CT1 Sample Questions — Batch 1: Descriptive Statistics & Correlation

This document provides rigorous, textbook-style solutions for the descriptive statistics and correlation questions (Q1, Q2, Q3, and Q14) from the IFS UNIT - 1 & 2 sample questions sheet. All formulas are rendered in LaTeX, with step-by-step mathematical derivations and formal academic inferences.

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Question 1: Comprehensive Descriptive Statistics, Covariance & Correlation

Given Data

Let $X$ denote Sales and $Y$ denote Expenditure. $X = \{15, 18, 25, 27, 30, 35\}, \quad n = 6$ $Y = \{50, 65, 82, 95, 110, 120\}, \quad n = 6$

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Step-by-Step Preliminary Calculations

First, we construct the summation and deviation tables for both variables to compute all subsequent metrics.

Table 1: Sales ($X$) Deviations and Moments ($\bar{X} = 25$)

$i$	$x_i$	$(x_i - \bar{X})$	$(x_i - \bar{X})^2$	$(x_i - \bar{X})^3$	$(x_i - \bar{X})^4$
1	15	$-10$	$100$	$-1000$	$10000$
2	18	$-7$	$49$	$-343$	$2401$
3	25	$0$	$0$	$0$	$0$
4	27	$2$	$4$	$8$	$16$
5	30	$5$	$25$	$125$	$625$
6	35	$10$	$100$	$1000$	$10000$
$\sum$	$150$	$0$	$278$	$-210$	$23042$

Table 2: Expenditure ($Y$) Deviations and Moments ($\bar{Y} = 87$)

$i$	$y_i$	$(y_i - \bar{Y})$	$(y_i - \bar{Y})^2$	$(y_i - \bar{Y})^3$	$(y_i - \bar{Y})^4$
1	50	$-37$	$1369$	$-50653$	$1874161$
2	65	$-22$	$484$	$-10648$	$234256$
3	82	$-5$	$25$	$-125$	$625$
4	95	$8$	$64$	$512$	$4096$
5	110	$23$	$529$	$12167$	$279841$
6	120	$33$	$1089$	$35937$	$1185921$
$\sum$	$522$	$0$	$3560$	$-12810$	$3578900$

Table 3: Cross-Product of Deviations

$i$	$(x_i - \bar{X})$	$(y_i - \bar{Y})$	$(x_i - \bar{X})(y_i - \bar{Y})$
1	$-10$	$-37$	$370$
2	$-7$	$-22$	$154$
3	$0$	$-5$	$0$
4	$2$	$8$	$16$
5	$5$	$23$	$115$
6	$10$	$33$	$330$
$\sum$	$0$	$0$	$985$

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Part A: Central Tendency

1. Arithmetic Mean ($\bar{X}, \bar{Y}$)

Definition: The sum of all observations divided by the total number of observations. $\bar{X} = \frac{1}{n} \sum_{i=1}^n x_i = \frac{150}{6} = 25$ $\bar{Y} = \frac{1}{n} \sum_{i=1}^n y_i = \frac{522}{6} = 87$

2. Median ($M$)

Definition: The value splitting the sorted dataset into two equal halves. Let the ordered (sorted) observations be $x_{(1)}, x_{(2)}, \dots, x_{(n)}$ in ascending order.

• If $n$ is odd: The median is the single middle value located at the $\frac{n+1}{2}$-th position: $M = x_{\left(\frac{n+1}{2}\right)}$
• If $n$ is even: The median is the arithmetic mean of the two middle values located at the $\frac{n}{2}$-th and $\left(\frac{n}{2}+1\right)$-th positions: $M = \frac{x_{\left(\frac{n}{2}\right)} + x_{\left(\frac{n}{2}+1\right)}}{2}$

Since $n = 6$ is even in our datasets:

