Q.Find variance and S.D. for the following set of numbers. 7, 11, 2, 4, 9, 6, 3, 7, 11, 2, 5, 8, 3, 6, 8, 8, 2, 6

Standard deviation is defined as the positive square root of variance.

By definition, standard deviation $\sigma = \sqrt{\text{variance}}$. Given variance $= 144$:

$\sigma = \sqrt{144} = 12$

Standard deviation is always taken as the non-negative root (a spread measure cannot be negative), which rules out option (b) $-12$.

The square of the standard deviation is called the variance, option (c).

Standard deviation ($\sigma$) measures the spread of data about the mean. By definition, variance ($\sigma^2$) is exactly the square of the standard deviation: $\text{Variance} = \sigma^2 = (\text{Standard Deviation})^2$.

So among the given options, the square of standard deviation is the Variance, not mean-deviation, range, or mean.

Standard deviation $=\sqrt{\text{Variance}}$; here $\sqrt{225}=15$.

By definition, variance is the square of the standard deviation, so standard deviation $=\sqrt{\text{variance}}$.

Given variance $=225$: standard deviation $=\sqrt{225}=15$.

Variance is defined as the square of the standard deviation.

By definition, variance $= (\text{standard deviation})^2$.

Given standard deviation $= 0.012$:

$\text{Variance} = (0.012)^2 = 0.012 \times 0.012 = 0.000144$

Given standard deviation $=1.6$, the variance is $2.56$.

By definition, variance $=(\text{standard deviation})^2$. Here standard deviation $=1.6$, so variance $=1.6\times1.6=2.56$.

This difference is called the range of the series, the simplest measure of dispersion.

Step 1. Range is defined as: Range $=$ Maximum value $-$ Minimum value.

Step 2. It measures how spread out the data is, from the smallest to the largest observation.