Descriptive Statistics: Mean, Median, Mode and Standard Deviation

Descriptive statistics summarise a data set with a few numbers: where it is centred (mean, median, mode) and how spread out it is (range, variance, standard deviation). These are the sample counterparts of the expectation and variance of a random variable. The calculations are straightforward, but the divisor $n$ versus $n - 1$, the effect of shifting and scaling data, combining groups, and the order of mean, median and mode in a skewed distribution all need care.

None within this subject. Basic arithmetic and summation notation suffice. Random Variables, Expectation and Variance gives the probabilistic view of the same ideas.

Throughout, the data are $x_1, x_2, \ldots, x_n$, and $x_{(1)} \le x_{(2)} \le \cdots \le x_{(n)}$ are the same values sorted.

Arithmetic mean. $\bar{x} = \frac{1}{n}\sum_{i=1}^{n} x_i$. For values $x_i$ with frequencies $f_i$, $\bar{x} = \sum f_i x_i / \sum f_i$. Key facts:

Median. The middle value of the sorted data: $x_{((n+1)/2)}$ if $n$ is odd, and the average $\frac{1}{2}\big(x_{(n/2)} + x_{(n/2 + 1)}\big)$ if $n$ is even (the usual convention). The median minimises $\sum |x_i - c|$ and is robust to outliers.

Mode. The most frequent value. A data set can have no mode (all values distinct, under the usual convention), one mode, or several (bimodal, multimodal).

Grouped data (class width $h$, total frequency $N$):

where for the median $L$ is the lower boundary of the class containing the $(N/2)$-th value, $C$ the cumulative frequency before it and $f$ its frequency; for the mode $L$ is the lower boundary of the modal class, $f_1$ its frequency and $f_0, f_2$ the frequencies of the classes before and after. These are interpolation formulas, not exact values.

Other means for positive data: geometric mean $G = (x_1 \cdots x_n)^{1/n}$ and harmonic mean $H = n / \sum (1/x_i)$. Always $\text{AM} \ge \text{GM} \ge \text{HM}$, with equality if and only if all values are equal. For two numbers, $G^2 = A \cdot H$.

Divisor convention. Sample formulas with divisor $n-1$ require $n\ge2$. Use the divisor $n$ when the data are the whole population, and $n - 1$ when the data are a sample used to estimate the population variance ($s^2$ is then an unbiased estimator). Check whether the question's "variance" uses divisor $n$ or $n-1$; if the options contain both values, decide which the question means. The standard deviation is the square root: $\sigma$ or $s$.

Shortcut formula (population): $\sigma^2 = \frac{1}{n}\sum x_i^2 - \bar{x}^2$. Equivalently, the variance is the mean of squares minus the square of the mean, mirroring $E[X^2] - (E[X])^2$.

Coefficient of variation: $\mathrm{CV}=\text{standard deviation}/\bar x$, often as a percentage, using $\sigma$ or $s$ consistently. The quotient requires a nonzero mean. Interpreting CV as relative spread normally requires a positive-mean ratio-scale quantity with a meaningful zero; arbitrary shifted scales such as Celsius do not support that interpretation. Positive rescaling preserves CV; shifting generally does not.

If every value is transformed as $y_i = a x_i + b$:

Adding a constant shifts location but leaves spread unchanged.

Combined variance (population form) of two groups with sizes $n_1, n_2$, means $\bar{x}_1, \bar{x}_2$, variances $\sigma_1^2, \sigma_2^2$ and combined mean $\bar{x}$, with $d_j = \bar{x}_j - \bar{x}$:

These orderings are typical, not guaranteed; counterexamples exist for unusual shapes. Pearson's empirical relation $\text{Mode} \approx 3\,\text{Median} - 2\,\text{Mean}$ holds approximately for moderately skewed unimodal data. It is not an identity. Pearson's coefficient of skewness $(\text{Mean} - \text{Mode})/\sigma$ is typically positive for ordinary right-skewed unimodal shapes; in unusual distributions its sign need not agree with moment skewness.

For paired observations $(x_i,y_i)$, $i=1,\ldots,n$, sample covariance requires $n\ge2$. Pearson's $r$ additionally requires nonzero variation in both variables; it is undefined if either is constant. The formulas are

The divisor cancels in $r$, so $r$ is the same whether $n$ or $n - 1$ is used throughout. Always $-1 \le r \le 1$; $r = \pm 1$ exactly when the points lie on a line with non-zero slope; $r$ is unchanged by shifting either variable and by multiplying it by a positive constant, and changes sign under a negative constant. $r$ measures linear association only: for $y = x^2$ on $x = -2, -1, 0, 1, 2$, $r = 0$ although $y$ is a function of $x$.

Data: $2, 4, 4, 4, 5, 5, 7, 9$ ($n = 8$). Mean $= 40/8 = 5$. Median $= (4 + 5)/2 = 4.5$. Mode $= 4$. Squared deviations sum to $9 + 1 + 1 + 1 + 0 + 0 + 4 + 16 = 32$, so the population variance is $32/8 = 4$ ($\sigma = 2$), and the sample variance is $32/7 \approx 4.571$ ($s \approx 2.138$). Shortcut check: $\frac{1}{8}\sum x_i^2 = 232/8 = 29$ and $29 - 25 = 4$. Range $= 7$.

Group A: 30 students, mean 60, standard deviation 10. Group B: 20 students, mean 70, standard deviation 5. Combined mean $= (1800 + 1400)/50 = 64$; $d_A = -4$, $d_B = 6$. Combined variance $= [30(100 + 16) + 20(25 + 36)]/50 = 4700/50 = 94$, standard deviation $\approx 9.70$.

Temperatures in Celsius have mean 20 and standard deviation 5. In Fahrenheit, $F = 1.8C + 32$: mean $68$, standard deviation $9$, variance $81$.

The mean of 10 observations is 15. One value, 12, was recorded as 21. The corrected total is $150 - 21 + 12 = 141$, so the corrected mean is $14.1$.

$N = 30$, $N/2 = 15$; cumulative frequencies are $5, 13, 25, 30$, so the median class is 20 to 30. Median $\approx 20 + \frac{15 - 13}{12}\cdot 10 \approx 21.67$. Modal class 20 to 30: Mode $\approx 20 + \frac{12 - 8}{24 - 8 - 5}\cdot 10 = 20 + 40/11 \approx 23.64$.

For $2$ and $8$: AM $= 5$, GM $= 4$, HM $= 2 \cdot 2 \cdot 8/(2 + 8) = 3.2$. Check $G^2 = 16 = 5 \times 3.2$.

Data $(1, 2), (2, 4), (3, 5), (4, 4), (5, 5)$. Means $\bar{x} = 3$, $\bar{y} = 4$. Deviations: $x - \bar{x} = -2, -1, 0, 1, 2$; $y - \bar{y} = -2, 0, 1, 0, 1$. Then $\sum (x - \bar{x})(y - \bar{y}) = 4 + 0 + 0 + 0 + 2 = 6$, $\sum (x - \bar{x})^2 = 10$, $\sum (y - \bar{y})^2 = 6$, and $r = 6/\sqrt{60} \approx 0.775$. The sample covariance is $6/4 = 1.5$. Converting $x$ to other units ($x' = 10x + 7$) leaves $r$ unchanged.