• For $X$: Sorted $X = \{15, 18, 25, 27, 30, 35\}$ $M_X = \frac{x_{(3)} + x_{(4)}}{2} = \frac{25 + 27}{2} = 26$
• For $Y$: Sorted $Y = \{50, 65, 82, 95, 110, 120\}$ $M_Y = \frac{y_{(3)} + y_{(4)}}{2} = \frac{82 + 95}{2} = 88.5$

3. Mode

Definition: The value that occurs with the highest frequency. Since all observations in $X$ and $Y$ are unique, there is no unique mode. Using the empirical relationship (Karl Pearson): $\text{Mode} \approx 3(\text{Median}) - 2(\text{Mean})$

• For $X$: $\text{Mode}_X = 3(26) - 2(25) = 78 - 50 = 28$
• For $Y$: $\text{Mode}_Y = 3(88.5) - 2(87) = 265.5 - 174 = 91.5$

Academic Inference (Central Tendency): Both Sales ($X$) and Expenditure ($Y$) exhibit stable, well-defined central locations. The average Sales is $\bar{X} = 25$ with a median of $26$, and the average Expenditure is $\bar{Y} = 87$ with a median of $88.5$. For both variables, the mean is slightly less than the median ($\bar{X} < M_X$ and $\bar{Y} < M_Y$). Under Karl Pearson's empirical approximation, the modes are located at $28$ and $91.5$, meaning that the mean, median, and mode satisfy the relation $\text{Mean} < \text{Median} < \text{Mode}$. This consistent ordering suggests that both distributions are mildly left-skewed.

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Part B: Measures of Dispersion

1. Range ($R$)

Definition: The difference between the maximum and minimum values in the dataset. $R_X = X_{\text{max}} - X_{\text{min}} = 35 - 15 = 20$ $R_Y = Y_{\text{max}} - Y_{\text{min}} = 120 - 50 = 70$

2. Quartiles and Quartile Deviation (QD)

Using the standard interpolation formula for continuous position estimation, the $p$-th percentile position is $P = \frac{k(n+1)}{4}$:

• First Quartile ($Q_1$): Position $= 1.75$
  • For $X$: $Q_{1,X} = x_{(1)} + 0.75(x_{(2)} - x_{(1)}) = 15 + 0.75(18 - 15) = 17.25$
  • For $Y$: $Q_{1,Y} = y_{(1)} + 0.75(y_{(2)} - y_{(1)}) = 50 + 0.75(65 - 50) = 61.25$
• Third Quartile ($Q_3$): Position $= 3 \times 1.75 = 5.25$
  • For $X$: $Q_{3,X} = x_{(5)} + 0.25(x_{(6)} - x_{(5)}) = 30 + 0.25(35 - 30) = 31.25$
  • For $Y$: $Q_{3,Y} = y_{(5)} + 0.25(y_{(6)} - y_{(5)}) = 110 + 0.25(120 - 110) = 112.5$

Quartile Deviation (Semi-Interquartile Range): $QD = \frac{Q_3 - Q_1}{2}$

• For $X$: $QD_X = \frac{31.25 - 17.25}{2} = \frac{14}{2} = 7$
• For $Y$: $QD_Y = \frac{112.5 - 61.25}{2} = \frac{51.25}{2} = 25.625$

3. Variance ($\sigma^2$) & Standard Deviation ($\sigma$)

We compute the population parameters ($\sigma^2, \sigma$):

• Population Variance ($\sigma^2$): $\sigma_X^2 = \frac{\sum (x_i - \bar{X})^2}{n} = \frac{278}{6} \approx 46.3333$ $\sigma_Y^2 = \frac{\sum (y_i - \bar{Y})^2}{n} = \frac{3560}{6} \approx 593.3333$
• Population Standard Deviation ($\sigma$): $\sigma_X = \sqrt{46.3333} \approx 6.8069$ $\sigma_Y = \sqrt{593.3333} \approx 24.3584$

Academic Inference (Absolute Dispersion): The absolute spread of Expenditure ($Y$) is significantly wider than that of Sales ($X$). The range of Expenditure ($R_Y = 70$) is $3.5$ times wider than Sales ($R_X = 20$). Similarly, the Quartile Deviation ($QD_Y = 25.625$ vs. $QD_X = 7$) and standard deviation ($\sigma_Y \approx 24.36$ vs. $\sigma_X \approx 6.81$) are multiple times larger. This reflects the scale differences of the raw data values; Expenditure values are much larger and naturally vary over a wider absolute numeric range, whereas Sales values are clustered tightly around a much lower center.

