Statistics revision notes

Statistics is the science of averages and their estimates, dealing with data collected for specific purposes to make decisions through analysis and interpretation.

1. Measures of Central Tendency

A representative value that gives a rough idea of where data points are centered.

2. Variability & Measures of Dispersion

Central tendency alone is insufficient to give complete information about data. Variability describes how scattered or bunched the data is around the central measure. A single number describing this variability is called a measure of dispersion. Types of measures of dispersion:

3. Variance

The mean of the squares of the deviations from the mean. It overcomes the mathematical difficulties of the Mean Deviation (which uses absolute values) by squaring the deviations to ensure non-negativity. $\rightarrow$ JEE Tip Variance is independent of the change of origin but depends on the change of scale.

4. Coefficient of Variation (CV)

A dimensionless, relative measure of dispersion used to compare the variability of two or more distributions. $CV = \frac{\sigma}{\bar{x}} \times 100$. A distribution with a lower CV is termed more consistent, uniform, or stable. $\rightarrow$ JEE Tip Always use CV, not standard deviation, to compare the consistency of two distinct datasets (e.g., comparing runs of two batsmen).

5. Historical Context (Theoretical Statistics)

1. Mean ($\bar{x}$)

• Ungrouped data: $\bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i$.
• Discrete frequency distribution: $\bar{x} = \frac{1}{N} \sum_{i=1}^{n} f_i x_i$, where $N = \sum f_i$.
• Step-deviation method (Shortcut): $\bar{x} = A + \frac{\sum f_i y_i}{N} \times h$, where $A$ is the assumed mean, $h$ is the class width, and $y_i = \frac{x_i - A}{h}$.

2. Median (M) for Continuous Frequency Distribution

3. Mean Deviation (M.D.)

• About a general value $a$: $M.D.(a) = \frac{1}{n} \sum |x_i - a|$.
• About Mean (Ungrouped): $M.D.(\bar{x}) = \frac{1}{n} \sum |x_i - \bar{x}|$.
• About Median (Ungrouped): $M.D.(M) = \frac{1}{n} \sum |x_i - M|$.
• About Mean (Grouped): $M.D.(\bar{x}) = \frac{1}{N} \sum f_i |x_i - \bar{x}|$.
• About Median (Grouped): $M.D.(M) = \frac{1}{N} \sum f_i |x_i - M|$.

4. Variance ($\sigma^2$)

• Ungrouped data (Basic definition): $\sigma^2 = \frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{x})^2$.
• Ungrouped data (Calculation shortcut): $\sigma^2 = \frac{1}{n} \sum_{i=1}^{n} x_i^2 - (\bar{x})^2$. $\rightarrow$ JEE Tip Always use this form in JEE calculations to save time.
• Discrete/Continuous Grouped data: $\sigma^2 = \frac{1}{N} \sum_{i=1}^{n} f_i (x_i - \bar{x})^2$.
• Grouped data (Calculation shortcut): $\sigma^2 = \frac{1}{N^2} \left[ N \sum f_i x_i^2 - \left(\sum f_i x_i\right)^2 \right]$.
• Step-deviation method: $\sigma_x^2 = h^2 \sigma_y^2 = \frac{h^2}{N^2} \left[ N \sum f_i y_i^2 - \left(\sum f_i y_i\right)^2 \right]$.

5. Standard Deviation ($\sigma$)

• It is simply the positive square root of variance: $\sigma = \sqrt{\sigma^2}$.

6. Combined Mean and Combined Variance (JEE Advanced Focus) If two groups of sizes $n_1$ and $n_2$ have means $\bar{x}_1, \bar{x}_2$ and variances $\sigma_1^2, \sigma_2^2$:

• Combined Mean: $\bar{x}_c = \frac{n_1 \bar{x}_1 + n_2 \bar{x}_2}{n_1 + n_2}$
• Combined Variance: $\sigma_c^2 = \frac{n_1(\sigma_1^2 + d_1^2) + n_2(\sigma_2^2 + d_2^2)}{n_1 + n_2}$, where $d_1 = \bar{x}_1 - \bar{x}_c$ and $d_2 = \bar{x}_2 - \bar{x}_c$.

• Mean Deviation: Cannot be subjected to further algebraic/calculus treatment because it involves absolute values (modulus function is not differentiable everywhere).
• Mean Deviation about Median: If the degree of variability in a series is very high, the median is not a reliable representative central tendency. Thus, M.D. about median cannot be fully relied upon in highly dispersed sets.
• Sum of Deviations: The sum of standard deviations without absolute values from the mean is always zero ($\sum (x_i - \bar{x}) = 0$). Hence, finding the simple average of deviations is useless for dispersion.
• Continuous Distributions: The formulas assume that the frequency in each class is entirely centered at its mid-point ($x_i$). This is an approximation.

• Sum of Squares Trap: The term $\sum x_i^2$ is the sum of the squares of the observations. This is strictly distinct from $(\sum x_i)^2$, which is the square of the sum. Do not confuse them in the variance formula $\frac{\sum x_i^2}{n} - (\frac{\sum x_i}{n})^2$.
• Modulus in Mean Deviation: Many students forget to drop the negative signs when calculating Mean Deviation. M.D. demands absolute values $|x_i - \bar{x}|$, not brackets.

1. Correcting Incorrect Observations: Given an incorrect mean and variance due to misread data points, calculating the exact correct mean and variance.
2. Linear Transformations: If a dataset undergoes operations like $x_{new} = \alpha x_{old} + \beta$, computing the new variance/SD.
3. Variance of Natural Numbers: Specifically the variance of the first $n$ natural numbers, arithmetic progressions, or geometric progressions.
4. Combined Variance: Questions providing data for boys vs. girls in a class and asking for the aggregate variance.

Correcting Incorrect Observations (Frequent JEE Profile)

Scenario: A student calculates the mean and variance of $n$ observations but later discovers that one observation $x_{wrong}$ was recorded incorrectly and should be $x_{correct}$.

• Step 1: Calculate Incorrect Sum of observations: $\sum x_{inc} = n \times \bar{x}_{inc}$
• Step 2: Calculate Correct Sum: $\sum x_{cor} = \sum x_{inc} - x_{wrong} + x_{correct}$
• Step 3: Calculate Correct Mean: $\bar{x}_{cor} = \frac{\sum x_{cor}}{n}$
• Step 4: Use the variance formula to find Incorrect Sum of Squares: $\sigma_{inc}^2 = \frac{\sum x_{inc}^2}{n} - (\bar{x}_{inc})^2 \implies \sum x_{inc}^2 = n(\sigma_{inc}^2 + \bar{x}_{inc}^2)$
• Step 5: Calculate Correct Sum of Squares: $\sum x_{cor}^2 = \sum x_{inc}^2 - (x_{wrong})^2 + (x_{correct})^2$
• Step 6: Calculate Correct Variance: $\sigma_{cor}^2 = \frac{\sum x_{cor}^2}{n} - (\bar{x}_{cor})^2$

Variance of First $n$ Natural Numbers

Let $x_i = i$ for $i = 1, 2, \ldots, n$. Mean: $\bar{x} = \frac{1}{n} \sum i = \frac{n(n+1)}{2n} = \frac{n+1}{2}$. Sum of squares: $\sum x_i^2 = \frac{n(n+1)(2n+1)}{6}$. Variance: $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2 = \frac{(n+1)(2n+1)}{6} - \left(\frac{n+1}{2}\right)^2 = \frac{n^2 - 1}{12}$. $\rightarrow$ JEE Tip Memorize $\sigma^2 = \frac{n^2 - 1}{12}$ for first $n$ natural numbers.