Formula for Mean Deviation (also called Mean Absolute Deviation):

Mean deviation is the average of the absolute deviations from a central value, typically the mean or median.

For a set of values $x_1, x_2, \ldots, x_n$ with mean $\bar{x}$:

$\text{Mean Deviation} = \frac{1}{n} \sum_{i=1}^{n} |x_i - \bar{x}|$

Where:

• $|x_i - \bar{x}|$ represents the absolute deviation of each value from the mean.

If the median is used instead of the mean, substitute the median value for $\bar{x}$ in the formula.

Formula for Standard Deviation:

Standard deviation measures the dispersion of data from its mean.

For a population with values $x_1, x_2, \ldots, x_N$ and population mean $\mu$:

$\sigma = \sqrt{\frac{1}{N} \sum_{i=1}^{N} (x_i - \mu)^2}$

For a sample with values $x_1, x_2, \ldots, x_n$ and sample mean $\bar{x}$:

$s = \sqrt{\frac{1}{n - 1} \sum_{i=1}^{n} (x_i - \bar{x})^2}$

Where:

• $\sigma$ is the population standard deviation.
• $s$ is the sample standard deviation.
• $x_i$ represents each data point.
• $\mu$ is the population mean, and $\bar{x}$ is the sample mean.

The main difference between the two formulas is the denominator: $N$ is used for the population, while $n - 1$ is used for the sample to account for sample bias, known as Bessel's correction.

Would you like more details or examples?

---

Here are some related questions for further understanding:

1. How is mean deviation different from standard deviation in practical use?
2. What is Bessel's correction, and why is it used in sample standard deviation?
3. How does the choice of using the median or mean affect the mean deviation?
4. Can we calculate variance from standard deviation?
5. What are the applications of standard deviation in real-life data analysis?

Tip: Standard deviation provides a measure of data spread in the same units as the data, making it more interpretable in context.