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Part C: Relative Dispersion & Coefficients

1. Coefficient of Quartile Deviation

$\text{Coeff of QD} = \frac{Q_3 - Q_1}{Q_3 + Q_1}$

• For $X$: $\text{Coeff of QD}_X = \frac{14}{48.5} \approx 0.2887$
• For $Y$: $\text{Coeff of QD}_Y = \frac{51.25}{173.75} \approx 0.2950$

2. Coefficient of Standard Deviation

$\text{Coeff of SD} = \frac{\sigma}{\text{Mean}}$

• For $X$: $\text{Coeff of SD}_X = \frac{6.8069}{25} \approx 0.2723$
• For $Y$: $\text{Coeff of SD}_Y = \frac{24.3584}{87} \approx 0.2800$

3. Coefficient of Variation (CV)

$\text{CV} = \left(\frac{\sigma}{\text{Mean}}\right) \times 100\%$

• For $X$: $\text{CV}_X = 27.23\%$
• For $Y$: $\text{CV}_Y = 28.00\%$

Academic Inference (Relative Dispersion): To eliminate the scale effect and compare the dispersion of the two variables on an equal, unitless basis, we analyze the Coefficient of Variation (CV) and other relative coefficients.

• Coefficient of QD: $\text{Coeff of QD}_X \approx 0.2887$ vs. $\text{Coeff of QD}_Y \approx 0.2950$.
• Coefficient of SD: $\text{Coeff of SD}_X \approx 0.2723$ vs. $\text{Coeff of SD}_Y \approx 0.2800$.
• Coefficient of Variation: $CV_X = 27.23\%$ vs. $CV_Y = 28.00\%$.

Since all relative coefficients for Expenditure ($Y$) are slightly larger than those for Sales ($X$), the relative dispersion of Expenditure is slightly higher than that of Sales. This indicates that Sales ($X$) represents a more stable and consistent dataset relative to its mean, while Expenditure ($Y$) shows slightly higher relative fluctuation and lower consistency.

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Part D: Measures of Distribution (Skewness & Kurtosis)

1. Central Moments ($\mu_k$)

$\mu_k = \frac{1}{n} \sum_{i=1}^n (x_i - \bar{X})^k$

• For $X$: $\mu_2 = \frac{278}{6} \approx 46.3333, \quad \mu_3 = \frac{-210}{6} = -35, \quad \mu_4 = \frac{23042}{6} \approx 3840.3333$
• For $Y$: $\mu_2 = \frac{3560}{6} \approx 593.3333, \quad \mu_3 = \frac{-12810}{6} = -2135, \quad \mu_4 = \frac{3578900}{6} \approx 596483.3333$

2. Skewness

• Karl Pearson's Coefficient of Skewness ($S_{k,P}$): $S_{k,P} = \frac{3(\text{Mean} - \text{Median})}{\sigma}$
  • For $X$: $S_{k,P}^X = \frac{3(25 - 26)}{6.8069} = \frac{-3}{6.8069} \approx -0.4407$ (Left-skewed)
  • For $Y$: $S_{k,P}^Y = \frac{3(87 - 88.5)}{24.3584} = \frac{-4.5}{24.3584} \approx -0.1847$ (Left-skewed)
• Moment-based Skewness ($\gamma_1$): $\gamma_1 = \frac{\mu_3}{\mu_2^{1.5}}$
  • For $X$: $\gamma_{1,X} = \frac{-35}{(46.3333)^{1.5}} = \frac{-35}{315.534} \approx -0.1109$
  • For $Y$: $\gamma_{1,Y} = \frac{-2135}{(593.3333)^{1.5}} = \frac{-2135}{14452.793} \approx -0.1477$

