_i)$; $\text{GM}(x_i^k) = \text{GM}(x_i)^k$; $\text{GM}(x_i/y_i) = \text{GM}(x_i)/\text{GM}(y_i)$; and the GM of the combined list of two groups of sizes $n_1, n_2$ is $(G_1^{n_1} G_2^{n_2})^{1/(n_1+n_2)}$. Logs turn every one of these into the corresponding rule for the arithmetic mean.

For $n$ positive observations $x_1, x_2, \ldots, x_n$, the harmonic mean is $n$ divided by the sum of the reciprocals. Two numbers: $\text{HM} = \dfrac{2mn}{m+n}$ — the closed form to reach for under time pressure ("HM of 60 and $x$ is 48" is $\tfrac{120x}{60+x} = 48 \Rightarrow x = 40$). With $\text{GM} = \sqrt{mn}$ this gives $\text{GM}^2 = \text{AM}\cdot\text{HM}$ for two numbers, and $\text{AM} \ge \text{GM} \ge \text{HM}$ always.

$\sum (x_i - \bar{x}) = 0$ for any dataset — this is a defining property of the mean. The empirical relation $\text{Mode} \approx 3\,\text{Median} - 2\,\text{Mean}$ holds approximately for moderately skewed unimodal distributions and is used to recover the third measure when two are known. For the same moderately-skewed regime, Mean Deviation $\approx \tfrac{4}{5}\,\text{Standard Deviation}$ — equivalently $4\,\text{SD} = 5\,\text{MD}$ (MD measured about the mean). Direction of skew fixes the ORDER: positive (right) skew pulls the mean toward the long right tail, so $\text{Mode} < \text{Median} < \text{Mean}$; negative (left) skew is the mirror, $\text{Mean} < \text{Median} < \text{Mode}$; a symmetric distribution has all three equal. The median always sits between the other two. And an exact inequality, not empirical: $\text{SD} \ge \text{MD}$ about the mean, for every dataset (the root-mean-square of the deviations is at least their mean absolute value).