Q.Find the mean and the variance of the following data:
Class-interval: 0-30 | 30-60 | 60-90 | 90-120 | 120-150 | 150-180 | 180-210
Frequency: 2 | 3 | 5 | 10 | 3 | 5 | 2 OR The Mean of 5 observations is 4.4 and their variance is 8.24. If three observations are 1, 2 and 6, find the other two observations.

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

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

$\sigma = \sqrt{144} = 12$ …

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: $\text{Variance} = \sigma^2 = (\text{Standard Deviation})^2$.

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

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

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Variance is defined as the square of the standard deviation.

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

Given standard deviation $= 0.012$: …

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

By definition, variance $=(\text{standard deviation})^2$. Here standard deviation $=1.6$ …

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.

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