3. Kurtosis

• Moment-based Kurtosis ($\beta_2$ & Excess Kurtosis $\gamma_2$): $\beta_2 = \frac{\mu_4}{\mu_2^2}, \quad \gamma_2 = \beta_2 - 3$
  • For $X$: $\beta_{2,X} = \frac{3840.3333}{2146.7778} \approx 1.7889 \implies \gamma_{2,X} = 1.7889 - 3 = -1.2111 \quad (\text{Platykurtic})$
  • For $Y$: $\beta_{2,Y} = \frac{596483.3333}{352044.4444} \approx 1.6943 \implies \gamma_{2,Y} = 1.6943 - 3 = -1.3057 \quad (\text{Platykurtic})$

Academic Inference (Distribution Shape):

• Skewness: Both Karl Pearson's coefficient ($S_{k,P}^X \approx -0.4407, S_{k,P}^Y \approx -0.1847$) and moment-based skewness ($\gamma_{1,X} \approx -0.1109, \gamma_{1,Y} \approx -0.1477$) are strictly negative. This confirms that both datasets are left-skewed (negatively skewed). The left tail is slightly elongated, showing a small density of lower-value records dragging the mean slightly below the median.
• Kurtosis: Both datasets exhibit an excess kurtosis significantly below zero ($\gamma_{2,X} \approx -1.2111, \gamma_{2,Y} \approx -1.3057$). This classifies both distributions as Platykurtic (thin-tailed, flatter peak than a standard normal curve), representing stable, bounded variables with very low outlier probability.

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Part E: Covariance and Correlation Co-efficient

1. Covariance ($\text{Cov}(X,Y)$)

Definition: A measure of the joint variability of two random variables.

• Population Covariance ($\sigma_{XY}$): $\sigma_{XY} = \frac{1}{n} \sum_{i=1}^n (x_i - \bar{X})(y_i - \bar{Y}) = \frac{985}{6} \approx 164.1667$

2. Pearson Correlation Coefficient ($r$)

Definition: The normalized measure of linear relationship strength. $r = \frac{\sum (x_i - \bar{X})(y_i - \bar{Y})}{\sqrt{\sum (x_i - \bar{X})^2 \sum (y_i - \bar{Y})^2}} = \frac{985}{\sqrt{278 \times 3560}} = \frac{985}{\sqrt{989680}} \approx \frac{985}{994.8266} \approx 0.9901$

Exam tip: A value of $r \approx 0.9901$ indicates an extremely strong, almost perfect positive linear correlation. When sales increase, expenditure increases in a highly predictable linear fashion.

Academic Inference (Bivariate Relationship): The large positive covariance ($\sigma_{XY} \approx 164.17$) and nearly perfect Pearson's correlation coefficient ($r \approx 0.9901$) reveal an extremely strong, positive linear association between Sales ($X$) and Expenditure ($Y$). This implies that $98.03\%$ of the variation in Expenditure can be linearly explained by Sales (coefficient of determination $r^2 \approx 0.9803$). In a corporate context, this signifies that as Sales scale up, company Expenditures rise in a highly systematic, predictable, and tightly coupled linear trajectory.

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Question 2: Five-Point Summary and IQR

Given Data

The dataset contains paired variables $X$ and $Y$. Since they are univariate sequences, we compute the 5-point summaries separately. $X = \{1, 3, 5, 7, 8, 10\}, \quad n = 6$ $Y = \{8, 12, 15, 17, 18, 20\}, \quad n = 6$

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Step-by-Step Calculation

1. Five-Point Summary for $X$

• Minimum: $1$
• $Q_1$ (First Quartile): Position $= \frac{1(n+1)}{4} = 1.75$th item $Q_1 = 1 + 0.75 \times (3 - 1) = 2.5$
• Median ($Q_2$): Position $= \frac{2(n+1)}{4} = 3.5$th item $\text{Median} = \frac{5 + 7}{2} = 6.0$
• $Q_3$ (Third Quartile): Position $= \frac{3(n+1)}{4} = 5.25$th item $Q_3 = 8 + 0.25 \times (10 - 8) = 8.5$
• Maximum: $10$

