Mathematics: Class 11 — Chapter 13: Statistics

Two datasets can have the exact same mean and median but completely different distributions. Consider the runs scored by two batsmen over ten matches:

• Batsman A: $30, 91, 0, 64, 42, 80, 30, 5, 117, 71$ (Mean = $53$, Median = $53$)
• Batsman B: $53, 46, 48, 50, 53, 53, 58, 60, 57, 52$ (Mean = $53$, Median = $53$)

Batsman B is highly consistent, with scores clustered tightly around the mean, whereas Batsman A is highly volatile, with scores widely scattered. To quantify this variability, we require measures of dispersion.

The simplest measure of dispersion is the Range, defined as the difference between the maximum and minimum values in a dataset:

• Range of Batsman A: $117 - 0 = 117$
• Range of Batsman B: $60 - 46 = 14$

While easy to calculate, the range is highly sensitive to outliers and does not reflect the dispersion of data points between the extremes.

Let $x_1, x_2, \dots, x_n$ be $n$ observations. The mean deviation about a central value $a$ is given by:

• Mean Deviation about Mean ($\bar{x}$): $\text{M.D.}(\bar{x}) = \frac{1}{n} \sum_{i=1}^{n} |x_i - \bar{x}|$
• Mean Deviation about Median ($M$): $\text{M.D.}(M) = \frac{1}{n} \sum_{i=1}^{n} |x_i - M|$

Mathematical Problem Statement: Find the mean deviation about the mean for the data: $6, 7, 10, 12, 13, 4, 8, 12$.

---

Step-by-Step Algorithmic Solution Trace:

1. Calculate the mean ($\bar{x}$): $$\bar{x} = \frac{6 + 7 + 10 + 12 + 13 + 4 + 8 + 12}{8} = \frac{72}{8} = 9$$
2. Find the absolute deviations $|x_i - \bar{x}|$: $$|6-9|, |7-9|, |10-9|, |12-9|, |13-9|, |4-9|, |8-9|, |12-9| = 3, 2, 1, 3, 4, 5, 1, 3$$
3. Sum the absolute deviations: $$\sum_{i=1}^{8} |x_i - \bar{x}| = 3 + 2 + 1 + 3 + 4 + 5 + 1 + 3 = 22$$
4. Compute the Mean Deviation: $$\text{M.D.}(\bar{x}) = \frac{22}{8} = 2.75$$

Final Evaluation Output: $\text{M.D.}(\bar{x}) = 2.75$

For grouped data, observations $x_i$ occur with corresponding frequencies $f_i$.

• Discrete Frequency Distribution: $$\text{M.D.}(\bar{x}) = \frac{1}{N} \sum_{i=1}^{n} f_i |x_i - \bar{x}|, \quad \text{where } N = \sum_{i=1}^{n} f_i$$ $$\text{M.D.}(M) = \frac{1}{N} \sum_{i=1}^{n} f_i |x_i - M|$$
• Continuous Frequency Distribution: Classes are converted to mid-points $x_i$, and the same discrete formulas are applied. The median $M$ is calculated using: $$M = l + \left(\frac{\frac{N}{2} - C}{f}\right) \times h$$ where $l$ is the lower limit of the median class, $C$ is the cumulative frequency of the preceding class, $f$ is the frequency of the median class, and $h$ is the class width.

• The absolute value mathematical operation $|x_i - a|$ is not easily subjected to further algebraic treatment.
• If the degree of variability in a series is very high, the median may not remain a representative central tendency, making the mean deviation about the median unreliable.

• Variance ($\sigma^2$): $$\sigma^2 = \frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{x})^2$$
• Standard Deviation ($\sigma$): $$\sigma = \sqrt{\frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{x})^2}$$

• Variance ($\sigma^2$): $$\sigma^2 = \frac{1}{N} \sum_{i=1}^{n} f_i (x_i - \bar{x})^2$$
• Standard Deviation ($\sigma$): $$\sigma = \sqrt{\frac{1}{N} \sum_{i=1}^{n} f_i (x_i - \bar{x})^2}$$
• Alternative Computational Formula: $$\sigma = \frac{1}{N} \sqrt{N \sum f_i x_i^2 - \left(\sum f_i x_i\right)^2}$$

When mid-points $x_i$ or frequencies $f_i$ are large, we use step-deviations $y_i = \frac{x_i - A}{h}$, where $A$ is an assumed mean and $h$ is the class width.

• Mean: $\bar{x} = A + h \bar{y}$
• Variance ($\sigma_x^2$): $$\sigma_x^2 = \frac{h^2}{N^2} \left[ N \sum_{i=1}^{n} f_i y_i^2 - \left(\sum_{i=1}^{n} f_i y_i\right)^2 \right]$$
• Standard Deviation ($\sigma_x$): $$\sigma_x = \frac{h}{N} \sqrt{N \sum_{i=1}^{n} f_i y_i^2 - \left(\sum_{i=1}^{n} f_i y_i\right)^2}$$