Five-Point Summary vector for $X$: $\begin{bmatrix} \text{Min} & Q_1 & \text{Median} & Q_3 & \text{Max} \end{bmatrix} = \begin{bmatrix} 1.0 & 2.5 & 6.0 & 8.5 & 10.0 \end{bmatrix}$

Interquartile Range (IQR): $IQR_X = Q_3 - Q_1 = 8.5 - 2.5 = 6.0$

Quartile Range: The interval span is $[Q_1, Q_3] = [2.5, 8.5]$.

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2. Five-Point Summary for $Y$

• Minimum: $8$
• $Q_1$ (First Quartile): Position $= 1.75$th item $Q_1 = 8 + 0.75 \times (12 - 8) = 11.0$
• Median ($Q_2$): Position $= 3.5$th item $\text{Median} = \frac{15 + 17}{2} = 16.0$
• $Q_3$ (Third Quartile): Position $= 5.25$th item $Q_3 = 18 + 0.25 \times (20 - 18) = 18.5$
• Maximum: $20$

Five-Point Summary vector for $Y$: $\begin{bmatrix} \text{Min} & Q_1 & \text{Median} & Q_3 & \text{Max} \end{bmatrix} = \begin{bmatrix} 8.0 & 11.0 & 16.0 & 18.5 & 20.0 \end{bmatrix}$

Interquartile Range (IQR): $IQR_Y = Q_3 - Q_1 = 18.5 - 11.0 = 7.5$

Quartile Range: The interval span is $[Q_1, Q_3] = [11.0, 18.5]$.

Academic Inference (Five-Point Summary): The five-point summary provides a complete, robust overview of the distribution boundaries and central half of the datasets. For $X$, the middle $50\%$ of the data lies in a narrow span between $2.5$ and $8.5$ ($IQR_X = 6.0$). For $Y$, the central $50\%$ of the observations span a slightly wider range from $11.0$ to $18.5$ ($IQR_Y = 7.5$). This demonstrates that the core density of the Expenditure ($Y$) dataset is slightly more spread out in absolute terms than the core density of the Sales ($X$) dataset.

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Question 3: Spearman's Rank Correlation for Three Judges

Given Data

Three judges rank 10 competitors ($N = 10$).

• $R_A$: $\{6, 5, 3, 10, 2, 4, 9, 7, 8, 1\}$
• $R_B$: $\{5, 8, 4, 7, 10, 2, 1, 6, 9, 3\}$
• $R_C$: $\{4, 9, 8, 1, 2, 3, 10, 5, 7, 6\}$

Spearman's rank correlation coefficient is given by: $\rho = 1 - \frac{6 \sum_{i=1}^N d_i^2}{N(N^2 - 1)}$

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Step-by-Step Derivation

Let us construct the table of difference-squared ($d_i^2$) for all pairs:

Competitor $i$	$R_A$	$R_B$	$R_C$	$d^2_{AB} = (R_A - R_B)^2$	$d^2_{BC} = (R_B - R_C)^2$	$d^2_{AC} = (R_A - R_C)^2$
1	$6$	$5$	$4$	$1$	$1$	$4$
2	$5$	$8$	$9$	$9$	$1$	$16$
3	$3$	$4$	$8$	$1$	$16$	$25$
4	$10$	$7$	$1$	$9$	$36$	$81$
5	$2$	$10$	$2$	$64$	$64$	$0$
6	$4$	$2$	$3$	$4$	$1$	$1$
7	$9$	$1$	$10$	$64$	$81$	$1$
8	$7$	$6$	$5$	$1$	$1$	$4$
9	$8$	$9$	$7$	$1$	$4$	$1$
10	$1$	$3$	$6$	$4$	$9$	$25$
$\sum$				$158$	$214$	$158$

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Calculating Coefficients

1. Judges A and B ($\rho_{AB}$)

$\rho_{AB} = 1 - \frac{6 \times 158}{10(10^2 - 1)} = 1 - \frac{948}{990} = 1 - 0.9576 = 0.0424$

2. Judges B and C ($\rho_{BC}$)

$\rho_{BC} = 1 - \frac{6 \times 214}{10(10^2 - 1)} = 1 - \frac{1284}{990} = 1 - 1.2970 = -0.2970$

3. Judges A and C ($\rho_{AC}$)

$\rho_{AC} = 1 - \frac{6 \times 158}{10(10^2 - 1)} = 1 - \frac{948}{990} = 1 - 0.9576 = 0.0424$

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Academic Inference

• Judges A and B / Judges A and C: Show a very weak, positive rank correlation ($\rho \approx 0.0424$). This suggests practically independent ranking scales with virtually no common pattern.
• Judges B and C: Show a moderate negative rank correlation ($\rho \approx -0.2970$). This reveals a systematic divergence in their musical preferences.
• Conclusion: Judges A and B (as well as A and C) have more or less the same taste in music because their coefficient is positive (albeit very low), whereas Judges B and C have opposing rankings as indicated by their negative coefficient.

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Question 14: Stronger Relationship: Calorie Intake vs Exercise Hours

Given Data

A nutritionist collects data for 4 individuals ($n = 4$):

• Calorie Intake ($X_1$): $\{2500, 2200, 2000, 1800\}$
• Exercise Hours ($X_2$): $\{2, 4, 6, 8\}$
• Weight Loss Progress ($Y$): $\{1, 2, 4, 6\}$

We determine the strength of relationships by computing Pearson's correlation coefficient ($r$) for both independent variables.

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Step-by-Step Derivation

First, calculate the means: $\bar{X}_1 = \frac{2500+2200+2000+1800}{4} = 2125$ $\bar{X}_2 = \frac{2+4+6+8}{4} = 5$ $\bar{Y} = \frac{1+2+4+6}{4} = 3.25$

Table 4: Deviation Products for Calorie Intake ($X_1$) vs Weight Loss ($Y$)

$i$	$(x_{1,i} - \bar{X}_1)$	$(y_i - \bar{Y})$	$(x_{1,i} - \bar{X}_1)^2$	$(y_i - \bar{Y})^2$	$(x_{1,i} - \bar{X}_1)(y_i - \bar{Y})$
1	$375$	$-2.25$	$140625$	$5.0625$	$-843.75$
2	$75$	$-1.25$	$5625$	$1.5625$	$-93.75$
3	$-125$	$0.75$	$15625$	$0.5625$	$-93.75$
4	$-325$	$2.75$	$105625$	$7.5625$	$-893.75$
$\sum$	$0$	$0$	$267500$	$14.75$	$-1925.0$

Table 5: Deviation Products for Exercise Hours ($X_2$) vs Weight Loss ($Y$)

$i$	$(x_{2,i} - \bar{X}_2)$	$(y_i - \bar{Y})$	$(x_{2,i} - \bar{X}_2)^2$	$(y_i - \bar{Y})^2$	$(x_{2,i} - \bar{X}_2)(y_i - \bar{Y})$
1	$-3$	$-2.25$	$9$	$5.0625$	$6.75$
2	$-1$	$-1.25$	$1$	$1.5625$	$1.25$
3	$1$	$0.75$	$1$	$0.5625$	$0.75$
4	$3$	$2.75$	$9$	$7.5625$	$8.25$
$\sum$	$0$	$0$	$20$	$14.75$	$17.0$

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Computing Pearson's Correlation Coefficient

1. Calorie Intake vs Weight Loss ($r_{X_1, Y}$)

$r_{X_1, Y} = \frac{\sum (x_{1,i} - \bar{X}_1)(y_i - \bar{Y})}{\sqrt{\sum (x_{1,i} - \bar{X}_1)^2 \sum (y_i - \bar{Y})^2}} = \frac{-1925}{\sqrt{267500 \times 14.75}} = \frac{-1925}{\sqrt{3945625}} = \frac{-1925}{1986.3144} \approx -0.9691$

2. Exercise Hours vs Weight Loss ($r_{X_2, Y}$)

$r_{X_2, Y} = \frac{\sum (x_{2,i} - \bar{X}_2)(y_i - \bar{Y})}{\sqrt{\sum (x_{2,i} - \bar{X}_2)^2 \sum (y_i - \bar{Y})^2}} = \frac{17.0}{\sqrt{20 \times 14.75}} = \frac{17.0}{\sqrt{295}} = \frac{17.0}{17.1756} \approx 0.9898$

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Academic Inference

• Calorie Intake ($X_1$) vs Weight Loss ($Y$): Shows an exceptionally strong, negative linear correlation ($r \approx -0.9691$). This indicates that as calorie intake decreases, weight loss progress increases, which aligns perfectly with standard thermodynamic models of energy balance.
• Exercise Hours ($X_2$) vs Weight Loss ($Y$): Shows an exceptionally strong, positive linear correlation ($r \approx 0.9898$). This indicates that as exercise hours increase, weight loss progress increases.
• Strength Comparison: Comparing the magnitudes of the correlations: $\left|r_{X_2, Y}\right| = 0.9898 > \left|r_{X_1, Y}\right| = 0.9691$
• Conclusion: Exercise Hours ($X_2$) has a stronger relationship with Weight Loss Progress than Calorie Intake ($X_1$) for this cohort, although both are highly predictive factors.

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Quick-Facts & Formula Sheet (Descriptive Statistics & Correlation - Population Model)

Metric / Parameter	Mathematical Formula
Arithmetic Mean	$\mu = \frac{1}{N} \sum_{i=1}^N x_i$
Median ($N$ is odd)	$M = x_{\left(\frac{N+1}{2}\right)}$
Median ($N$ is even)	$M = \frac{x_{\left(\frac{N}{2}\right)} + x_{\left(\frac{N}{2}+1\right)}}{2}$
Quartile Position	$P(Q_k) = \frac{k(N+1)}{4}$
Quartile Deviation (QD)	$QD = \frac{Q_3 - Q_1}{2}$
Coeff of QD	$\text{Coeff of QD} = \frac{Q_3 - Q_1}{Q_3 + Q_1}$
Variance	$\sigma^2 = \frac{1}{N} \sum_{i=1}^N (x_i - \mu)^2$
Standard Deviation	$\sigma = \sqrt{\sigma^2}$
Coeff of SD	$\text{Coeff of SD} = \frac{\sigma}{\mu}$
Coeff of Variation (CV)	$CV = \left( \frac{\sigma}{\mu} \right) \times 100\%$
Pearson's Skewness ($S_{k,P}$)	$S_{k,P} = \frac{3(\mu - M_d)}{\sigma}$
Moment-Based Skewness	$\mu_k = \frac{1}{N}\sum(x_i-\mu)^k \implies \gamma_1 = \frac{\mu_3}{\mu_2^{1.5}}$
Pearson's Kurtosis ($\beta_2$)	$\beta_2 = \frac{\mu_4}{\mu_2^2}$
Excess Kurtosis ($\gamma_2$)	$\gamma_2 = \beta_2 - 3$
Covariance	$\sigma_{XY} = \frac{1}{N}\sum(x_i - \mu_X)(y_i - \mu_Y)$
Pearson Correlation ($r$)	$r = \frac{\sigma_{XY}}{\sigma_X \sigma_Y} = \frac{\sum(x_i-\mu_X)(y_i-\mu_Y)}{\sqrt{\sum(x_i-\mu_X)^2 \sum(y_i-\mu_Y)^2}}$
Spearman's Rank Correlation	$\rho = 1 - \frac{6 \sum d_i^2}{N(N^2 - 1)}$